Linear differential equations of second order are of crucial importance in the study of differential equations for two main reasons. The first is that linear equations have a rich theoretical structure that underlies a number of systematic methods of solution. Further, a substantial portion of this structure and of these methods is understandable at a fairly elementary mathematical level. In order to present the key ideas in the simplest possible context, we describe them in this chapter for second-order equations. The second reason to study second-order linear differential equations is that they are vital to any serious investigation of the classical areas of mathematical physics. One cannot go very far in the development of fluid mechanics, heat conduction, wave motion, or electromagnetic phenomena without finding it necessary to solve second-order linear differential equations. We illustrate this at the end of this chapter with a discussion of the oscillations of some basic mechanical and electrical systems.
3.1 Homogeneous Differential Equations
with Constant Coefficients
Many second-order ordinary differential equations have the form
{:(1)(d^(2)y)/(dt^(2))=f(t,y,(dy)/(dt)):}\begin{equation*}
\frac{d^{2} y}{d t^{2}}=f\left(t, y, \frac{d y}{d t}\right) \tag{1}
\end{equation*}
where ff is some given function. Usually, we will denote the independent variable by tt since time is often the independent variable in physical problems, but sometimes we will use xx instead. We will use yy, or occasionally some other letter, to designate the dependent variable. Equation (1) is said to be linear if the function ff has the form
{:(2)f(t,y,(dy)/(dt))=g(t)-p(t)(dy)/(dt)-q(t)y:}\begin{equation*}
f\left(t, y, \frac{d y}{d t}\right)=g(t)-p(t) \frac{d y}{d t}-q(t) y \tag{2}
\end{equation*}
that is, if ff is linear in yy and dy//dtd y / d t. In equation (2) g,pg, p, and qq are specified functions of the independent variable tt but do not depend on yy. In this case we usually rewrite equation (1) as
In discussing equation (3) and in trying to solve it, we will restrict ourselves to intervals in which p,qp, q, and gg are continuous functions. ^(1){ }^{1}
If equation (1) is not of the form (3) or (4), then it is called nonlinear. Analytical investigations of nonlinear equations are relatively difficult, so we will have little to say about them in this book. Numerical or geometical approaches are often more appropriate, and these are discussed in Chapters 8 and 9.
An initial value problem consists of a differential equation such as equations (1), (3), or (4) together with a pair of initial conditions
where y_(0)y_{0} and y_(0)^(')y_{0}^{\prime} are given numbers prescribing values for yy and y^(')y^{\prime} at the initial point t_(0)t_{0}. Observe that the initial conditions for a second-order differential equation identify not only a particular point (t_(0),y_(0))\left(t_{0}, y_{0}\right) through which the graph of the solution must pass, but also the slope y_(0)^(')y_{0}^{\prime} of the graph at that point. It is reasonable to expect that two initial conditions are needed for a second-order differential equation because, roughly speaking, two integrations are required to find a solution and each integration introduces an arbitrary constant. Presumably, two initial conditions will suffice to determine values for these two constants.
A second-order linear differential equation is said to be homogeneous if the term g(t)g(t) in equation (3), or the term G(t)G(t) in equation (4), is zero for all tt. Otherwise, the equation is called nonhomogeneous. Alternatively, the nonhomogeneous term g(t)g(t), or G(t)G(t), is sometimes called the forcing function because in many applications it describes an externally applied force. We begin our discussion with homogeneous equations, which we will write in the form
Later, in Sections 3.5 and 3.6, we will show that once the homogeneous equation has been solved, it is always possible to solve the corresponding nonhomogeneous equation (4), or at least to express the solution in terms of an integral. Thus the problem of solving the homogeneous equation is the more fundamental one.
In this chapter we will concentrate our attention on equations in which the functions PP, QQ, and RR are constants. In this case, equation (7) becomes
{:(8)ay^('')+by^(')+cy=0:}\begin{equation*}
a y^{\prime \prime}+b y^{\prime}+c y=0 \tag{8}
\end{equation*}
where a,ba, b, and cc are given constants. It turns out that equation (8) can always be solved easily in terms of the elementary functions of calculus. On the other hand, it is usually much more difficult to solve equation (7) if the coefficients are not constants, and a treatment of that case is deferred until Chapter 5. Before taking up equation (8), let us first gain some experience by looking at a simple example that in many ways is typical.
Observe that equation (9) is just equation (8) with a=1,b=0a=1, b=0, and c=-1c=-1. In words, equation (9) says that we seek a function with the property that the second derivative of the function is the same as the function itself. Do any of the functions that you studied in calculus have this property? A little thought will probably produce at least one such function, namely, y_(1)(t)=e^(t)y_{1}(t)=e^{t}, the exponential function. A little more thought may also produce a second function, y_(2)(t)=e^(-t)y_{2}(t)=e^{-t}. Some further experimentation reveals that constant multiples of these two solutions are also solutions.
For example, the functions 2e^(t)2 e^{t} and 5e^(-t)5 e^{-t} also satisfy equation (9), as you can verify by calculating their second derivatives. In the same way, the functions c_(1)y_(1)(t)=c_(1)e^(t)c_{1} y_{1}(t)=c_{1} e^{t} and c_(2)y_(2)(t)=c_(2)e^(-t)c_{2} y_{2}(t)=c_{2} e^{-t} satisfy the differential equation (9) for all values of the constants c_(1)c_{1} and c_(2)c_{2}.
Next, it is vital to notice that the sum of any two solutions of equation (9) is also a solution. In particular, since c_(1)y_(1)(t)c_{1} y_{1}(t) and c_(2)y_(2)(t)c_{2} y_{2}(t) are solutions of equation (9) for any values of c_(1)c_{1} and c_(2)c_{2}, so is the function
Again, this can be verified by calculating the second derivative y^('')y^{\prime \prime} from equation (11). We have y^(')=c_(1)e^(t)-c_(2)e^(-t)y^{\prime}=c_{1} e^{t}-c_{2} e^{-t} and y^('')=c_(1)e^(t)+c_(2)e^(-t)y^{\prime \prime}=c_{1} e^{t}+c_{2} e^{-t}; thus y^('')y^{\prime \prime} is the same as yy, and equation (9) is satisfied.
Let us summarize what we have done so far in this example. Once we notice that the functions y_(1)(t)=e^(t)y_{1}(t)=e^{t} and y_(2)(t)=e^(-t)y_{2}(t)=e^{-t} are solutions of equation (9), it follows that the general linear combination (11) of these functions is also a solution. Since the coefficients c_(1)c_{1} and c_(2)c_{2} in equation (11) are arbitrary, this expression represents an infinite two-parameter family of solutions of the differential equation (9).
We now turn to the task of picking out a particular member of this infinite family of solutions that also satisfies the given pair of initial conditions (10). In other words, we seek the solution that passes through the point (0,2)(0,2) and at that point has the slope -1 . First, to ensure the solution passes through the point (0,2)(0,2), we set t=0t=0 and y=2y=2 in equation (11); this gives the equation
whose coefficients a,ba, b, and cc are arbitrary (real) constants? In the first place, in the example the solutions were exponential functions. Further, once we had identified two solutions, we were able to use a linear combination of them to satisfy the given initial conditions as well as the differential equation itself.
It turns out that by exploiting these two ideas, we can solve equation (8) for any values of its coefficients and also satisfy any given set of initial conditions for yy and y^(')y^{\prime}.
We start by seeking exponential solutions of the form y=e^(rt)y=e^{r t}, where rr is a parameter to be determined. Then it follows that y^(')=re^(rt)y^{\prime}=r e^{r t} and y^('')=r^(2)e^(rt)y^{\prime \prime}=r^{2} e^{r t}. By substituting these expressions for y,y^(')y, y^{\prime}, and y^('')y^{\prime \prime} in equation (8), we obtain
Since e^(rt)!=0e^{r t} \neq 0, this condition is satisfied only when the other factor is zero:
{:(17)ar^(2)+br+c=0:}\begin{equation*}
a r^{2}+b r+c=0 \tag{17}
\end{equation*}
Equation (17) is called the characteristic equation for the differential equation (8). Its significance lies in the fact that if rr is a root of the polynomial equation (17), then y=e^(rt)y=e^{r t} is a solution of the differential equation (8). Since equation (17) is a quadratic equation with real coefficients, it has two roots, which may be real and different, complex conjugates, or real but repeated. We consider the first case here and the latter two cases in Sections 3.3 and 3.4, respectively.
Assuming that the roots of the characteristic equation (17) are real and different, let them be denoted by r_(1)r_{1} and r_(2)r_{2}, where r_(1)!=r_(2)r_{1} \neq r_{2}. Then y_(1)(t)=e^(r_(1)t)y_{1}(t)=e^{r_{1} t} and y_(2)(t)=e^(r_(2)t)y_{2}(t)=e^{r_{2} t} are two solutions of equation (8). Just as in Example 1, it now follows that
The fact that r_(1)r_{1} is a root of equation (17) means that ar_(1)^(2)+br_(1)+c=0a r_{1}^{2}+b r_{1}+c=0. Since r_(2)r_{2} is also a root of the characteristic equation (17), it follows that ar_(2)^(2)+br_(2)+c=0a r_{2}^{2}+b r_{2}+c=0 as well. This completes the verification that yy as given by equation (18) is indeed a solution of equation (8).
Now suppose that we want to find the particular member of the family of solutions (18) that satisfies the initial conditions (6)
Since the roots of the characteristic equation (17) are assumed to be different, r_(1)-r_(2)!=0r_{1}-r_{2} \neq 0 so that the expressions in equation (24) always make sense. Thus, no matter what initial conditions are assigned-that is, regardless of the values of t_(0),y_(0)t_{0}, y_{0}, and y_(0)^(')y_{0}^{\prime} in equations (6)-it is always possible to determine c_(1)c_{1} and c_(2)c_{2} so that the initial conditions are satisfied. Moreover, there is only one possible choice of c_(1)c_{1} and c_(2)c_{2} for each set of initial conditions. With the values of c_(1)c_{1} and c_(2)c_{2} given by equation (24), the expression (18) is the solution of the initial value problem
It is possible to show, on the basis of the fundamental theorem cited in the next section, that all solutions of equation (8) are included in the expression (18), at least for the case in which the roots of equation (17) are real and different. Therefore, we call equation (18) the general solution of equation (8). The fact that any possible initial conditions can be satisfied by the proper choice of the constants in equation (18) makes more plausible the idea that this expression does include all solutions of equation (8).
The general solution of the differential equation was found in Example 2 and is given by equation (27). To satisfy the first initial condition, we set t=0t=0 and y=2y=2 in equation (27); thus c_(1)c_{1} and c_(2)c_{2} must satisfy
To use the second initial condition, we must first differentiate equation (27). This gives y^(')=-2c_(1)e^(-2t)-3c_(2)e^(-3t)y^{\prime}=-2 c_{1} e^{-2 t}-3 c_{2} e^{-3 t}. Then, setting t=0t=0 and y^(')=3y^{\prime}=3, we obtain
By solving equations (29) and (30), we find that c_(1)=9c_{1}=9 and c_(2)=-7c_{2}=-7. Using these values in the expression (27), we obtain the solution
of the initial value problem (28). The graph of the solution is shown in Figure 3.1.1.
FIGURE 3.1.1 Solution of the initial value problem (28): y^('')+5y^(')+6y=0,y(0)=2,quady^(')(0)=3y^{\prime \prime}+5 y^{\prime}+6 y=0, y(0)=2, \quad y^{\prime}(0)=3.
The solution of these equations is c_(1)=-(1)/(2),c_(2)=(5)/(2)c_{1}=-\frac{1}{2}, c_{2}=\frac{5}{2}, so the solution of the initial value problem (32) is
FIGURE 3.1.2 Solution of the initial value problem (32): 4y^('')-8y^(')+3y=0,y(0)=2,quady^(')(0)=1//24 y^{\prime \prime}-8 y^{\prime}+3 y=0, y(0)=2, \quad y^{\prime}(0)=1 / 2.
EXAMPLE 5
The solution (31) of the initial value problem (28) initially increases (because its initial slope is positive), but eventually approaches zero (because both terms involve negative exponential functions). Therefore, the solution must have a maximum point, and the graph in Figure 3.1.1 confirms this. Determine the location of this maximum point.
Solution:
The coordinates of the maximum point can be estimated from the graph, but to find them more precisely, we seek the point where the solution has a horizontal tangent line. By differentiating the solution (31), y=9e^(-2t)-7e^(-3t)y=9 e^{-2 t}-7 e^{-3 t}, with respect to tt, we obtain
Setting y^(')y^{\prime} equal to zero and multiplying by e^(3t)e^{3 t}, we find that the critical value t_(m)t_{m} satisfies e^(t)=7//6e^{t}=7 / 6; hence
In this example the initial slope is 3 , but the solution of the given differential equation behaves in a similar way for any other positive initial slope. In Problem 19 you are asked to determine how the coordinates of the maximum point depend on the initial slope.
Returning to the equation ay^('')+by^(')+cy=0a y^{\prime \prime}+b y^{\prime}+c y=0 with arbitrary coefficients, recall that when r_(1)!=r_(2)r_{1} \neq r_{2}, its general solution (18) is the sum of two exponential functions. Therefore, the solution has a relatively simple geometrical behavior: as tt increases, the magnitude of the solution either tends to zero (when both exponents are negative) or else exhibits unbounded growth (when at least one exponent is positive). These two cases are illustrated by the solutions of Examples 3 and 4, which are shown in Figures 3.1.1 and 3.1.2, respectively. Note that whether a growing solution approaches +oo+\infty or -oo-\infty as t rarr oot \rightarrow \infty is determined by the sign of the coefficient of the exponential for the larger root of the characteristic equation. (See Problem 21.) There is also a third case that occurs less often: the solution approaches a constant when one exponent is zero and the other is negative.
In Sections 3.3 and 3.4, respectively, we return to the problem of solving the equation ay^('')+by^(')+cy=0a y^{\prime \prime}+b y^{\prime}+c y=0 when the roots of the characteristic equation either are complex conjugates or are real and equal. In the meantime, in Section 3.2, we provide a systematic discussion of the mathematical structure of the solutions of all second-order linear homogeneous equations.
Problems
In each of Problems 1 through 6, find the general solution of the given differential equation.
In each of Problems 7 through 12, find the solution of the given initial value problem. Sketch the graph of the solution and describe its behavior as tt increases.
( 7. y^('')+y^(')-2y=0,quad y(0)=1,quady^(')(0)=1y^{\prime \prime}+y^{\prime}-2 y=0, \quad y(0)=1, \quad y^{\prime}(0)=1
G 8. y^('')+4y^(')+3y=0,quad y(0)=2,quady^(')(0)=-1y^{\prime \prime}+4 y^{\prime}+3 y=0, \quad y(0)=2, \quad y^{\prime}(0)=-1
(G) 9. y^('')+3y^(')=0,quad y(0)=-2,quady^(')(0)=3y^{\prime \prime}+3 y^{\prime}=0, \quad y(0)=-2, \quad y^{\prime}(0)=3
G 10. 2y^('')+y^(')-4y=0,quad y(0)=0,quady^(')(0)=12 y^{\prime \prime}+y^{\prime}-4 y=0, \quad y(0)=0, \quad y^{\prime}(0)=1
G 11. y^('')+8y^(')-9y=0,quad y(1)=1,quady^(')(1)=0y^{\prime \prime}+8 y^{\prime}-9 y=0, \quad y(1)=1, \quad y^{\prime}(1)=0
G 12. 4y^('')-y=0,quad y(-2)=1,quady^(')(-2)=-14 y^{\prime \prime}-y=0, \quad y(-2)=1, \quad y^{\prime}(-2)=-1
13. Find a differential equation whose general solution is y=c_(1)e^(2t)+c_(2)e^(-3t)y=c_{1} e^{2 t}+c_{2} e^{-3 t}.
(G) 14. Find the solution of the initial value problem
Then determine the maximum value of the solution and also find the point where the solution is zero.
16. Solve the initial value problem y^('')-y^(')-2y=0,y(0)=alphay^{\prime \prime}-y^{\prime}-2 y=0, y(0)=\alpha, y^(')(0)=2y^{\prime}(0)=2. Then find alpha\alpha so that the solution approaches zero as t rarr oot \rightarrow \infty.
In each of Problems 17 and 18, determine the values of alpha\alpha, if any, for which all solutions tend to zero as t rarr oot \rightarrow \infty; also determine the values of alpha\alpha, if any, for which all (nonzero) solutions become unbounded as t rarr oot \rightarrow \infty.
17. y^('')-(2alpha-1)y^(')+alpha(alpha-1)y=0y^{\prime \prime}-(2 \alpha-1) y^{\prime}+\alpha(\alpha-1) y=0
18. y^('')+(3-alpha)y^(')-2(alpha-1)y=0y^{\prime \prime}+(3-\alpha) y^{\prime}-2(\alpha-1) y=0
19. Consider the initial value problem (see Example 5)
where beta > 0\beta>0.
a. Solve the initial value problem.
b. Determine the coordinates t_(m)t_{m} and y_(m)y_{m} of the maximum point of the solution as functions of beta\beta.
c. Determine the smallest value of beta\beta for which y_(m) >= 4y_{m} \geq 4.
d. Determine the behavior of t_(m)t_{m} and y_(m)y_{m} as beta rarr oo\beta \rightarrow \infty.
20. Consider the equation ay^('')+by^(')+cy=da y^{\prime \prime}+b y^{\prime}+c y=d, where a,b,ca, b, c, and dd are constants.
a. Find all equilibrium, or constant, solutions of this differential equation.
b. Let y_(e)y_{e} denote an equilibrium solution, and let Y=y-y_(e)Y=y-y_{e}. Thus YY is the deviation of a solution yy from an equilibrium solution. Find the differential equation satisfied by YY.
21. Consider the equation ay^('')+by^(')+cy=0a y^{\prime \prime}+b y^{\prime}+c y=0, where a,ba, b, and cc are constants with a > 0a>0. Find conditions on a,ba, b, and cc such that the roots of the characteristic equation are:
a. real, different, and negative.
b. real with opposite signs.
c. real, different, and positive.
In each case, determine the behavior of the solution as t rarr oot \rightarrow \infty.
3.2 Solutions of Linear Homogeneous
Equations; the Wronskian
In the preceding section we showed how to solve some differential equations of the form
where a,ba, b, and cc are constants. Now we build on those results to provide a clearer picture of the structure of the solutions of all second-order linear homogeneous equations. In turn, this understanding will assist us in finding the solutions of other problems that we will encounter later.
To discuss general properties of linear differential equations, it is helpful to introduce a differential operator notation. Let pp and qq be continuous functions on an open interval II-that is, for alpha < t < beta\alpha<t<\beta. The cases for alpha=-oo\alpha=-\infty, or beta=oo\beta=\infty, or both, are included. Then, for any function phi\phi that is twice differentiable on II, we define the differential operator LL by the equation
It is important to understand that the result of applying the operator LL to a function phi\phi is another function, which we refer to as L[phi]L[\phi]. The value of L[phi]L[\phi] at a point tt is
The operator LL is often written as L=D^(2)+pD+qL=D^{2}+p D+q, where DD is the derivative operator, that is, D[phi]=phi^(')D[\phi]=\phi^{\prime}.
In this section we study the second-order linear homogeneous differential equation L[phi](t)=0L[\phi](t)=0. Since it is customary to use the symbol yy to denote phi(t)\phi(t), we will usually write this equation in the form
where t_(0)t_{0} is any point in the interval II, and y_(0)y_{0} and y_(0)^(')y_{0}^{\prime} are given real numbers. We would like to know whether the initial value problem (2), (3) always has a solution, and whether it may have more than one solution. We would also like to know whether anything can be said about the form and structure of solutions that might be helpful in finding solutions of particular problems. Answers to these questions are contained in the theorems in this section.
The fundamental theoretical result for initial value problems for second-order linear equations is stated in Theorem 3.2.1, which is analogous to Theorem 2.4.1 for first-order linear equations. The result applies equally well to nonhomogeneous equations, so the theorem is stated in that form.
Theorem 3.2.1 | (Existence and Uniqueness Theorem)
where p,qp, q, and gg are continuous on an open interval II that contains the point t_(0)t_{0}. This problem has exactly one solution y=phi(t)y=\phi(t), and the solution exists throughout the interval II.
We emphasize that the theorem says three things:
The initial value problem has a solution; in other words, a solution exists.
The initial value problem has only one solution; that is, the solution is unique.
The solution phi\phi is defined throughout the interval II where the coefficients are continuous and is at least twice differentiable there.
For some problems some of these assertions are easy to prove. For instance, we found in Example 1 of Section 3.1 that the initial value problem
The fact that we found a solution certainly establishes that a solution exists for this initial value problem. Further, the solution (6) is twice differentiable, indeed differentiable any number of times, throughout the interval (-oo,oo)(-\infty, \infty) where the coefficients in the differential equation are continuous. On the other hand, it is not obvious, and is more difficult to show, that the initial value problem (5) has no solutions other than the one given by equation (6). Nevertheless, Theorem 3.2.1 states that this solution is indeed the only solution of the initial value problem (5).
For most problems of the form (4), it is not possible to write down a useful expression for the solution. This is a major difference between first-order and second-order linear differential equations. Therefore, all parts of the theorem must be proved by general methods that do not involve having such an expression. The proof of Theorem 3.2.1 is fairly difficult, and we do not discuss it here. ^(2){ }^{2} We will, however, accept Theorem 3.2.1 as true and make use of it whenever necessary.
EXAMPLE 1
Find the longest interval in which the solution of the initial value problem
If the given differential equation is written in the form of equation (4), then
p(t)=(1)/(t-3),quad q(t)=-(t+3)/(t(t-3)),quad" and "quad g(t)=0p(t)=\frac{1}{t-3}, \quad q(t)=-\frac{t+3}{t(t-3)}, \quad \text { and } \quad g(t)=0
The only points of discontinuity of the coefficients are t=0t=0 and t=3t=3. Therefore, the longest open interval, containing the initial point t=1t=1, in which all the coefficients are continuous is 0 < t < 30<t<3. Thus this is the longest interval in which Theorem 3.2.1 guarantees that the solution exists.
EXAMPLE 2
Find the unique solution of the initial value problem
where pp and qq are continuous in an open interval II containing t_(0)t_{0}.
Solution:
The function y=phi(t)=0y=\phi(t)=0 for all tt in II certainly satisfies the differential equation and initial conditions. By the uniqueness part of Theorem 3.2.1, it is the only solution of the given problem.
Let us now assume that y_(1)y_{1} and y_(2)y_{2} are two solutions of equation (2); in other words,
and similarly for y_(2)y_{2}. Then, just as in the examples in Section 3.1, we can generate more solutions by forming linear combinations of y_(1)y_{1} and y_(2)y_{2}. We state this result as a theorem.
Theorem 3.2.2 | (Principle of Superposition)
If y_(1)y_{1} and y_(2)y_{2} are two solutions of the differential equation (2),
then the linear combination c_(1)y_(1)+c_(2)y_(2)c_{1} y_{1}+c_{2} y_{2} is also a solution for any values of the constants c_(1)c_{1} and c_(2)c_{2}.
A special case of Theorem 3.2.2 occurs if either c_(1)c_{1} or c_(2)c_{2} is zero. Then we conclude that any constant multiple of a solution of equation (2) is also a solution.
Since L[y_(1)]=0L\left[y_{1}\right]=0 and L[y_(2)]=0L\left[y_{2}\right]=0, it follows that L[c_(1)y_(1)+c_(2)y_(2)]=0L\left[c_{1} y_{1}+c_{2} y_{2}\right]=0 also. Therefore, regardless of the values of c_(1)c_{1} and c_(2)c_{2}, the function yy as given by equation (7) satisfies the differential equation (2), and the proof of Theorem 3.2.2 is complete.
Theorem 3.2.2 states that, beginning with only two solutions of equation (2), we can construct an infinite family of solutions by means of equation (7). The next question is whether all solutions of equation (2) are included in equation (7) or whether there may be other solutions of a different form. We begin to address this question by examining whether the constants c_(1)c_{1} and c_(2)c_{2} in equation (7) can be chosen so as to satisfy the initial conditions (3). These initial conditions require c_(1)c_{1} and c_(2)c_{2} to satisfy the equations
If W!=0W \neq 0, then equations (8) have a unique solution (c_(1),c_(2))\left(c_{1}, c_{2}\right) regardless of the values of y_(0)y_{0} and y_(0)^(')y_{0}^{\prime}. This solution is given by
With these values for c_(1)c_{1} and c_(2)c_{2}, the linear combination y=c_(1)y_(1)(t)+c_(2)y_(2)(t)y=c_{1} y_{1}(t)+c_{2} y_{2}(t) satisfies the initial conditions (3) as well as the differential equation (2). Note that the denominator in the expressions for c_(1)c_{1} and c_(2)c_{2} is the nonzero determinant WW.
On the other hand, if W=0W=0, then the denominators appearing in equations (10) and (11) are zero. In this case equations (8) have no solution unless y_(0)y_{0} and y_(0)^(')y_{0}^{\prime} have values that also make the numerators in equations (10) and (11) equal to zero. Thus, if W=0W=0, there are many initial conditions that cannot be satisfied no matter how c_(1)c_{1} and c_(2)c_{2} are chosen.
The determinant WW is called the Wronskian ^(3){ }^{3} determinant, or simply the Wronskian, of the solutions y_(1)y_{1} and y_(2)y_{2}. Sometimes we use the more extended notation W[y_(1),y_(2)](t_(0))W\left[y_{1}, y_{2}\right]\left(t_{0}\right) to stand for the expression on the right-hand side of equation (9), thereby emphasizing that the Wronskian depends on the functions y_(1)y_{1} and y_(2)y_{2}, and that it is evaluated at the point t_(0)t_{0}. The preceding argument establishes the following result.
Theorem 3.2.3
Suppose that y_(1)y_{1} and y_(2)y_{2} are two solutions of equation (2)
In Example 2 of Section 3.1 we found that y_(1)(t)=e^(-2t)y_{1}(t)=e^{-2 t} and y_(2)(t)=e^(-3t)y_{2}(t)=e^{-3 t} are solutions of the differential equation
Since WW is nonzero for all values of tt, the functions y_(1)y_{1} and y_(2)y_{2} can be used to construct solutions of the given differential equation, together with initial conditions prescribed at any value of tt. One such initial value problem was solved in Example 3 of Section 3.1.
The next theorem justifies the term "general solution" that we introduced in Section 3.1 for the linear combination c_(1)y_(1)+c_(2)y_(2)c_{1} y_{1}+c_{2} y_{2}. ^(3){ }^{3} Wronskian determinants are named for Jósef Maria Hoëné-Wronski (1776-1853), who was born in Poland but spent most of his life in France. Wronski was a gifted but troubled man, and his life was marked by frequent heated disputes with other individuals and institutions.
Theorem 3.2.4
Suppose that y_(1)y_{1} and y_(2)y_{2} are two solutions of the second-order linear differential equation (2),
with arbitrary coefficients c_(1)c_{1} and c_(2)c_{2} includes every solution of equation (2) if and only if there is a point t_(0)t_{0} where the Wronskian of y_(1)y_{1} and y_(2)y_{2} is not zero.
Let the function phi\phi be any solution of equation (2). To prove the theorem, we must determine whether phi\phi is included in the linear combinations c_(1)y_(1)+c_(2)y_(2)c_{1} y_{1}+c_{2} y_{2}. That is, we must determine whether there are values of the constants c_(1)c_{1} and c_(2)c_{2} that make the linear combination the same as phi\phi. Let t_(0)t_{0} be a point where the Wronskian of y_(1)y_{1} and y_(2)y_{2} is nonzero. Then evaluate phi\phi and phi^(')\phi^{\prime} at this point and call these values y_(0)y_{0} and y_(0)^(')y_{0}^{\prime}, respectively; that is,
The function phi\phi is certainly a solution of this initial value problem. Further, because we are assuming that W[y_(1),y_(2)](t_(0))W\left[y_{1}, y_{2}\right]\left(t_{0}\right) is nonzero, it is possible (by Theorem 3.2.3) to choose c_(1)c_{1} and c_(2)c_{2} such that y=c_(1)y_(1)(t)+c_(2)y_(2)(t)y=c_{1} y_{1}(t)+c_{2} y_{2}(t) is also a solution of the initial value problem (9). In fact, the proper values of c_(1)c_{1} and c_(2)c_{2} are given by equations (10) or (11). The uniqueness part of Theorem 3.2.1 guarantees that these two solutions of the same initial value problem are actually the same function; thus, for the proper choice of c_(1)c_{1} and c_(2)c_{2},
and therefore phi\phi is included in the family of functions c_(1)y_(1)+c_(2)y_(2)c_{1} y_{1}+c_{2} y_{2}. Finally, since phi\phi is an arbitrary solution of equation (2), it follows that every solution of this equation is included in this family.
Now suppose that there is no point t_(0)t_{0} where the Wronskian is nonzero. Thus W[y_(1),y_(2)](t_(0))=0W\left[y_{1}, y_{2}\right]\left(t_{0}\right)=0 for every point t_(0)t_{0}. Then (by Theorem 3.2.3) there are values of y_(0)y_{0} and y_(0)^(')y_{0}^{\prime} such that no values of c_(1)c_{1} and c_(2)c_{2} satisfy the system (8). Select a pair of such values for y_(0)y_{0} and y_(0)^(')y_{0}^{\prime} and choose the solution phi(t)\phi(t) of equation (2) that satisfies the initial condition (3). Observe that this initial value problem is guaranteed to have a solution by Theorem 3.2.1. However, this solution is not included in the family y=c_(1)y_(1)+c_(2)y_(2)y=c_{1} y_{1}+c_{2} y_{2}. Thus, in cases where W[y_(1),y_(2)](t_(0))=0W\left[y_{1}, y_{2}\right]\left(t_{0}\right)=0 for every t_(0)t_{0}, the linear combinations of y_(1)y_{1} and y_(2)y_{2} do not include all solutions of equation (2). This completes the proof of Theorem 3.2.4.
Theorem 3.2.4 states that the Wronskian of y_(1)y_{1} and y_(2)y_{2} is not everywhere zero if and only if the linear combination c_(1)y_(1)+c_(2)y_(2)c_{1} y_{1}+c_{2} y_{2} contains all solutions of equation (2). It is therefore natural (and we have already done this in the preceding section) to call the expression
with arbitrary constant coefficients the general solution of equation (2). The solutions y_(1)y_{1} and y_(2)y_{2} are said to form a fundamental set of solutions of equation (2) if and only if their Wronskian is nonzero.
We can restate the result of Theorem 3.2.4 in slightly different language: to find the general solution, and therefore all solutions, of an equation of the form (2), we need only find two solutions of the given equation whose Wronskian is nonzero. We did precisely this in several examples in Section 3.1, although there we did not calculate the Wronskians. You should now go back and do that, thereby verifying that all the solutions we called "general solutions" in Section 3.1 do satisfy the necessary Wronskian condition.
Now that you have a little experience verifying the nonzero Wronskian condition for the examples from Section 3.1, the following example handles all second-order linear differential equations whose characteristic polynomial has two distinct real roots.
EXAMPLE 4
Suppose that y_(1)(t)=e^(r_(1)t)y_{1}(t)=e^{r_{1} t} and y_(2)(t)=e^(r_(2)t)y_{2}(t)=e^{r_{2} t} are two solutions of an equation of the form (2). Show that if r_(1)!=r_(2)r_{1} \neq r_{2}, then y_(1)y_{1} and y_(2)y_{2} form a fundamental set of solutions of equation (2).
Solution:
We calculate the Wronskian of y_(1)y_{1} and y_(2)y_{2} :
Since the exponential function is never zero, and since we are assuming that r_(2)-r_(1)!=0r_{2}-r_{1} \neq 0, it follows that WW is nonzero for every value of tt. Consequently, y_(1)y_{1} and y_(2)y_{2} form a fundamental set of solutions of equation (2).
EXAMPLE 5
Show that y_(1)(t)=t^(1//2)y_{1}(t)=t^{1 / 2} and y_(2)(t)=t^(-1)y_{2}(t)=t^{-1} form a fundamental set of solutions of
{:(14)2t^(2)y^('')+3ty^(')-y=0","quad t > 0:}\begin{equation*}
2 t^{2} y^{\prime \prime}+3 t y^{\prime}-y=0, \quad t>0 \tag{14}
\end{equation*}
Solution:
We will show how to solve equation (14) later (see Problem 25 in Section 3.3). However, at this stage we can verify by direct substitution that y_(1)y_{1} and y_(2)y_{2} are solutions of the differential equation (14). Since y_(1)^(')(t)=(1)/(2)t^(-1//2)y_{1}^{\prime}(t)=\frac{1}{2} t^{-1 / 2} and y_(1)^('')(t)=-(1)/(4)t^(-3//2)y_{1}^{\prime \prime}(t)=-\frac{1}{4} t^{-3 / 2}, we have
Since W!=0W \neq 0 for t > 0t>0, we conclude that y_(1)y_{1} and y_(2)y_{2} form a fundamental set of solutions there. Thus the general solution of differential equation (14) is y(t)=c_(1)t^(1//2)+c_(2)t^(-1)y(t)=c_{1} t^{1 / 2}+c_{2} t^{-1} for t > 0t>0.
In several cases we have been able to find a fundamental set of solutions, and therefore the general solution, of a given differential equation. However, this is often a difficult task, and the question arises as to whether a differential equation of the form (2) always has a fundamental set of solutions. The following theorem provides an affirmative answer to this question.
whose coefficients pp and qq are continuous on some open interval II. Choose some point t_(0)t_{0} in II. Let y_(1)y_{1} be the solution of equation (2) that also satisfies the initial conditions
Then y_(1)y_{1} and y_(2)y_{2} form a fundamental set of solutions of equation (2).
First observe that the existence of the functions y_(1)y_{1} and y_(2)y_{2} is ensured by the existence part of Theorem 3.2.1. To show that they form a fundamental set of solutions, we need only calculate their Wronskian at t_(0)t_{0} :
Since their Wronskian is not zero at the point t_(0)t_{0}, the functions y_(1)y_{1} and y_(2)y_{2} do form a fundamental set of solutions, thus completing the proof of Theorem 3.2.5.
Note that the potentially difficult part of this proof, demonstrating the existence of a pair of solutions, is taken care of by reference to Theorem 3.2.1. Note also that Theorem 3.2.5 does not address the question of how to find the solutions y_(1)y_{1} and y_(2)y_{2} by solving the specified initial value problems. Nevertheless, it may be reassuring to know that a fundamental set of solutions always exists.
EXAMPLE 6
Find the fundamental set of solutions y_(1)y_{1} and y_(2)y_{2} specified by Theorem 3.2.5 for the differential equation
In Section 3.1 we noted that two solutions of equation (16) are y_(1)(t)=e^(t)y_{1}(t)=e^{t} and y_(2)(t)=e^(-t)y_{2}(t)=e^{-t}. The Wronskian of these solutions is W[y_(1),y_(2)](t)=-2!=0W\left[y_{1}, y_{2}\right](t)=-2 \neq 0, so they form a fundamental set of solutions. However, they are not the fundamental solutions indicated by Theorem 3.2.5 because they do not satisfy the initial conditions mentioned in that theorem at the point t=0t=0.
To find the fundamental solutions specified by the theorem, we need to find the solutions satisfying the proper initial conditions. Let us denote by y_(3)(t)y_{3}(t) the solution of equation (16) that satisfies the initial conditions
these functions also form a fundamental set of solutions, as stated by Theorem 3.2.5. Therefore, the general solution of equation (16) can be written as
as well as in the form (18). We have used k_(1)k_{1} and k_(2)k_{2} for the arbitrary constants in equation (20) because they are not the same as the constants c_(1)c_{1} and c_(2)c_{2} in equation (18). One purpose of this example is to make it clear that a given differential equation has more than one fundamental set of solutions; indeed, it has infinitely many (see Problem 16). As a rule, you should choose the set that is most convenient.
In the next section we will encounter equations that have complex-valued solutions. The following theorem is fundamental in dealing with such equations and their solutions.
Theorem 3.2.6
Consider again the second-order linear differential equation (2),
where pp and qq are continuous real-valued functions. If y=u(t)+iv(t)y=u(t)+i v(t) is a complex-valued solution of differential equation (2), then its real part uu and its imaginary part vv are also solutions of this equation.
To prove this theorem, we substitute u(t)+iv(t)u(t)+i v(t) for yy in L[y]L[y], obtaining
Then, by separating equation (21) into its real and imaginary parts - and this is where we need to know that p(t)p(t) and q(t)q(t) are real-valued-we find that
Recall that a complex number is zero if and only if its real and imaginary parts are both zero. We know that L[y]=0L[y]=0 because yy is a solution of equation (2). Therefore, both L[u]=0L[u]=0 and L[v]=0L[v]=0; consequently, the two real-valued functions uu and vv are also solutions of equation (2), so the theorem is established. We will see examples of the use of Theorem 3.2.6 in Section 3.3.
Incidentally, the complex conjugate bar(y)\bar{y} of a solution yy is also a solution. While this can be proved by an argument similar to the one just used to prove Theorem 3.2.6, it is also a consequence of Theorem 3.2.2 since bar(y)=u(t)-iv(t)\bar{y}=u(t)-i v(t) is a linear combination of two solutions.
Now let us examine further the properties of the Wronskian of two solutions of a secondorder linear homogeneous differential equation. The following theorem, perhaps surprisingly, gives a simple explicit formula for the Wronskian of any two solutions of any such equation, even if the solutions themselves are not known.
Theorem 3.2.7 | (Abel's Theorem) ^(4){ }^{4}
If y_(1)y_{1} and y_(2)y_{2} are solutions of the second-order linear differential equation
where cc is a certain constant that depends on y_(1)y_{1} and y_(2)y_{2}, but not on tt. Further, W[y_(1),y_(2)](t)W\left[y_{1}, y_{2}\right](t) either is zero for all tt in II (if c=0c=0 ) or else is never zero in II (if c!=0c \neq 0 ).
To prove Abel's theorem, we start by noting that y_(1)y_{1} and y_(2)y_{2} satisfy
Equation (27) can be solved immediately since it is both a first-order linear differential equation (Section 2.1) and a separable differential equation (Section 2.2). Thus
where cc is a constant.
The value of cc depends on which pair of solutions of equation (22) is involved. However, since the exponential function is never zero, W(t)W(t) is not zero unless c=0c=0, in which case W(t)W(t) is zero for all tt. This completes the proof of Theorem 3.2.7.
Note that the Wronskians of any two fundamental sets of solutions of the same differential equation can differ only by a multiplicative constant, and that the Wronskian of any fundamental set of solutions can be determined, up to a multiplicative constant, without solving the differential equation. Further, since under the conditions of Theorem 3.2.7 the Wronskian WW is either always zero or never zero, you can determine which case actually occurs by evaluating WW at any single convenient value of tt.
EXAMPLE 7
In Example 5 we verified that y_(1)(t)=t^(1//2)y_{1}(t)=t^{1 / 2} and y_(2)(t)=t^(-1)y_{2}(t)=t^{-1} are solutions of the equation
{:(29)2t^(2)y^('')+3ty^(')-y=0","quad t > 0:}\begin{equation*}
2 t^{2} y^{\prime \prime}+3 t y^{\prime}-y=0, \quad t>0 \tag{29}
\end{equation*}
Verify that the Wronskian of y_(1)y_{1} and y_(2)y_{2} is given by Abel's formula (23).
Solution:
From the example just cited we know that W[y_(1),y_(2)](t)=-(3)/(2)t^(-3//2)W\left[y_{1}, y_{2}\right](t)=-\frac{3}{2} t^{-3 / 2}. To use equation (23), we must write the differential equation (29) in the standard form with the coefficient of y^('')y^{\prime \prime} equal to 1 . Thus we obtain
Equation (30) gives the Wronskian of any pair of solutions of equation (29). For the particular solutions given in this example, we must choose c=-(3)/(2)c=-\frac{3}{2}.
Summary. We can summarize the discussion in this section as follows: to find the general solution of the differential equation
we must first find two functions y_(1)y_{1} and y_(2)y_{2} that satisfy the differential equation in alpha < t < beta\alpha<t<\beta. Then we must make sure that there is a point in the interval where the Wronskian WW of y_(1)y_{1} and y_(2)y_{2} is nonzero. Under these circumstances y_(1)y_{1} and y_(2)y_{2} form a fundamental set of solutions, and the general solution is
where c_(1)c_{1} and c_(2)c_{2} are arbitrary constants. If initial conditions are prescribed at a given point in alpha < t < beta\alpha<t<\beta, then c_(1)c_{1} and c_(2)c_{2} can be chosen so as to satisfy these conditions.
Problems
In each of Problems 1 through 5, find the Wronskian of the given pair of functions.
In each of Problems 6 through 9, determine the longest interval in which the given initial value problem is certain to have a unique twicedifferentiable solution. Do not attempt to find the solution.
6. quad ty^('')+3y=t,quad y(1)=1,quady^(')(1)=2\quad t y^{\prime \prime}+3 y=t, \quad y(1)=1, \quad y^{\prime}(1)=2
7. t(t-4)y^('')+3ty^(')+4y=2,quad y(3)=0,quady^(')(3)=-1t(t-4) y^{\prime \prime}+3 t y^{\prime}+4 y=2, \quad y(3)=0, \quad y^{\prime}(3)=-1
8. y^('')+(cos t)y^(')+3(ln |t|)y=0,quad y(2)=3,quady^(')(2)=1y^{\prime \prime}+(\cos t) y^{\prime}+3(\ln |t|) y=0, \quad y(2)=3, \quad y^{\prime}(2)=1
9. (x-2)y^('')+y^(')+(x-2)(tan x)y=0,quad y(3)=1,quady^(')(3)=2(x-2) y^{\prime \prime}+y^{\prime}+(x-2)(\tan x) y=0, \quad y(3)=1, \quad y^{\prime}(3)=2
10. Verify that y_(1)(t)=t^(2)y_{1}(t)=t^{2} and y_(2)(t)=t^(-1)y_{2}(t)=t^{-1} are two solutions of the differential equation t^(2)y^('')-2y=0t^{2} y^{\prime \prime}-2 y=0 for t > 0t>0. Then show that y=c_(1)t^(2)+c_(2)t^(-1)y=c_{1} t^{2}+c_{2} t^{-1} is also a solution of this equation for any c_(1)c_{1} and c_(2)c_{2}.
11. Verify that y_(1)(t)=1y_{1}(t)=1 and y_(2)(t)=t^(1//2)y_{2}(t)=t^{1 / 2} are solutions of the differential equation yy^('')+(y^('))^(2)=0y y^{\prime \prime}+\left(y^{\prime}\right)^{2}=0 for t > 0t>0. Then show that y=c_(1)+c_(2)t^(1//2)y=c_{1}+c_{2} t^{1 / 2} is not, in general, a solution of this equation. Explain why this result does not contradict Theorem 3.2.2.
12. Show that if y=phi(t)y=\phi(t) is a solution of the differential equation y^('')+p(t)y^(')+q(t)y=g(t)y^{\prime \prime}+p(t) y^{\prime}+q(t) y=g(t), where g(t)g(t) is not always zero, then y=c phi(t)y=c \phi(t), where cc is any constant other than 1 , is not a solution. Explain why this result does not contradict the remark following Theorem 3.2.2.
13. Can y=sin(t^(2))y=\sin \left(t^{2}\right) be a solution on an interval containing t=0t=0 of an equation y^('')+p(t)y^(')+q(t)y=0y^{\prime \prime}+p(t) y^{\prime}+q(t) y=0 with continuous coefficients? Explain your answer.
14. If the Wronskian WW of ff and gg is 3e^(4t)3 e^{4 t}, and if f(t)=e^(2t)f(t)=e^{2 t}, find g(t)g(t).
15. If the Wronskian of ff and gg is t cos t-sin tt \cos t-\sin t, and if u=f+3g,v=f-gu=f+3 g, v=f-g, find the Wronskian of uu and vv.
16. Assume that y_(1)y_{1} and y_(2)y_{2} are a fundamental set of solutions of y^('')+p(t)y^(')+q(t)y=0y^{\prime \prime}+p(t) y^{\prime}+q(t) y=0 and let y_(3)=a_(1)y_(1)+a_(2)y_(2)y_{3}=a_{1} y_{1}+a_{2} y_{2} and y_(4)=b_(1)y_(1)+b_(2)y_(2)y_{4}=b_{1} y_{1}+b_{2} y_{2}, where a_(1),a_(2),b_(1)a_{1}, a_{2}, b_{1}, and b_(2)b_{2} are any constants. Show that
Are y_(3)y_{3} and y_(4)y_{4} also a fundamental set of solutions? Why or why not?
In each of Problems 17 and 18, find the fundamental set of solutions specified by Theorem 3.2.5 for the given differential equation and initial point.
17. y^('')+y^(')-2y=0,quadt_(0)=0y^{\prime \prime}+y^{\prime}-2 y=0, \quad t_{0}=0
18. y^('')+4y^(')+3y=0,quadt_(0)=1y^{\prime \prime}+4 y^{\prime}+3 y=0, \quad t_{0}=1
In each of Problems 19 through 21, verify that the functions y_(1)y_{1} and y_(2)y_{2} are solutions of the given differential equation. Do they constitute a fundamental set of solutions?
19. y^('')+4y=0;quady_(1)(t)=cos(2t),quady_(2)(t)=sin(2t)y^{\prime \prime}+4 y=0 ; \quad y_{1}(t)=\cos (2 t), \quad y_{2}(t)=\sin (2 t)
20. y^('')-2y^(')+y=0;quady_(1)(t)=e^(t),quady_(2)(t)=te^(t)y^{\prime \prime}-2 y^{\prime}+y=0 ; \quad y_{1}(t)=e^{t}, \quad y_{2}(t)=t e^{t}
21. x^(2)y^('')-x(x+2)y^(')+(x+2)y=0,x > 0x^{2} y^{\prime \prime}-x(x+2) y^{\prime}+(x+2) y=0, x>0; y_(1)(x)=x,quady_(2)(x)=xe^(x)y_{1}(x)=x, \quad y_{2}(x)=x e^{x}
22. Consider the equation y^('')-y^(')-2y=0y^{\prime \prime}-y^{\prime}-2 y=0.
a. Show that y_(1)(t)=e^(-t)y_{1}(t)=e^{-t} and y_(2)(t)=e^(2t)y_{2}(t)=e^{2 t} form a fundamental set of solutions.
b. Let y_(3)(t)=-2e^(2t),y_(4)(t)=y_(1)(t)+2y_(2)(t)y_{3}(t)=-2 e^{2 t}, y_{4}(t)=y_{1}(t)+2 y_{2}(t), and y_(5)(t)=2y_(1)(t)-2y_(3)(t)y_{5}(t)=2 y_{1}(t)-2 y_{3}(t). Are y_(3)(t),y_(4)(t)y_{3}(t), y_{4}(t), and y_(5)(t)y_{5}(t) also solutions of the given differential equation?
c. Determine whether each of the following pairs forms a fundamental set of solutions: {y_(1)(t),y_(3)(t)};{y_(2)(t),y_(3)(t)}\left\{y_{1}(t), y_{3}(t)\right\} ;\left\{y_{2}(t), y_{3}(t)\right\};
In each of Problems 23 through 25, find the Wronskian of two solutions of the given differential equation without solving the equation.
23. t^(2)y^('')-t(t+2)y^(')+(t+2)y=0t^{2} y^{\prime \prime}-t(t+2) y^{\prime}+(t+2) y=0
24. (cos t)y^('')+(sin t)y^(')-ty=0(\cos t) y^{\prime \prime}+(\sin t) y^{\prime}-t y=0
25. (1-x^(2))y^('')-2xy^(')+alpha(alpha+1)y=0\left(1-x^{2}\right) y^{\prime \prime}-2 x y^{\prime}+\alpha(\alpha+1) y=0, Legendre's equation
26. Show that if pp is differentiable and p(t) > 0p(t)>0, then the Wronskian W(t)W(t) of two solutions of [p(t)y^(')]^(')+q(t)y=0\left[p(t) y^{\prime}\right]^{\prime}+q(t) y=0 is W(t)=c//p(t)W(t)=c / p(t), where cc is a constant.
27. If the differential equation ty^('')+2y^(')+te^(t)y=0t y^{\prime \prime}+2 y^{\prime}+t e^{t} y=0 has y_(1)y_{1} and y_(2)y_{2} as a fundamental set of solutions and if W[y_(1),y_(2)](1)=2W\left[y_{1}, y_{2}\right](1)=2, find the value of W[y_(1),y_(2)](5)W\left[y_{1}, y_{2}\right](5).
28. If the Wronskian of any two solutions of y^('')+p(t)y^(')+q(t)y=0y^{\prime \prime}+p(t) y^{\prime}+q(t) y=0 is constant, what does this imply about the coefficients pp and qq ?
In Problems 29 and 30, assume that pp and qq are continuous and that the functions y_(1)y_{1} and y_(2)y_{2} are solutions of the differential equation y^('')+p(t)y^(')+q(t)y=0y^{\prime \prime}+p(t) y^{\prime}+q(t) y=0 on an open interval II.
29. Prove that if y_(1)y_{1} and y_(2)y_{2} are zero at the same point in II, then they cannot be a fundamental set of solutions on that interval.
30. Prove that if y_(1)y_{1} and y_(2)y_{2} have a common point of inflection t_(0)t_{0} in II, then they cannot be a fundamental set of solutions on II unless both pp and qq are zero at t_(0)t_{0}.
31. Exact Equations. The equation
where f(x)f(x) is to be determined in terms of P(x),Q(x)P(x), Q(x), and R(x)R(x). The latter equation can be integrated once immediately, resulting in a first-order linear equation for yy that can be solved as in Section 2.1. By equating the coefficients of the preceding equations and then eliminating f(x)f(x), show that a necessary condition for exactness is
It can be shown that this is also a sufficient condition.
In each of Problems 32 through 34, use the result of Problem 31 to determine whether the given equation is exact. If it is, then solve the equation.
32. y^('')+xy^(')+y=0y^{\prime \prime}+x y^{\prime}+y=0
33. xy^('')-(cos x)y^(')+(sin x)y=0,quad x > 0x y^{\prime \prime}-(\cos x) y^{\prime}+(\sin x) y=0, \quad x>0
34. x^(2)y^('')+xy^(')-y=0,x > 0x^{2} y^{\prime \prime}+x y^{\prime}-y=0, x>0
35. The Adjoint Equation. If a second-order linear homogeneous equation is not exact, it can be made exact by multiplying by an appropriate integrating factor mu(x)\mu(x). Thus we require that mu(x)\mu(x) be such that
This equation is known as the adjoint of the original equation and is important in the advanced theory of differential equations. In general,
the problem of solving the adjoint differential equation is as difficult as that of solving the original equation, so only occasionally is it possible to find an integrating factor for a second-order equation.
In each of Problems 36 and 37, use the result of Problem 35 to find the adjoint of the given differential equation.
36. x^(2)y^('')+xy^(')+(x^(2)-nu^(2))y=0x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-\nu^{2}\right) y=0, Bessel's equation
37. y^('')-xy=0y^{\prime \prime}-x y=0, Airy's equation
38. A second-order linear equation P(x)y^('')+Q(x)y^(')+R(x)y=0P(x) y^{\prime \prime}+Q(x) y^{\prime}+R(x) y=0 is said to be self-adjoint if its adjoint is the same as the original equation. Show that a necessary condition for this equation to be self-adjoint is that P^(')(x)=Q(x)P^{\prime}(x)=Q(x). Determine whether each of the equations in Problems 36 and 37 is self-adjoint.
3.3 Complex Roots of the Characteristic Equation
We continue our discussion of the second-order linear differential equation
{:(1)ay^('')+by^(')+cy=0",":}\begin{equation*}
a y^{\prime \prime}+b y^{\prime}+c y=0, \tag{1}
\end{equation*}
where a,ba, b, and cc are given real numbers. In Section 3.1 we found that if we seek solutions of the form y=e^(rt)y=e^{r t}, then rr must be a root of the characteristic equation
{:(2)ar^(2)+br+c=0:}\begin{equation*}
a r^{2}+b r+c=0 \tag{2}
\end{equation*}
We showed in Section 3.1 that if the roots r_(1)r_{1} and r_(2)r_{2} are real and different, which occurs whenever the discriminant b^(2)-4acb^{2}-4 a c is positive, then the general solution of equation (1) is
Our first task is to explore what is meant by these expressions, which involve evaluating the exponential function for a complex exponent. For example, if lambda=-1,mu=2\lambda=-1, \mu=2, and t=3t=3, then from equation (5),
What does it mean to raise the number ee to a complex power? The answer is provided by an important relation known as Euler's formula.
Euler's Formula. To assign a meaning to the expressions in equations (5), we need to give a definition of the complex exponential function. Of course, we want the definition to reduce to the familiar real exponential function when the exponent is real. There are several ways to discover how this extension of the exponential function should be defined. Here we use a method based on infinite series; an alternative is outlined in Problem 20.
Recall from calculus that the Taylor series for e^(t)e^{t} about t=0t=0 is
To simplify this series, we write (it)^(n)=i^(n)t^(n)(i t)^{n}=i^{n} t^{n} and make use of the facts that i^(2)=-1i^{2}=-1, i^(3)=-i,i^(4)=1i^{3}=-i, i^{4}=1, and so forth. When nn is even, there is an integer kk with n=2kn=2 k; in this case i^(n)=i^(2k)=(-1)^(k)i^{n}=i^{2 k}=(-1)^{k}. And when nn is odd, n=2k+1n=2 k+1, so i^(n)=i^(2k+1)=i(-1)^(k)i^{n}=i^{2 k+1}=i(-1)^{k}. This suggests separating the terms in the right-hand side of (8) into its real and imaginary parts. The result is ^(5){ }^{5}
The first series in equation (9) is precisely the Taylor series for cos t\cos t about t=0t=0, and the second is the Taylor series for sin t\sin t about t=0t=0. Thus we have
{:(10)e^(it)=cos t+i sin t:}\begin{equation*}
e^{i t}=\cos t+i \sin t \tag{10}
\end{equation*}
Equation (10) is known as Euler's formula and is an extremely important mathematical relationship.
Although our derivation of equation (10) is based on the unverified assumption that the series (7) can be used for complex as well as real values of the independent variable, our intention is to use this derivation only to make equation (10) seem plausible. We now put matters on a firm foundation by adopting equation (10) as the definition of e^(it)e^{i t}. In other words, whenever we write e^(it)e^{i t}, we mean the expression on the right-hand side of equation (10).
There are some variations of Euler's formula that are also worth noting. If we replace tt by -t-t in equation (10) and recall that cos(-t)=cos t\cos (-t)=\cos t and sin(-t)=-sin t\sin (-t)=-\sin t, then we have
{:(11)e^(-it)=cos t-i sin t:}\begin{equation*}
e^{-i t}=\cos t-i \sin t \tag{11}
\end{equation*}
Further, if tt is replaced by mu t\mu t in equation (10), then we obtain a generalized version of Euler's formula, namely,
Next, we want to extend the definition of the exponential function to arbitrary complex exponents of the form (lambda+i mu)t(\lambda+i \mu) t. Since we want the usual properties of the exponential function to hold for complex exponents, we certainly want exp((lambda+i mu)t)\exp ((\lambda+i \mu) t) to satisfy
We now take equation (14) as the definition of exp[(lambda+i mu)t]\exp [(\lambda+i \mu) t]. The value of the exponential function with a complex exponent is a complex number whose real and imaginary parts are given by the terms on the right-hand side of equation (14). Observe that the real and imaginary parts of exp((lambda+i mu)t)\exp ((\lambda+i \mu) t) are expressed entirely in terms of elementary real-valued functions. For example, the quantity in equation (6) has the value
With the definitions (10) and (14), it is straightforward to show that the usual laws of exponents are valid for the complex exponential function. You can also use equation (14) to verify that the differentiation formula
You can verify that the Wronskian W[y_(1),y_(2)](t)=-6ie^(-t)W\left[y_{1}, y_{2}\right](t)=-6 i e^{-t}, which is not zero, so the general solution of equation (15) can be expressed as a linear combination of y_(1)(t)y_{1}(t) and y_(2)(t)y_{2}(t) with arbitrary coefficients.
However, the initial value problem (16), (17) has only real coefficients, and it is often desirable to express the solution of such a problem in terms of real-valued functions. To do this we can make use of Theorem 3.2.6, which states that the real and imaginary parts of a complex-valued solution of equation (16) are also solutions of the same differential equation. Thus, starting from y_(1)(t)y_{1}(t), we obtain
as real-valued solutions ^(6){ }^{6} of equation (16). On calculating the Wronskian of u(t)u(t) and v(t)v(t), we find that W[u,v](t)=3e^(-t)W[u, v](t)=3 e^{-t}, which is not zero; thus u(t)u(t) and v(t)v(t) form a fundamental set of solutions, and the general solution of equation (16) can be written as
where c_(1)c_{1} and c_(2)c_{2} are arbitrary constants.
To satisfy the initial conditions (17), we first substitute t=0t=0 and y=2y=2 in the solution (20) with the result that c_(1)=2c_{1}=2. Then, by differentiating equation (21), setting t=0t=0, and setting y^(')=8y^{\prime}=8, we obtain -(1)/(2)c_(1)+3c_(2)=8-\frac{1}{2} c_{1}+3 c_{2}=8 so that c_(2)=3c_{2}=3. Thus the solution of the initial value problem (16), (17) is
The graph of this solution is shown in Figure 3.3.1.
From the graph we see that the solution of this problem oscillates, with period 2pi//32 \pi / 3 and a decaying amplitude. The sine and cosine factors control the oscillatory nature of the solution, and the negative exponential factor in each term causes the magnitude of the oscillations to decrease toward zero as time increases. ^(6){ }^{6} If you are not completely sure that u(t)u(t) and v(t)v(t) are solutions of the given differential equation, you should substitute these functions into equation (16) and confirm that they satisfy it. (See Problem 23.)
FIGURE 3.3.1 Solution of the initial value problem (16), (17):
Complex Roots; The General Case. The functions y_(1)(t)y_{1}(t) and y_(2)(t)y_{2}(t), given by equations (5) and with the meaning expressed by equation (14), are solutions of equation (1) when the roots of the characteristic equation (2) are complex numbers lambda+-i mu\lambda \pm i \mu. However, the solutions y_(1)y_{1} and y_(2)y_{2} are complex-valued functions, whereas in general we would prefer to have real-valued solutions because the differential equation itself has real coefficients. Just as in Example 1, we can use Theorem 3.2.6 to find a fundamental set of real-valued solutions by choosing the real and imaginary parts of either y_(1)(t)y_{1}(t) or y_(2)(t)y_{2}(t). In this way we obtain the solutions
Thus, as long as mu!=0\mu \neq 0, the Wronskian WW is not zero, so uu and vv form a fundamental set of solutions. (Of course, if mu=0\mu=0, then the roots are real and equal and the discussions in this section, and in Section 3.1, are not applicable.) Consequently, if the roots of the characteristic equation are complex numbers lambda+-i mu\lambda \pm i \mu, with mu!=0\mu \neq 0, then the general solution of equation (1) is
where c_(1)c_{1} and c_(2)c_{2} are arbitrary constants. Note that the solution (25) can be written down as soon as the values of lambda\lambda and mu\mu are known. Let us now consider some further examples.
The characteristic equation is 16r^(2)-8r+145=016 r^{2}-8 r+145=0 and its roots are r=(1)/(4)+-3ir=\frac{1}{4} \pm 3 i. Thus the general solution of the differential equation is
as the solution of the initial value problem (26). The graph of this solution is shown in Figure 3.3.2.
In this case we observe that the solution is a growing oscillation. Again the trigonometric factors in equation (28) determine the oscillatory part of the solution (again with period 2pi//32 \pi / 3 ), while the exponential factor (with a positive exponent this time) causes the magnitude of the oscillation to increase with time.
FIGURE 3.3.2 Solution of the initial value problem (26): 16y^('')-8y^(')+145 y=0,y(0)=-2,y^(')(0)=116 y^{\prime \prime}-8 y^{\prime}+145 y=0, y(0)=-2, y^{\prime}(0)=1.
FIGURE 3.3.3 Solutions of equation (29): y^('')+9y=0y^{\prime \prime}+9 y=0, with two sets of initial conditions: y(0)=1,y^(')(0)=2y(0)=1, y^{\prime}(0)=2 (dashed, green) and y(0)=2,y^(')(0)=8y(0)=2, y^{\prime}(0)=8 (solid, blue). Both solutions have the same period, but different amplitudes and phase shifts.
Note that if the real part of the roots is zero, as in this example, then there is no exponential factor in the solution. Figure 3.3.3 shows the graph of two solutions of equation (28) with different initial conditions. In each case the solution is a pure oscillation with period 2pi//32 \pi / 3 but whose amplitude and phase shift are determined by the initial conditions. Since there is no exponential factor in the solution (30), the amplitude of each oscillation remains constant in time.
Problems
In each of Problems 1 through 4, use Euler's formula to write the given expression in the form a+iba+i b.
exp(2-3i)\exp (2-3 i)
e^(i pi)e^{i \pi}
e^(2-(pi//2)i)e^{2-(\pi / 2) i}
2^(1-i)2^{1-i}
In each of Problems 5 through 11, find the general solution of the given differential equation.
5. y^('')-2y^(')+2y=0y^{\prime \prime}-2 y^{\prime}+2 y=0
6. y^('')-2y^(')+6y=0y^{\prime \prime}-2 y^{\prime}+6 y=0
7. y^('')+2y^(')+2y=0y^{\prime \prime}+2 y^{\prime}+2 y=0
8. y^('')+6y^(')+13 y=0y^{\prime \prime}+6 y^{\prime}+13 y=0
9. y^('')+2y^(')+1.25 y=0y^{\prime \prime}+2 y^{\prime}+1.25 y=0
10. 9y^('')+9y^(')-4y=09 y^{\prime \prime}+9 y^{\prime}-4 y=0
11. y^('')+4y^(')+6.25 y=0y^{\prime \prime}+4 y^{\prime}+6.25 y=0
In each of Problems 12 through 15, find the solution of the given initial value problem. Sketch the graph of the solution and describe its behavior for increasing tt.
(G 12. y^('')+4y=0,quad y(0)=0,quady^(')(0)=1y^{\prime \prime}+4 y=0, \quad y(0)=0, \quad y^{\prime}(0)=1
(G)13. y^('')-2y^(')+5y=0,quad y(pi//2)=0,quady^(')(pi//2)=2y^{\prime \prime}-2 y^{\prime}+5 y=0, \quad y(\pi / 2)=0, \quad y^{\prime}(\pi / 2)=2
(G) 14. y^('')+y=0,quad y(pi//3)=2,quady^(')(pi//3)=-4y^{\prime \prime}+y=0, \quad y(\pi / 3)=2, \quad y^{\prime}(\pi / 3)=-4
(G) 15. y^('')+2y^(')+2y=0,quad y(pi//4)=2,quady^(')(pi//4)=-2y^{\prime \prime}+2 y^{\prime}+2 y=0, \quad y(\pi / 4)=2, \quad y^{\prime}(\pi / 4)=-2
(N) 16. Consider the initial value problem
a. Find the solution u(t)u(t) of this problem.
b. Find the smallest TT such that |u(t)| <= 0.1|u(t)| \leq 0.1 for all t > Tt>T.
(N) 18. Consider the initial value problem
a. Find the solution y(t)y(t) of this problem.
b. Find alpha\alpha such that y=0y=0 when t=1t=1.
c. Find, as a function of alpha\alpha, the smallest positive value of tt for which y=0y=0.
d. Determine the limit of the expression found in part c as alpha rarr oo\alpha \rightarrow \infty.
19. Show that W[e^(lambda t)cos(mu t),e^(lambda t)sin(mu t)]=mue^(2lambda t)W\left[e^{\lambda t} \cos (\mu t), e^{\lambda t} \sin (\mu t)\right]=\mu e^{2 \lambda t}.
20. In this problem we outline a different derivation of Euler's formula.
a. Show that y_(1)(t)=cos ty_{1}(t)=\cos t and y_(2)(t)=sin ty_{2}(t)=\sin t are a fundamental set of solutions of y^('')+y=0y^{\prime \prime}+y=0; that is, show that they are solutions and that their Wronskian is not zero.
b. Show (formally) that y=e^(it)y=e^{i t} is also a solution of y^('')+y=0y^{\prime \prime}+y=0. Therefore,
for some constants c_(1)c_{1} and c_(2)c_{2}. Why is this so?
c. Set t=0t=0 in equation (31) to show that c_(1)=1c_{1}=1.
d. Assuming that equation (15) is true, differentiate equation (31) and then set t=0t=0 to conclude that c_(2)=ic_{2}=i. Use the values of c_(1)c_{1} and c_(2)c_{2} in equation (31) to arrive at Euler's formula.
21. Using Euler's formula, show that
If e^(rt)e^{r t} is given by equation (14), show that e^((r_(1)+r_(2))t)=e^(r_(1)t)e^(r_(2)t)e^{\left(r_{1}+r_{2}\right) t}=e^{r_{1} t} e^{r_{2} t} for any complex numbers r_(1)r_{1} and r_(2)r_{2}.
where b^(2)-4ac < 0b^{2}-4 a c<0 and the characteristic equation has complex roots lambda+-i mu\lambda \pm i \mu. Substitute the functions
u(t)=e^(lambda t)cos(mu t)" and "v(t)=e^(lambda t)sin(mu t)u(t)=e^{\lambda t} \cos (\mu t) \text { and } v(t)=e^{\lambda t} \sin (\mu t)
for yy in the differential equation and thereby confirm that they are solutions.
24. If the functions y_(1)y_{1} and y_(2)y_{2} are a fundamental set of solutions of y^('')+p(t)y^(')+q(t)y=0y^{\prime \prime}+p(t) y^{\prime}+q(t) y=0, show that between consecutive zeros of y_(1)y_{1} there is one and only one zero of y_(2)y_{2}. Note that this result is illustrated by the solutions y_(1)(t)=cos ty_{1}(t)=\cos t and y_(2)(t)=sin ty_{2}(t)=\sin t of the equation y^('')+y=0y^{\prime \prime}+y=0.
Hint: Suppose that t_(1)t_{1} and t_(2)t_{2} are two zeros of y_(1)y_{1} between which there are no zeros of y_(2)y_{2}. Apply Rolle's theorem to y_(1)//y_(2)y_{1} / y_{2} to reach a contradiction.
Change of Variables. Sometimes a differential equation with variable coefficients,
can be put in a more suitable form for finding a solution by making a change of the independent variable. We explore these ideas in Problems 25 through 36. In particular, in Problem 25 we show that a class of equations known as Euler equations can be transformed into equations with constant coefficients by a simple change of the independent variable. Problems 26 through 31 are examples of this type of equation. Problem 32 determines conditions under which the more general equation (32) can be transformed into a differential equation with constant coefficients. Problems 33 through 36 give specific applications of this procedure.
25. Euler Equations. An equation of the form
{:(33)t^(2)(d^(2)y)/(dt^(2))+alpha t(dy)/(dt)+beta y=0","quad t > 0:}\begin{equation*}
t^{2} \frac{d^{2} y}{d t^{2}}+\alpha t \frac{d y}{d t}+\beta y=0, \quad t>0 \tag{33}
\end{equation*}
where alpha\alpha and beta\beta are real constants, is called an Euler equation.
a. Let x=ln tx=\ln t and calculate dy//dtd y / d t and d^(2)y//dt^(2)d^{2} y / d t^{2} in terms of dy//dxd y / d x and d^(2)y//dx^(2)d^{2} y / d x^{2}.
b. Use the results of part a to transform equation (33) into
Observe that differential equation (34) has constant coefficients. If y_(1)(x)y_{1}(x) and y_(2)(x)y_{2}(x) form a fundamental set of solutions of equation (34), then y_(1)(ln t)y_{1}(\ln t) and y_(2)(ln t)y_{2}(\ln t) form a fundamental set of solutions of equation (33).
In each of Problems 26 through 31, use the method of Problem 25 to solve the given equation for t > 0t>0.
26. t^(2)y^('')+ty^(')+y=0t^{2} y^{\prime \prime}+t y^{\prime}+y=0
27. t^(2)y^('')+4ty^(')+2y=0t^{2} y^{\prime \prime}+4 t y^{\prime}+2 y=0
28. t^(2)y^('')-4ty^(')-6y=0t^{2} y^{\prime \prime}-4 t y^{\prime}-6 y=0
29. t^(2)y^('')-4ty^(')+6y=0t^{2} y^{\prime \prime}-4 t y^{\prime}+6 y=0
30. t^(2)y^('')+3ty^(')-3y=0t^{2} y^{\prime \prime}+3 t y^{\prime}-3 y=0
31. t^(2)y^('')+7ty^(')+10 y=0t^{2} y^{\prime \prime}+7 t y^{\prime}+10 y=0
32. In this problem we determine conditions on pp and qq that enable equation (32) to be transformed into an equation with constant coefficients by a change of the independent variable. Let x=u(t)x=u(t) be the new independent variable, with the relation between xx and tt to be specified later.
a. Show that
c. In order for equation (35) to have constant coefficients, the coefficients of d^(2)y//dx^(2),dy//dxd^{2} y / d x^{2}, d y / d x, and yy must all be proportional. If q(t) > 0q(t)>0, then we can choose the constant of proportionality to be 1 ; hence, after integrating with respect to tt,
{:(36)x=u(t)=int(q(t))^(1//2)dt:}\begin{equation*}
x=u(t)=\int(q(t))^{1 / 2} d t \tag{36}
\end{equation*}
d. With xx chosen as in part c , show that the coefficient of dy//dxd y / d x in equation (35) is also a constant, provided that the expression
is a constant. Thus equation (32) can be transformed into an equation with constant coefficients by a change of the independent variable, provided that the function (q^(')+2pq)//q^(3//2)\left(q^{\prime}+2 p q\right) / q^{3 / 2} is a constant.
e. How must the analysis and results in d be modified if q(t) < 0q(t)<0 ?
In each of Problems 33 through 36, try to transform the given equation into one with constant coefficients by the method of Problem 32. If this is possible, find the general solution of the given equation.
33. y^('')+ty^(')+e^(-t^(2))y=0,quad-oo < t < ooy^{\prime \prime}+t y^{\prime}+e^{-t^{2}} y=0, \quad-\infty<t<\infty
34. y^('')+3ty^(')+t^(2)y=0,quad-oo < t < ooy^{\prime \prime}+3 t y^{\prime}+t^{2} y=0, \quad-\infty<t<\infty
35. ty^('')+(t^(2)-1)y^(')+t^(3)y=0,quad0 < t < oot y^{\prime \prime}+\left(t^{2}-1\right) y^{\prime}+t^{3} y=0, \quad 0<t<\infty
36. y^('')+ty^(')-e^(-t^(2))y=0y^{\prime \prime}+t y^{\prime}-e^{-t^{2}} y=0
3.4 Repeated Roots; Reduction of Order
In Sections 3.1 and 3.3 we showed how to solve the equation
{:(1)ay^('')+by^(')+cy=0:}\begin{equation*}
a y^{\prime \prime}+b y^{\prime}+c y=0 \tag{1}
\end{equation*}
when the roots of the characteristic equation
{:(2)ar^(2)+br+c=0:}\begin{equation*}
a r^{2}+b r+c=0 \tag{2}
\end{equation*}
either are real and different or are complex conjugates. Now we consider the third possibility, namely, that the two roots r_(1)r_{1} and r_(2)r_{2} are equal. This case is transitional between the other two and occurs when the discriminant b^(2)-4acb^{2}-4 a c is zero. Then it follows from the quadratic formula that
so r_(1)=r_(2)=-2r_{1}=r_{2}=-2. Therefore, one solution of equation (5) is y_(1)(t)=e^(-2t)y_{1}(t)=e^{-2 t}. To find the general solution of equation (5), we need a second solution that is not a constant multiple of y_(1)y_{1}. This second solution can be found in several ways (see Problems 15 through 17); here we use a method originated by d'Alembert ^(7){ }^{7} in the eighteenth century. Recall that since y_(1)(t)y_{1}(t) is a solution of equation (1), so is cy_(1)(t)c y_{1}(t) for any constant cc. The basic idea is to generalize this observation by replacing cc by a function v(t)v(t) and then trying to determine v(t)v(t) so that the product v(t)y_(1)(t)v(t) y_{1}(t) is also a solution of equation (1).
To carry out this program, we substitute y=v(t)y_(1)(t)y=v(t) y_{1}(t) in equation (5) and use the resulting equation to find v(t)v(t). Starting with
where c_(1)c_{1} and c_(2)c_{2} are arbitrary constants. Finally, substituting for v(t)v(t) in equation (6), we obtain
{:(11)y=c_(1)te^(-2t)+c_(2)e^(-2t).:}\begin{equation*}
y=c_{1} t e^{-2 t}+c_{2} e^{-2 t} . \tag{11}
\end{equation*}
The second term on the right-hand side of equation (11) corresponds to the original solution y_(1)(t)=exp(-2t)y_{1}(t)=\exp (-2 t), but the first term arises from a second solution, namely, y_(2)(t)=t exp(-2t)y_{2}(t)=t \exp (-2 t). We can verify that these two solutions form a fundamental set by calculating their Wronskian:
form a fundamental set of solutions of equation (5), and the general solution of that equation is given by equation (11). Note that both y_(1)(t)y_{1}(t) and y_(2)(t)y_{2}(t) tend to zero as t rarr oot \rightarrow \infty; consequently, all solutions of equation (5) behave in this way. The graphs of typical solutions are shown in Figure 3.4.1.
FIGURE 3.4.1 Three solutions of equation (5): y^('')+4y^(')+4y=0y^{\prime \prime}+4 y^{\prime}+4 y=0, with different sets of initial conditions: y(0)=2,y^(')(0)=1y(0)=2, y^{\prime}(0)=1 (blue, dashed); y(0)=1,y^(')(0)=1y(0)=1, y^{\prime}(0)=1 (green, solid); y(0)=1//2,y^(')(0)=1(y(0)=1 / 2, y^{\prime}(0)=1( red )).
The procedure used in Example 1 can be extended to a general equation whose characteristic equation has repeated roots. That is, we assume that the coefficients in equation (1) satisfy b^(2)-4ac=0b^{2}-4 a c=0, in which case
y_(1)(t)=e^(-bt//(2a))y_{1}(t)=e^{-b t /(2 a)}
is a solution. To find a second solution, we assume that
{:(13)y=v(t)y_(1)(t)=v(t)e^(-bt//(2a)):}\begin{equation*}
y=v(t) y_{1}(t)=v(t) e^{-b t /(2 a)} \tag{13}
\end{equation*}
and substitute for yy in equation (1) to determine v(t)v(t). We have
{:(14)y^(')=v^(')(t)e^(-bt//(2a))-(b)/(2a)v(t)e^(-bt//(2a)):}\begin{equation*}
y^{\prime}=v^{\prime}(t) e^{-b t /(2 a)}-\frac{b}{2 a} v(t) e^{-b t /(2 a)} \tag{14}
\end{equation*}
and
{:(15)y^('')=v^('')(t)e^(-bt//(2a))-(b)/(a)v^(')(t)e^(-bt//(2a))+(b^(2))/(4a^(2))v(t)e^(-bt//(2a)).:}\begin{equation*}
y^{\prime \prime}=v^{\prime \prime}(t) e^{-b t /(2 a)}-\frac{b}{a} v^{\prime}(t) e^{-b t /(2 a)}+\frac{b^{2}}{4 a^{2}} v(t) e^{-b t /(2 a)} . \tag{15}
\end{equation*}
Canceling the factor e^(-b//(2a))e^{-b /(2 a)}, which is nonzero, and rearranging the remaining terms, we find that
{:(17)av^('')(t)+(-b+b)v^(')(t)+((b^(2))/(4a)-(b^(2))/(2a)+c)v(t)=0:}\begin{equation*}
a v^{\prime \prime}(t)+(-b+b) v^{\prime}(t)+\left(\frac{b^{2}}{4 a}-\frac{b^{2}}{2 a}+c\right) v(t)=0 \tag{17}
\end{equation*}
The term involving v^(')(t)v^{\prime}(t) is obviously zero. Further, the coefficient of v(t)v(t) is c-b^(2)//(4a)c-b^{2} /(4 a), which is also zero because b^(2)-4ac=0b^{2}-4 a c=0 in the problem that we are considering. Thus, just as in Example 1, equation (17) reduces to
v^('')(t)=0v^{\prime \prime}(t)=0
so
v(t)=c_(1)+c_(2)tv(t)=c_{1}+c_{2} t
Hence, from equation (13), we have
{:(18)y=c_(1)e^(-bt//(2a))+c_(2)te^(-bt//(2a)).:}\begin{equation*}
y=c_{1} e^{-b t /(2 a)}+c_{2} t e^{-b t /(2 a)} . \tag{18}
\end{equation*}
Thus yy is a linear combination of the two solutions
{:(19)y_(1)(t)=e^(-bt//(2a))","quady_(2)(t)=te^(-bt//(2a)).:}\begin{equation*}
y_{1}(t)=e^{-b t /(2 a)}, \quad y_{2}(t)=t e^{-b t /(2 a)} . \tag{19}
\end{equation*}
The Wronskian of these two solutions is
{:(20)W(y_(1),y_(2))(t)=|[e^(-bt//(2a)),te^(-bt//(2a))],[-(b)/(2a)e^(-bt//(2a)),(1-(bt)/(2a))e^(-bt//(2a))]|=e^(-bt//a):}W\left(y_{1}, y_{2}\right)(t)=\left|\begin{array}{cc}
e^{-b t /(2 a)} & t e^{-b t /(2 a)} \tag{20}\\
-\frac{b}{2 a} e^{-b t /(2 a)} & \left(1-\frac{b t}{2 a}\right) e^{-b t /(2 a)}
\end{array}\right|=e^{-b t / a}
Since W[y_(1),y_(2)](t)W\left[y_{1}, y_{2}\right](t) is never zero, the solutions y_(1)y_{1} and y_(2)y_{2} given by equation (19) are a fundamental set of solutions. Further, equation (18) is the general solution of equation (1) when the roots of the characteristic equation are equal. In other words, in this case there is one exponential solution corresponding to the repeated root and a second solution that is obtained by multiplying the exponential solution by tt.
The graph of this solution is shown by the blue curve in Figure 3.4.2.
FIGURE 3.4.2 Solutions of y^('')-y^(')+y//4=0,y(0)=2y^{\prime \prime}-y^{\prime}+y / 4=0, y(0)=2, with y^(')(0)=1//3y^{\prime}(0)=1 / 3 (blue) and with y^(')(0)=2(y^{\prime}(0)=2( red )).
Let us now modify the initial value problem (16) by changing the initial slope; to be specific, let the second initial condition be y^(')(0)=2y^{\prime}(0)=2. The solution of this modified problem is
y=2e^(t//2)+te^(t//2)y=2 e^{t / 2}+t e^{t / 2}
and its graph is shown by the red curve in Figure 3.4.2. The graphs shown in this figure suggest that there is a critical initial slope, with a value between 1//31 / 3 and 2 , that separates solutions that increase as t rarr oot \rightarrow \infty from those that ultimately decrease as t rarr oot \rightarrow \infty. In Problem 12 you are asked to determine this critical initial slope.
The asymptotic behavior of solutions is similar in this case to that when the roots are real and different. If the exponents are either positive or negative, then the magnitude of the solution grows or decays accordingly; the linear factor tt has little influence. A decaying solution is shown in Figure 3.4.1 and growing solutions in Figure 3.4.2. However, if the repeated root is zero, then the differential equation is y^('')=0y^{\prime \prime}=0 and the general solution is a linear function of tt.
Summary. We can now summarize the results that we have obtained for second-order linear homogeneous equations with constant coefficients
{:(24)ay^('')+by^(')+cy=0:}\begin{equation*}
a y^{\prime \prime}+b y^{\prime}+c y=0 \tag{24}
\end{equation*}
Let r_(1)r_{1} and r_(2)r_{2} be the roots of the corresponding characteristic equation
{:(25)ar^(2)+br+c=0:}\begin{equation*}
a r^{2}+b r+c=0 \tag{25}
\end{equation*}
If r_(1)r_{1} and r_(2)r_{2} are real but not equal, then the general solution of differential equation (24) is
If r_(1)=r_(2)r_{1}=r_{2}, then the general solution is
{:(28)y=c_(1)e^(r_(1)t)+c_(2)te^(r_(1)t):}\begin{equation*}
y=c_{1} e^{r_{1} t}+c_{2} t e^{r_{1} t} \tag{28}
\end{equation*}
Reduction of Order. It is worth noting that the procedure used in this section for equations with constant coefficients is more generally applicable. Suppose that we know one solution y_(1)(t)y_{1}(t), not everywhere zero, of
Despite its appearance, equation (32) is actually a first-order differential equation for the function v^(')v^{\prime} and can be solved either as a first-order linear equation or as a separable equation. Once v^(')v^{\prime} has been found, then vv is obtained by an integration. Finally, yy is determined from equation (30). This procedure is called the method of reduction of order, because the crucial step is the solution of a first-order differential equation for v^(')v^{\prime} rather than the original secondorder differential equation for yy. Although it is possible to write down a formula for v(t)v(t), we will instead illustrate how this method works by an example.
EXAMPLE 3
Given that y_(1)(t)=t^(-1)y_{1}(t)=t^{-1} is a solution of
{:(33)2t^(2)y^('')+3ty^(')-y=0","quad t > 0:}\begin{equation*}
2 t^{2} y^{\prime \prime}+3 t y^{\prime}-y=0, \quad t>0 \tag{33}
\end{equation*}
v(t)=(2)/(3)ct^(3//2)+kv(t)=\frac{2}{3} c t^{3 / 2}+k
It follows that
{:(35)y=v(t)t^(-1)=(2)/(3)ct^(1//2)+kt^(-1)",":}\begin{equation*}
y=v(t) t^{-1}=\frac{2}{3} c t^{1 / 2}+k t^{-1}, \tag{35}
\end{equation*}
where cc and kk are arbitrary constants. The second term on the right-hand side of equation (35) is a multiple of y_(1)(t)y_{1}(t) and can be dropped, but the first term provides a new solution y_(2)(t)=t^(1//2)y_{2}(t)=t^{1 / 2}. You can verify that the Wronskian of y_(1)y_{1} and y_(2)y_{2} is
{:(36)W[y_(1),y_(2)](t)=(3)/(2)t^(-3//2)!=0" for "t > 0:}\begin{equation*}
W\left[y_{1}, y_{2}\right](t)=\frac{3}{2} t^{-3 / 2} \neq 0 \text { for } t>0 \tag{36}
\end{equation*}
Consequently, y_(1)y_{1} and y_(2)y_{2} form a fundamental set of solutions of equation (33) for t > 0t>0.
Problems
In each of Problems 1 through 8, find the general solution of the given differential equation.
In each of Problems 9 through 11 , solve the given initial value problem. Sketch the graph of the solution and describe its behavior for increasing tt.
9. 9y^('')-12y^(')+4y=0,quad y(0)=2,quady^(')(0)=-19 y^{\prime \prime}-12 y^{\prime}+4 y=0, \quad y(0)=2, \quad y^{\prime}(0)=-1
10. y^('')-6y^(')+9y=0,quad y(0)=0,quady^(')(0)=2y^{\prime \prime}-6 y^{\prime}+9 y=0, \quad y(0)=0, \quad y^{\prime}(0)=2
11. y^('')+4y^(')+4y=0,quad y(-1)=2,quady^(')(-1)=1y^{\prime \prime}+4 y^{\prime}+4 y=0, \quad y(-1)=2, \quad y^{\prime}(-1)=1
12. Consider the following modification of the initial value problem in Example 2:
Find the solution as a function of bb, and then determine the critical value of bb that separates solutions that remain positive for all t > 0t>0 from those that eventually become negative.
a. Solve the initial value problem and plot the solution.
b. Determine the coordinates (t_(M),y_(M))\left(t_{M}, y_{M}\right) of the maximum point.
c. Change the second initial condition to y^(')(0)=b > 0y^{\prime}(0)=b>0 and find the solution as a function of bb.
d. Find the coordinates (t_(M),y_(M))\left(t_{M}, y_{M}\right) of the maximum point in terms of bb. Describe the dependence of t_(M)t_{M} and y_(M)y_{M} on bb as bb increases.
14. Consider the equation ay^('')+by^(')+cy=0a y^{\prime \prime}+b y^{\prime}+c y=0. If the roots of the corresponding characteristic equation are real, show that a solution to the differential equation either is everywhere zero or else can take on the value zero at most once.
Problems 15 through 17 indicate other ways of finding the second solution when the characteristic equation has repeated roots.
15. a. Consider the equation y^('')+2ay^(')+a^(2)y=0y^{\prime \prime}+2 a y^{\prime}+a^{2} y=0. Show that the roots of the characteristic equation are r_(1)=r_(2)=-ar_{1}=r_{2}=-a so that one solution of the equation is e^(-at)e^{-a t}.
b. Use Abel's formula [equation (23) of Section 3.2] to show that the Wronskian of any two solutions of the given equation is
W(t)=y_(1)(t)y_(2)^(')(t)-y_(1)^(')(t)y_(2)(t)=c_(1)e^(-2at),W(t)=y_{1}(t) y_{2}^{\prime}(t)-y_{1}^{\prime}(t) y_{2}(t)=c_{1} e^{-2 a t},
where c_(1)c_{1} is a constant.
c. Let y_(1)(t)=e^(-at)y_{1}(t)=e^{-a t} and use the result of part b to obtain a differential equation satisfied by a second solution y_(2)(t)y_{2}(t). By solving this equation, show that y_(2)(t)=te^(-at)y_{2}(t)=t e^{-a t}.
16. Suppose that r_(1)r_{1} and r_(2)r_{2} are roots of ar^(2)+br+c=0a r^{2}+b r+c=0 and that r_(1)!=r_(2)r_{1} \neq r_{2}; then exp(r_(1)t)\exp \left(r_{1} t\right) and exp(r_(2)t)\exp \left(r_{2} t\right) are solutions of the differential equation ay^('')+by^(')+cy=0a y^{\prime \prime}+b y^{\prime}+c y=0. Show that
is also a solution of the equation for r_(2)!=r_(1)r_{2} \neq r_{1}. Then think of r_(1)r_{1} as fixed, and use l'Hôpital's rule to evaluate the limit of phi(t;r_(1),r_(2))\phi\left(t ; r_{1}, r_{2}\right) as r_(2)rarrr_(1)r_{2} \rightarrow r_{1}, thereby obtaining the second solution in the case of equal roots.
17. a. If ar^(2)+br+c=0a r^{2}+b r+c=0 has equal roots r_(1)r_{1}, show that
Since the right-hand side of equation (37) is zero when r=r_(1)r=r_{1}, it follows that exp(r_(1)t)\exp \left(r_{1} t\right) is a solution of L[y]=ay^('')+by^(')+cy=0L[y]=a y^{\prime \prime}+b y^{\prime}+c y=0. b. Differentiate equation (37) with respect to rr, and interchange differentiation with respect to rr and with respect to tt, thus showing that
Since the right-hand side of equation (36) is zero when r=r_(1)r=r_{1}, conclude that t exp(r_(1)t)t \exp \left(r_{1} t\right) is also a solution of L[y]=0L[y]=0.
In each of Problems 18 through 22, use the method of reduction of order to find a second solution of the given differential equation.
18. t^(2)y^('')-4ty^(')+6y=0,t > 0;quady_(1)(t)=t^(2)t^{2} y^{\prime \prime}-4 t y^{\prime}+6 y=0, t>0 ; \quad y_{1}(t)=t^{2}
19. t^(2)y^('')+2ty^(')-2y=0,quad t > 0;quady_(1)(t)=tt^{2} y^{\prime \prime}+2 t y^{\prime}-2 y=0, \quad t>0 ; \quad y_{1}(t)=t
20. t^(2)y^('')+3ty^(')+y=0,t > 0;quady_(1)(t)=t^(-1)t^{2} y^{\prime \prime}+3 t y^{\prime}+y=0, t>0 ; \quad y_{1}(t)=t^{-1}
21. xy^('')-y^(')+4x^(3)y=0,quad x > 0;quady_(1)(x)=sin(x^(2))x y^{\prime \prime}-y^{\prime}+4 x^{3} y=0, \quad x>0 ; \quad y_{1}(x)=\sin \left(x^{2}\right)
22. x^(2)y^('')+xy^(')+(x^(2)-0.25)y=0,quad x > 0;quady_(1)(x)=x^(-1//2)sin xx^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-0.25\right) y=0, \quad x>0 ; \quad y_{1}(x)=x^{-1 / 2} \sin x
23. The differential equation
arises in the study of the turbulent flow of a uniform stream past a circular cylinder. Verify that y_(1)(x)=exp(-deltax^(2)//2)y_{1}(x)=\exp \left(-\delta x^{2} / 2\right) is one solution, and then find the general solution in the form of an integral.
24. The method of Problem 15 can be extended to second-order equations with variable coefficients. If y_(1)y_{1} is a known nonvanishing solution of y^('')+p(t)y^(')+q(t)y=0y^{\prime \prime}+p(t) y^{\prime}+q(t) y=0, show that a second solution y_(2)y_{2}
satisfies (y_(2)//y_(1))^(')=W[y_(1),y_(2)]//y_(1)^(2)\left(y_{2} / y_{1}\right)^{\prime}=W\left[y_{1}, y_{2}\right] / y_{1}^{2}, where W[y_(1),y_(2)]W\left[y_{1}, y_{2}\right] is the Wronskian of y_(1)y_{1} and y_(2)y_{2}. Then use Abel's formula (equation (23) of Section 3.2) to determine y_(2)y_{2}.
In each of Problems 25 through 27, use the method of Problem 24 to find a second independent solution of the given equation.
25. t^(2)y^('')+3ty^(')+y=0,quad t > 0;quady_(1)(t)=t^(-1)t^{2} y^{\prime \prime}+3 t y^{\prime}+y=0, \quad t>0 ; \quad y_{1}(t)=t^{-1}
26. ty^('')-y^(')+4t^(3)y=0,quad t > 0;quady_(1)(t)=sin(t^(2))t y^{\prime \prime}-y^{\prime}+4 t^{3} y=0, \quad t>0 ; \quad y_{1}(t)=\sin \left(t^{2}\right)
27. x^(2)y^('')+xy^(')+(x^(2)-0.25)y=0,quad x > 0;quady_(1)(x)=x^(-1//2)sin xx^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-0.25\right) y=0, \quad x>0 ; \quad y_{1}(x)=x^{-1 / 2} \sin x
Behavior of Solutions as t rarr oot \rightarrow \infty. Problems 28 through 30 are concerned with the behavior of solutions as t rarr oot \rightarrow \infty.
28. If a,ba, b, and cc are positive constants, show that all solutions of ay^('')+by^(')+cy=0a y^{\prime \prime}+b y^{\prime}+c y=0 approach zero as t rarr oot \rightarrow \infty.
29. a. If a > 0a>0 and c > 0c>0, but b=0b=0, show that the result of Problem 28 is no longer true, but that all solutions are bounded as t rarr oot \rightarrow \infty.
b. If a > 0a>0 and b > 0b>0, but c=0c=0, show that the result of Problem 28 is no longer true, but that all solutions approach a constant that depends on the initial conditions as t rarr oot \rightarrow \infty. Determine this constant for the initial conditions y(0)=y_(0),y^(')(0)=y_(0)^(')y(0)=y_{0}, y^{\prime}(0)=y_{0}^{\prime}.
30. Show that y=sin ty=\sin t is a solution of
y^('')+(ksin^(2)t)y^(')+(1-k cos t sin t)y=0y^{\prime \prime}+\left(k \sin ^{2} t\right) y^{\prime}+(1-k \cos t \sin t) y=0
for any value of the constant kk. If 0 < k < 20<k<2, show that 1-k cos t sin t > 01-k \cos t \sin t>0 and ksin^(2)t >= 0k \sin ^{2} t \geq 0. Thus observe that even though the coefficients of this variable-coefficient differential equation are nonnegative (and the coefficient of y^(')y^{\prime} is zero only at the points t=0,pi,2pi,dots)t=0, \pi, 2 \pi, \ldots), it has a solution that does not approach zero as t rarr oot \rightarrow \infty. Compare this situation with the result of Problem 28. Thus we observe a not unusual situation in the study of differential equations: equations that are apparently very similar can have quite different properties.
Euler Equations. In each of Problems 31 through 34, use the substitution introduced in Problem 25 in Section 3.3 to solve the given differential equation.
31. t^(2)y^('')-3ty^(')+4y=0,quad t > 0t^{2} y^{\prime \prime}-3 t y^{\prime}+4 y=0, \quad t>0
32. t^(2)y^('')+2ty^(')+0.25 y=0,quad t > 0t^{2} y^{\prime \prime}+2 t y^{\prime}+0.25 y=0, \quad t>0
33. t^(2)y^('')+3ty^(')+y=0,quad t > 0t^{2} y^{\prime \prime}+3 t y^{\prime}+y=0, \quad t>0
34. 4t^(2)y^('')-8ty^(')+9y=0,quad t > 04 t^{2} y^{\prime \prime}-8 t y^{\prime}+9 y=0, \quad t>0
3.5 Nonhomogeneous Equations; Method of
Undetermined Coefficients
We now turn our attention to the nonhomogeneous second-order linear differential equation
in which g(t)=0g(t)=0 and pp and qq are the same as in equation (1), is called the homogeneous differential equation corresponding to equation (1). The following two results describe the structure of solutions of the nonhomogeneous equation (1) and provide a foundation for constructing its general solution.
Theorem 3.5.1
If Y_(1)Y_{1} and Y_(2)Y_{2} are two solutions of the nonhomogeneous linear differential equation (1), then their difference Y_(1)-Y_(2)Y_{1}-Y_{2} is a solution of the corresponding homogeneous differential equation (2). If, in addition, y_(1)y_{1} and y_(2)y_{2} form a fundamental set of solutions of equation (2), then
Equation (6) states that Y_(1)-Y_(2)Y_{1}-Y_{2} is a solution of equation (2). Finally, since by Theorem 3.2.4 all solutions of equation (2) can be expressed as linear combinations of a fundamental set of solutions, it follows that the solution Y_(1)-Y_(2)Y_{1}-Y_{2} can be so written. Hence equation (3) holds and the proof is complete.
Theorem 3.5.2
The general solution of the nonhomogeneous equation (1) can be written in the form
where y_(1)y_{1} and y_(2)y_{2} form a fundamental set of solutions of the corresponding homogeneous equation (2), c_(1)c_{1} and c_(2)c_{2} are arbitrary constants, and YY is any solution of the nonhomogeneous equation (1).
The proof of Theorem 3.5.2 follows quickly from Theorem 3.5.1. Note that equation (3) holds if we identify Y_(1)Y_{1} with an arbitrary solution phi\phi of equation (1) and Y_(2)Y_{2} with the specific solution YY. From equation (3) we thereby obtain
which is equivalent to equation (7). Since phi\phi is an arbitrary solution of equation (1), the expression on the right-hand side of equation (7) includes all solutions of equation (1); thus it is natural to call it the general solution of equation (1).
In somewhat different words, Theorem 3.5.2 states that to solve the nonhomogeneous equation (1), we must do three things:
Find the general solution c_(1)y_(1)(t)+c_(2)y_(2)(t)c_{1} y_{1}(t)+c_{2} y_{2}(t) of the corresponding homogeneous equation. This solution is frequently called the complementary solution and may be denoted by y_(c)(t)y_{c}(t).
Find any solution Y(t)Y(t) of the nonhomogeneous equation. Often this solution is referred to as a particular solution.
Form the sum of the functions found in steps 1 and 2 .
We have already discussed how to find y_(c)(t)y_{c}(t), at least when the homogeneous equation (2) has constant coefficients. Therefore, in the remainder of this section and Section 3.6, we
will focus on finding a particular solution Y(t)Y(t) of the nonhomogeneous linear differential equation (1). There are two methods that we wish to discuss. They are known as the method of undetermined coefficients (discussed here) and the method of variation of parameters (see Section 3.6). Each has some advantages and some possible shortcomings.
Method of Undetermined Coefficients. The method of undetermined coefficients requires us to make an initial assumption about the form of the particular solution Y(t)Y(t), but with the coefficients left unspecified. We then substitute the assumed expression into the nonhomogeneous differential equation (1) and attempt to determine the coefficients so as to satisfy that equation. If we are successful, then we have found a solution of the differential equation (1) and can use it for the particular solution Y(t)Y(t). If we cannot determine the coefficients, then this means that there is no solution of the form that we assumed. In this case we may modify the initial assumption and try again.
The main advantage of the method of undetermined coefficients is that it is straightforward to execute once the assumption is made about the form of Y(t)Y(t). Its major limitation is that it is useful primarily for equations for which we can easily write down the correct form of the particular solution in advance. For this reason, this method is usually used only for problems in which the homogeneous equation has constant coefficients and the nonhomogeneous term is restricted to a relatively small class of functions. In particular, we consider only nonhomogeneous terms that consist of polynomials, exponential functions, sines, and cosines. Despite this limitation, the method of undetermined coefficients is useful for solving many problems that have important applications. However, the algebraic details may become tedious, and a computer algebra system can be very helpful in practical applications. We will illustrate the method of undetermined coefficients by several simple examples and then summarize some rules for using it.
We seek a function YY such that the combination Y^('')(t)-3Y^(')(t)-4Y(t)Y^{\prime \prime}(t)-3 Y^{\prime}(t)-4 Y(t) is equal to 3e^(2t)3 e^{2 t}. Since the exponential function reproduces itself through differentiation, the most plausible way to achieve the desired result is to assume that Y(t)Y(t) is some multiple of e^(2t)e^{2 t},
Y(t)=Ae^(2t)Y(t)=A e^{2 t}
where the coefficient AA is yet to be determined. To find AA, we calculate the first two derivatives of YY :
Y^(')(t)=2Ae^(2t),quadY^('')(t)=4Ae^(2t)Y^{\prime}(t)=2 A e^{2 t}, \quad Y^{\prime \prime}(t)=4 A e^{2 t}
and substitute for y,y^(')y, y^{\prime}, and y^('')y^{\prime \prime} in the nonhomogeneous differential equation (9). We obtain
{:(11)y^('')-3y^(')-4y=2sin t:}\begin{equation*}
y^{\prime \prime}-3 y^{\prime}-4 y=2 \sin t \tag{11}
\end{equation*}
Solution:
By analogy with Example 1, let us assume that Y(t)=A sin tY(t)=A \sin t, where AA is a constant to be determined. On substituting this guess in equation (11) we obtain
Y^('')-3Y^(')-4Y=-A sin t-3A cos t-4A sin t=2sin tY^{\prime \prime}-3 Y^{\prime}-4 Y=-A \sin t-3 A \cos t-4 A \sin t=2 \sin t
or, moving all terms to the left-hand side and collecting the terms involving sin t\sin t and cos t\cos t, we arrive at,
{:(12)(2+5A)sin t+3A cos t=0.:}\begin{equation*}
(2+5 A) \sin t+3 A \cos t=0 . \tag{12}
\end{equation*}
We want equation (12) to hold for all tt. Thus it must hold for two specific points, such as t=0t=0 and t=(pi)/(2)t=\frac{\pi}{2}. At these points equation (12) reduces to 3A=03 A=0 and 2+5A=02+5 A=0, respectively. These contradictory requirements mean that there is no choice of the constant AA that makes equation (12) true for t=0t=0 and t=(pi)/(2)t=\frac{\pi}{2}, much less for all tt. Thus we conclude that our assumption concerning Y(t)Y(t) is inadequate.
The appearance of the cosine term in equation (12) suggests that we modify our original assumption to include a cosine term in Y(t)Y(t); that is,
Y(t)=A sin t+B cos t,Y(t)=A \sin t+B \cos t,
where AA and BB are the undetermined coefficients. Then
Y^(')(t)=A cos t-B sin t,quadY^('')(t)=-A sin t-B cos t.Y^{\prime}(t)=A \cos t-B \sin t, \quad Y^{\prime \prime}(t)=-A \sin t-B \cos t .
By substituting these expressions for y,y^(')y, y^{\prime}, and y^('')y^{\prime \prime} in equation (11) and collecting terms, we obtain
{:(13)Y^('')-3Y^(')-4Y=(-A+3B-4A)sin t+(-B-3A-4B)cos t=2sin t:}\begin{equation*}
Y^{\prime \prime}-3 Y^{\prime}-4 Y=(-A+3 B-4 A) \sin t+(-B-3 A-4 B) \cos t=2 \sin t \tag{13}
\end{equation*}
Now, working exactly as with the first guess, move all terms to the left-hand side and evaluate t=0t=0 and t=(pi)/(2)t=\frac{\pi}{2} to find that AA and BB must satisfy the equations
Solving these algebraic equations for AA and BB, we obtain A=-(5)/(17)A=-\frac{5}{17} and B=(3)/(17)B=\frac{3}{17}; hence a particular solution of equation (11) is
Y(t)=-(5)/(17)sin t+(3)/(17)cos t.Y(t)=-\frac{5}{17} \sin t+\frac{3}{17} \cos t .
The method illustrated in the preceding examples can also be used when the right-hand side of the equation is a polynomial. Thus, to find a particular solution of
we initially assume that Y(t)Y(t) is a polynomial of the same degree as the nonhomogeneous term; that is, Y(t)=At^(2)+Bt+CY(t)=A t^{2}+B t+C.
To summarize our conclusions up to this point: if the nonhomogeneous term g(t)g(t) in differential equation (1) is an exponential function e^(alpha t)e^{\alpha t}, then assume that Y(t)Y(t) is proportional to the same exponential function; if g(t)g(t) is sin(beta t)\sin (\beta t) or cos(beta t)\cos (\beta t), then assume that Y(t)Y(t) is a linear combination of sin(beta t)\sin (\beta t) and cos(beta t)\cos (\beta t); if g(t)g(t) is a polynomial of degree nn, then assume that Y(t)Y(t) is a polynomial of degree nn. The same principle extends to the case where g(t)g(t) is a product of any two, or all three, of these types of functions, as the next example illustrates.
Now suppose that g(t)g(t) is the sum of two terms, g(t)=g_(1)(t)+g_(2)(t)g(t)=g_{1}(t)+g_{2}(t), and suppose that Y_(1)Y_{1} and Y_(2)Y_{2} are solutions of the equations
{:(16)ay^('')+by^(')+cy=g_(1)(t):}\begin{equation*}
a y^{\prime \prime}+b y^{\prime}+c y=g_{1}(t) \tag{16}
\end{equation*}
and
{:(17)ay^('')+by^(')+cy=g_(2)(t):}\begin{equation*}
a y^{\prime \prime}+b y^{\prime}+c y=g_{2}(t) \tag{17}
\end{equation*}
respectively. Then Y_(1)+Y_(2)Y_{1}+Y_{2} is a solution of the equation
{:(18)ay^('')+by^(')+cy=g(t).:}\begin{equation*}
a y^{\prime \prime}+b y^{\prime}+c y=g(t) . \tag{18}
\end{equation*}
To prove this statement, substitute Y_(1)(t)+Y_(2)(t)Y_{1}(t)+Y_{2}(t) for yy in equation (18) and make use of equations (16) and (17). A similar conclusion holds if g(t)g(t) is the sum of any finite number of terms. The practical significance of this result is that for an equation whose nonhomogeneous function g(t)g(t) can be expressed as a sum, you can consider instead several simpler equations and then add together the results. The following example is an illustration of this procedure.
Solutions of these three equations have been found in Examples 1, 2, and 3, respectively. Therefore, a particular solution of equation (19) is their sum, namely,
The procedure illustrated in these examples enables us to solve a fairly large class of problems in a reasonably efficient manner. However, there is one difficulty that sometimes occurs. The next example illustrates how it arises.
Since the left-hand side of equation (21) is zero, there is no choice of AA for which 0=2e^(-t)0=2 e^{-t}. Therefore, there is no particular solution of equation (20) of the assumed form. The reason for this possibly unexpected result becomes clear if we solve the homogeneous equation
that corresponds to equation (20). The two functions in a fundamental set of solutions of equation (22) are y_(1)(t)=e^(-t)y_{1}(t)=e^{-t} and y_(2)(t)=e^(4t)y_{2}(t)=e^{4 t}. Thus our assumed particular solution of equation (20) is actually a solution of the homogeneous equation (22); consequently, it cannot possibly be a solution of the nonhomogeneous equation (20). To find a solution of equation (20), we must therefore consider functions of a somewhat different form.
At this stage, we have several possible alternatives. One is simply to try to guess the proper form of the particular solution of equation (20). Another is to solve this equation in some different way and then to use the result to guide our assumptions if this situation arises again in the future; see Problems 22 and 27 for other solution methods. Still another possibility is to seek a simpler equation where this difficulty occurs and to use its solution to suggest how we might proceed with equation (20). Adopting the latter approach, we look for a first-order equation analogous to equation (20). One possibility is the linear equation
If we try to find a particular solution of equation (23) of the form Ae^(-t)A e^{-t}, we will fail because e^(-t)e^{-t} is a solution of the homogeneous equation y^(')+y=0y^{\prime}+y=0. However, from Section 2.1 we already know how to solve equation (23). An integrating factor is mu(t)=e^(t)\mu(t)=e^{t}, and by multiplying by mu(t)\mu(t) and then integrating both sides, we obtain the solution
{:(24)y=2te^(-t)+ce^(-t).:}\begin{equation*}
y=2 t e^{-t}+c e^{-t} . \tag{24}
\end{equation*}
The second term on the right-hand side of equation (24) is the general solution of the homogeneous equation y^(')+y=0y^{\prime}+y=0, but the first term is a solution of the full nonhomogeneous equation (23). Observe that it involves the exponential factor e^(-t)e^{-t} multiplied by the factor tt. This is the clue that we were looking for.
We now return to equation (20) and assume a particular solution of the form Y(t)=Ate^(-t)Y(t)=A t e^{-t}. Then
{:(25)Y^(')(t)=Ae^(-t)-Ate^(-t)","quadY^('')(t)=-2Ae^(-t)+Ate^(-t).:}\begin{equation*}
Y^{\prime}(t)=A e^{-t}-A t e^{-t}, \quad Y^{\prime \prime}(t)=-2 A e^{-t}+A t e^{-t} . \tag{25}
\end{equation*}
Substituting these expressions for y,y^(')y, y^{\prime}, and y^('')y^{\prime \prime} in equation (20), we obtain
Y^('')-3Y^(')-4Y=(-2A-3A)e^(-t)+(A+3A-4A)te^(-t)=2e^(-t)Y^{\prime \prime}-3 Y^{\prime}-4 Y=(-2 A-3 A) e^{-t}+(A+3 A-4 A) t e^{-t}=2 e^{-t}
The coefficient of te^(-t)t e^{-t} is zero, and from the terms involving e^(t)e^{t} we have -5A=2-5 A=2, so A=-(2)/(5)A=-\frac{2}{5}. Thus a particular solution of equation (20) is
{:(26)Y(t)=-(2)/(5)te^(-t).:}\begin{equation*}
Y(t)=-\frac{2}{5} t e^{-t} . \tag{26}
\end{equation*}
The outcome of Example 5 suggests a modification of the principle stated previously: if the assumed form of the particular solution duplicates a solution of the corresponding homogeneous equation, then modify the assumed particular solution by multiplying it by tt. Occasionally, this modification will be insufficient to remove all duplication with the solutions of the homogeneous equation, in which case it is necessary to multiply by tt a second time. For a second-order equation, it will never be necessary to carry the process further than this.
Summary. We now summarize the steps involved in finding the solution of an initial value problem consisting of a nonhomogeneous linear differential equation of the form
{:(27)ay^('')+by^(')+cy=g(t):}\begin{equation*}
a y^{\prime \prime}+b y^{\prime}+c y=g(t) \tag{27}
\end{equation*}
where the coefficients a,ba, b, and cc are constants, together with a given set of initial conditions.
Find the general solution of the corresponding homogeneous equation.
Make sure that the function g(t)g(t) in equation (27) belongs to the class of functions discussed in this section; that is, be sure it involves nothing more than exponential functions, sines, cosines, polynomials, or sums or products of such functions. If this is not the case, use the method of variation of parameters (discussed in Section 3.6).
If g(t)=g_(1)(t)+cdots+g_(n)(t)g(t)=g_{1}(t)+\cdots+g_{n}(t)-that is, if g(t)g(t) is a sum of nn terms-then form nn subproblems, each of which contains only one of the terms g_(1)(t),dots,g_(n)(t)g_{1}(t), \ldots, g_{n}(t). The i^("th ")i^{\text {th }} subproblem consists of the equation
where ii runs from 1 to nn.
4. For the i^("th ")i^{\text {th }} subproblem, assume a particular solution Y_(i)(t)Y_{i}(t) consisting of the appropriate exponential function, sine, cosine, polynomial, or combination thereof. If there is any duplication in the assumed form of Y_(i)(t)Y_{i}(t) with the solutions of the homogeneous equation (found in step 1), then multiply Y_(i)(t)Y_{i}(t) by tt, or (if necessary) by t^(2)t^{2}, so as to remove the duplication. See Table 3.5.1.
TABLE 3.5.1 The Particular Solution of ay^('')+by^(')+cy=g_(i)(t)a y^{\prime \prime}+b y^{\prime}+c y=g_{i}(t)
Notes: Here, ss is the smallest nonnegative integer (s=0,1(s=0,1, or 2 ) that will ensure that no term in Y_(i)(t)Y_{i}(t) is a solution of the corresponding homogeneous equation. Equivalently, for the three cases, ss is the number of times 0 is a root of the characteristic equation, alpha\alpha is a root of the characteristic equation, and alpha+i beta\alpha+i \beta is a root of the characteristic equation, respectively.
5. Find a particular solution Y_(i)(t)Y_{i}(t) for each of the subproblems. Then Y_(1)(t)+cdots+Y_(n)(t)Y_{1}(t)+\cdots+Y_{n}(t) is a particular solution of the full nonhomogeneous equation (27).
6. Form the sum of the general solution of the homogeneous equation (step 1) and the particular solution of the nonhomogeneous equation (step 5). This is the general solution of the nonhomogeneous equation.
7. When initial conditions are provided, use them to determine the values of the arbitrary constants remaining in the general solution.
For some problems this entire procedure is easy to carry out by hand, but often the algebraic calculations are lengthy. Once you understand clearly how the method works, a computer algebra system can be of great assistance in executing the details.
The method of undetermined coefficients is self-correcting in the sense that if you assume too little for Y(t)Y(t), then a contradiction is soon reached that usually points the way to the modification that is needed in the assumed form. On the other hand, if you assume too many terms, then some unnecessary work is done and some coefficients turn out to be zero, but at least the correct answer is obtained.
Proof of the Method of Undetermined Coefficients. In the preceding discussion we have described the method of undetermined coefficients on the basis of several examples. To prove that the procedure always works as stated, we now give a general argument, in which we consider three cases corresponding to different forms for the nonhomogeneous term g(t)g(t).
Case 1: g(t)=P_(n)(t)=a_(0)t^(n)+a_(1)t^(n-1)+cdots+a_(n)g(t)=P_{n}(t)=a_{0} t^{n}+a_{1} t^{n-1}+\cdots+a_{n}. In this case equation (27) becomes
{:(28)ay^('')+by^(')+cy=a_(0)t^(n)+a_(1)t^(n-1)+cdots+a_(n):}\begin{equation*}
a y^{\prime \prime}+b y^{\prime}+c y=a_{0} t^{n}+a_{1} t^{n-1}+\cdots+a_{n} \tag{28}
\end{equation*}
Equating the coefficients of like powers of tt, beginning with t^(n)t^{n}, leads to the following sequence of equations:
{:[cA_(0)=a_(0)","],[cA_(1)+nbA_(0)=a_(1)","],[vdots],[cA_(n)+bA_(n-1)+2aA_(n-2)=a_(n).]:}\begin{aligned}
c A_{0} & =a_{0}, \\
c A_{1}+n b A_{0} & =a_{1}, \\
& \vdots \\
c A_{n}+b A_{n-1}+2 a A_{n-2} & =a_{n} .
\end{aligned}
Provided that c!=0c \neq 0, the solution of the first equation is A_(0)=a_(0)//cA_{0}=a_{0} / c, and the remaining equations determine A_(1),dots,A_(n)A_{1}, \ldots, A_{n} successively.
If c=0c=0 but b!=0b \neq 0, then the polynomial on the left-hand side of equation (30) is of degree n-1n-1, and we cannot satisfy equation (30). To be sure that aY^('')(t)+bY^(')(t)a Y^{\prime \prime}(t)+b Y^{\prime}(t) is a polynomial of degree nn, we must choose Y(t)Y(t) to be a polynomial of degree n+1n+1. Hence we assume that
There is no constant term in this expression for Y(t)Y(t), but there is no need to include such a term since a constant is a solution of the homogeneous equation when c=0c=0. Since b!=0b \neq 0, we find A_(0)=a_(0)//(b(n+1))A_{0}=a_{0} /(b(n+1)), and the other coefficients A_(1),dots,A_(n)A_{1}, \ldots, A_{n} can be determined similarly.
If both cc and bb are zero, then the characteristic equation is ar^(2)=0a r^{2}=0 and r=0r=0 is a repeated root. Thus y_(1)=e^(0t)=1y_{1}=e^{0 t}=1 and y_(2)=te^(0t)=ty_{2}=t e^{0 t}=t form a fundamental set of solutions of the corresponding homogeneous equation. This leads us to assume that
The term aY^('')(t)a Y^{\prime \prime}(t) gives rise to a term of degree nn, and we can proceed as before. Again the constant and linear terms in Y(t)Y(t) are omitted since, in this case, they are both solutions of the homogeneous equation.
Case 2: g(t)=e^(alpha t)P_(n)(t)g(t)=e^{\alpha t} P_{n}(t). The problem of determining a particular solution of
Substituting for y,y^(')y, y^{\prime}, and y^('')y^{\prime \prime} in equation (31), canceling the factor e^(alpha t)e^{\alpha t}, and collecting terms, we obtain
{:(32)au^('')(t)+(2a alpha+b)u^(')(t)+(aalpha^(2)+b alpha+c)u(t)=P_(n)(t).:}\begin{equation*}
a u^{\prime \prime}(t)+(2 a \alpha+b) u^{\prime}(t)+\left(a \alpha^{2}+b \alpha+c\right) u(t)=P_{n}(t) . \tag{32}
\end{equation*}
The determination of a particular solution of equation (32) is precisely the same problem, except for the names of the constants, as solving equation (28). Therefore, if aalpha^(2)+b alpha+ca \alpha^{2}+b \alpha+c is not zero, we assume that u(t)=A_(0)t^(n)+cdots+A_(n)u(t)=A_{0} t^{n}+\cdots+A_{n}; hence a particular solution of equation (31) is of the form
On the other hand, if aalpha^(2)+b alpha+ca \alpha^{2}+b \alpha+c is zero but 2a alpha+b2 a \alpha+b is not, we must take u(t)u(t) to be of the form t(A_(0)t^(n)+cdots+A_(n))t\left(A_{0} t^{n}+\cdots+A_{n}\right). The corresponding form for Y(t)Y(t) is tt times the expression on the right-hand side of equation (33). Note that if aalpha^(2)+b alpha+ca \alpha^{2}+b \alpha+c is zero, then e^(alpha t)e^{\alpha t} is a solution of the homogeneous equation.
If both aalpha^(2)+b alpha+ca \alpha^{2}+b \alpha+c and 2a alpha+b2 a \alpha+b are zero (and this implies that both e^(alpha t)e^{\alpha t} and te^(alpha t)t e^{\alpha t} are solutions of the homogeneous equation), then the correct form for u(t)u(t) is t^(2)(A_(0)t^(n)+cdots+A_(n))t^{2}\left(A_{0} t^{n}+\cdots+A_{n}\right). Hence Y(t)Y(t) is t^(2)t^{2} times the expression on the right-hand side of equation (33).
Case 3: g(t)=e^(alpha t)P_(n)(t)cos(beta t)g(t)=e^{\alpha t} P_{n}(t) \cos (\beta t) or e^(alpha t)P_(n)(t)sin(beta t)e^{\alpha t} P_{n}(t) \sin (\beta t). These two cases are similar, so we consider only the latter. We can reduce this problem to the preceding one by noting that, as a consequence of Euler's formula, sin(beta t)=(e^(i beta t)-e^(-i beta t))//(2i)\sin (\beta t)=\left(e^{i \beta t}-e^{-i \beta t}\right) /(2 i). Hence g(t)g(t) is of the form
Usually, the latter form is preferred because it does not involve the use of complex-valued coefficients. If alpha+-i beta\alpha \pm i \beta satisfy the characteristic equation corresponding to the homogeneous equation, we must, of course, multiply each of the polynomials by tt to increase their degrees by 1 .
If the nonhomogeneous function involves both cos(beta t)\cos (\beta t) and sin(beta t)\sin (\beta t), it is usually convenient to treat these terms together, since each one individually may give rise to the same form for a particular solution. For example, if g(t)=t sin t+2cos tg(t)=t \sin t+2 \cos t, the form for Y(t)Y(t) would be
Y(t)=(A_(0)t+A_(1))sin t+(B_(0)t+B_(1))cos tY(t)=\left(A_{0} t+A_{1}\right) \sin t+\left(B_{0} t+B_{1}\right) \cos t
provided that sin t\sin t and cos t\cos t are not solutions of the homogeneous equation.
Problems
In each of Problems 1 through 10, find the general solution of the given differential equation.
y^('')+y^(')+4y=2sinh t quady^{\prime \prime}+y^{\prime}+4 y=2 \sinh t \quad Hint: sinh t=(e^(t)-e^(-t))//2\sinh t=\left(e^{t}-e^{-t}\right) / 2
In each of Problems 11 through 15 , find the solution of the given initial value problem.
11. y^('')+y^(')-2y=2t,quad y(0)=0,quady^(')(0)=1y^{\prime \prime}+y^{\prime}-2 y=2 t, \quad y(0)=0, \quad y^{\prime}(0)=1
12. y^('')+4y=t^(2)+3e^(t),quad y(0)=0,quady^(')(0)=2y^{\prime \prime}+4 y=t^{2}+3 e^{t}, \quad y(0)=0, \quad y^{\prime}(0)=2
13. y^('')-2y^(')+y=te^(t)+4,quad y(0)=1,quady^(')(0)=1y^{\prime \prime}-2 y^{\prime}+y=t e^{t}+4, \quad y(0)=1, \quad y^{\prime}(0)=1
14. y^('')+4y=3sin(2t),quad y(0)=2,quady^(')(0)=-1y^{\prime \prime}+4 y=3 \sin (2 t), \quad y(0)=2, \quad y^{\prime}(0)=-1
15. y^('')+2y^(')+5y=4e^(-t)cos(2t),quad y(0)=1,quady^(')(0)=0y^{\prime \prime}+2 y^{\prime}+5 y=4 e^{-t} \cos (2 t), \quad y(0)=1, \quad y^{\prime}(0)=0
In each of Problems 16 through 21:
a. Determine a suitable form for Y(t)Y(t) if the method of undetermined coefficients is to be used.
(N) b. Use a computer algebra system to find a particular solution of the given equation.
16. y^('')+3y^(')=2t^(4)+t^(2)e^(-3t)+sin(3t)y^{\prime \prime}+3 y^{\prime}=2 t^{4}+t^{2} e^{-3 t}+\sin (3 t)
17. y^('')-5y^(')+6y=e^(t)cos(2t)+e^(2t)(3t+4)sin ty^{\prime \prime}-5 y^{\prime}+6 y=e^{t} \cos (2 t)+e^{2 t}(3 t+4) \sin t
18. y^('')+2y^(')+2y=3e^(-t)+2e^(-t)cos t+4e^(-t)t^(2)sin ty^{\prime \prime}+2 y^{\prime}+2 y=3 e^{-t}+2 e^{-t} \cos t+4 e^{-t} t^{2} \sin t
19. y^('')+4y=t^(2)sin(2t)+(6t+7)cos(2t)y^{\prime \prime}+4 y=t^{2} \sin (2 t)+(6 t+7) \cos (2 t)
20. y^('')+3y^(')+2y=e^(t)(t^(2)+1)sin(2t)+3e^(-t)cos t+4e^(t)y^{\prime \prime}+3 y^{\prime}+2 y=e^{t}\left(t^{2}+1\right) \sin (2 t)+3 e^{-t} \cos t+4 e^{t}
21. y^('')+2y^(')+5y=3te^(-t)cos(2t)-2te^(-2t)cos ty^{\prime \prime}+2 y^{\prime}+5 y=3 t e^{-t} \cos (2 t)-2 t e^{-2 t} \cos t
22. Consider the equation
from Example 5. Recall that y_(1)(t)=e^(-t)y_{1}(t)=e^{-t} and y_(2)(t)=e^(4t)y_{2}(t)=e^{4 t} are solutions of the corresponding homogeneous equation. Adapting the method of reduction of order (Section 3.4), seek a solution of the nonhomogeneous equation of the form Y(t)=v(t)y_(1)(t)=v(t)e^(-t)Y(t)=v(t) y_{1}(t)=v(t) e^{-t}, where v(t)v(t) is to be determined.
a. Substitute Y(t),Y^(')(t)Y(t), Y^{\prime}(t), and Y^('')(t)Y^{\prime \prime}(t) into equation (34) and show that v(t)v(t) must satisfy v^('')-5v^(')=2v^{\prime \prime}-5 v^{\prime}=2.
b. Let w(t)=v^(')(t)w(t)=v^{\prime}(t) and show that w(t)w(t) satisfies w^(')-5w=2w^{\prime}-5 w=2. Solve this equation for w(t)w(t).
c. Integrate w(t)w(t) to find v(t)v(t) and then show that
Y(t)=-(2)/(5)te^(-t)+(1)/(5)c_(1)e^(4t)+c_(2)e^(-t)Y(t)=-\frac{2}{5} t e^{-t}+\frac{1}{5} c_{1} e^{4 t}+c_{2} e^{-t}
The first term on the right-hand side is the desired particular solution of the nonhomogeneous equation. Note that it is a product of tt and e^(-t)e^{-t}.
23. Determine the general solution of
y^('')+lambda^(2)y=sum_(m=1)^(N)a_(m)sin(m pi t),y^{\prime \prime}+\lambda^{2} y=\sum_{m=1}^{N} a_{m} \sin (m \pi t),
where lambda > 0\lambda>0 and lambda!=m pi\lambda \neq m \pi for m=1,dots,Nm=1, \ldots, N.
(N) 24. In many physical problems the nonhomogeneous term may be specified by different formulas in different time periods. As an example, determine the solution y=phi(t)y=\phi(t) of
satisfying the initial conditions y(0)=0y(0)=0 and y^(')(0)=1y^{\prime}(0)=1. Assume that yy and y^(')y^{\prime} are also continuous at t=pit=\pi. Plot the nonhomogeneous term and the solution as functions of time. Hint: First solve the initial value problem for t <= pit \leq \pi; then solve for t > pit>\pi, determining the constants in the latter solution from the continuity conditions at t=pit=\pi.
Behavior of Solutions as t rarr oot \rightarrow \infty. In Problems 25 and 26, we continue the discussion started with Problems 28 through 30 of Section 3.4. Consider the differential equation
{:(35)ay^('')+by^(')+cy=g(t)",":}\begin{equation*}
a y^{\prime \prime}+b y^{\prime}+c y=g(t), \tag{35}
\end{equation*}
where a,ba, b, and cc are positive.
25. If Y_(1)(t)Y_{1}(t) and Y_(2)(t)Y_{2}(t) are solutions of equation (35), show that Y_(1)(t)-Y_(2)(t)rarr0Y_{1}(t)-Y_{2}(t) \rightarrow 0 as t rarr oot \rightarrow \infty. Is this result true if b=0b=0 ?
26. If g(t)=dg(t)=d, a constant, show that every solution of equation (35) approaches d//cd / c as t rarr oot \rightarrow \infty. What happens if c=0c=0 ? What if b=0b=0 also?
27. In this problem we indicate an alternative procedure ^(8){ }^{8} for solving the differential equation
where bb and cc are constants, and DD denotes differentiation with respect to tt. Let r_(1)r_{1} and r_(2)r_{2} be the zeros of the characteristic polynomial of the corresponding homogeneous equation. These roots may be real and different, real and equal, or conjugate complex numbers.
a. Verify that equation (36) can be written in the factored form
where r_(1)+r_(2)=-br_{1}+r_{2}=-b and r_(1)r_(2)=cr_{1} r_{2}=c.
b. Let u=(D-r_(2))yu=\left(D-r_{2}\right) y. Then show that the solution of equation (36) can be found by solving the following two firstorder equations:
In each of Problems 28 through 30, use the method of Problem 27 to solve the given differential equation.
28. y^('')-3y^(')-4y=3e^(2t)quady^{\prime \prime}-3 y^{\prime}-4 y=3 e^{2 t} \quad (see Example 1)
29. y^('')+2y^(')+y=2e^(-t)quady^{\prime \prime}+2 y^{\prime}+y=2 e^{-t} \quad (see Problem 6)
30. y^('')+2y^(')=3+4sin(2t)y^{\prime \prime}+2 y^{\prime}=3+4 \sin (2 t) (see Problem 5)
3.6 Variation of Parameters
In this section we describe a second method of finding a particular solution of a nonhomogeneous equation. This method, variation of parameters, is due to Lagrange and complements the method of undetermined coefficients rather well. The main advantage of variation of parameters is that it is a general method; in principle at least, it can be applied to any equation, and it requires no detailed assumptions about the form of the solution. In fact, later in this section we use this method to derive a formula for a particular solution of an arbitrary second-order linear nonhomogeneous differential equation. On the other hand, the method of variation of parameters eventually requires us to evaluate certain integrals involving the nonhomogeneous term in the differential equation, and this may present difficulties. Before looking at this method in the general case, we illustrate its use in an example.
EXAMPLE 1
Find the general solution of
{:(1)y^('')+4y=8tan t quad-pi//2 < t < pi//2:}\begin{equation*}
y^{\prime \prime}+4 y=8 \tan t \quad-\pi / 2<t<\pi / 2 \tag{1}
\end{equation*}
Solution:
Observe that this problem is not a good candidate for the method of undetermined coefficients, as described in Section 3.5, because the nonhomogeneous term g(t)=8tan tg(t)=8 \tan t involves a quotient (rather than a sum or a product) of sin t\sin t and cos t\cos t. Therefore, the method of undetermined coefficients cannot be applied; we need a different approach.
Observe also that the homogeneous equation corresponding to equation (1) is
The basic idea in the method of variation of parameters is similar to the method of reduction of order introduced at the end of Section 3.4. In the general solution (3), replace the constants c_(1)c_{1} and c_(2)c_{2} by functions u_(1)(t)u_{1}(t) and u_(2)(t)u_{2}(t), respectively, and then determine these functions so that the resulting expression
is a solution of the nonhomogeneous equation (1).
To determine u_(1)u_{1} and u_(2)u_{2}, we need to substitute for yy from equation (4) in differential equation (1). However, even without carrying out this substitution, we can anticipate that the result will be a single equation involving some combination of u_(1),u_(2)u_{1}, u_{2}, and their first two derivatives. Since there is only one equation and two unknown functions, we can expect that there are many possible choices of u_(1)u_{1} and u_(2)u_{2} that will meet our needs. Alternatively, we may be able to impose a second condition of our own choosing, thereby obtaining two equations for the two unknown functions u_(1)u_{1} and u_(2)u_{2}. We will soon show (following Lagrange) that it is possible to choose this second condition in a way that makes the computation markedly more efficient. ^(9){ }^{9}
Returning now to equation (4), we differentiate it and rearrange the terms, thereby obtaining
Keeping in mind the possibility of choosing a second condition on u_(1)u_{1} and u_(2)u_{2}, let us require the sum of the last two terms on the right-hand side of equation (5) to be zero; that is, we require that
Although the ultimate effect of the condition (6) is not yet clear, the removal of the terms involving u_(1)^(')u_{1}^{\prime} and u_(2)^(')u_{2}^{\prime} has simplified the expression for y^(')y^{\prime}. Further, by differentiating equation (7), we obtain
Summarizing our results to this point, we want to choose u_(1)u_{1} and u_(2)u_{2} so as to satisfy equations (6) and (9). These equations can be viewed as a pair of linear algebraic equations for the two unknown quantities u_(1)^(')(t)u_{1}^{\prime}(t) and u_(2)^(')(t)u_{2}^{\prime}(t). Equations (6) and (9) can be solved in various ways. For example, solving equation (6) for u_(2)^(')(t)u_{2}^{\prime}(t), we have
Then, substituting for u_(2)^(')(t)u_{2}^{\prime}(t) in equation (9) and simplifying, we obtain
{:(11)u_(1)^(')(t)=-(8tan t sin(2t))/(2)=-8sin^(2)t.:}\begin{equation*}
u_{1}^{\prime}(t)=-\frac{8 \tan t \sin (2 t)}{2}=-8 \sin ^{2} t . \tag{11}
\end{equation*}
Further, putting this expression for u_(1)^(')(t)u_{1}^{\prime}(t) back in equation (10) and using the double-angle formulas, we find that
Having obtained u_(1)^(')(t)u_{1}^{\prime}(t) and u_(2)^(')(t)u_{2}^{\prime}(t), we next integrate so as to find u_(1)(t)u_{1}(t) and u_(2)(t)u_{2}(t). The result is
{:(13)u_(1)(t)=4sin t cos t-4t+c_(1):}\begin{equation*}
u_{1}(t)=4 \sin t \cos t-4 t+c_{1} \tag{13}
\end{equation*}
The terms in equation (15) involving the arbitrary constants c_(1)c_{1} and c_(2)c_{2} are the general solution of the corresponding homogeneous equation, while the other three terms are a particular solution of the nonhomogeneous equation (1). Thus equation (15) is the general solution of equation (1).
The particular solution identified at the end of Example 1 corresponds to choosing both c_(1)c_{1}, and c_(2)c_{2} to be zero in equation (15). Any other choice of c_(1)c_{1} and c_(2)c_{2} is also a particular solution of the same nonhomogeneous differential equation. Notice, in particular, that choosing c_(1)=0c_{1}=0 and c_(2)=2c_{2}=2 in equation (15) yields a particular solution with only two terms:
We conclude this first look at the method of variation of parameters with the observation that the particular solution involves terms that might be difficult to anticipate. This explains why the method of undetermined coefficients is not a good candidate for this problem, and why the method of variation of parameters is needed.
In the preceding example the method of variation of parameters worked well in determining a particular solution, and hence the general solution, of equation (1). The next question is whether this method can be applied effectively to an arbitrary equation. Therefore, we consider
This is a major assumption. So far we have shown how to solve equation (18) only if it has constant coefficients. If equation (18) has coefficients that depend on tt, then usually the methods described in Chapter 5 must be used to obtain y_(c)(t)y_{c}(t).
The crucial idea, as illustrated in Example 1, is to replace the constants c_(1)c_{1} and c_(2)c_{2} in equation (17) by functions u_(1)(t)u_{1}(t) and u_(2)(t)u_{2}(t), respectively; thus we have
Then we try to determine u_(1)(t)u_{1}(t) and u_(2)(t)u_{2}(t) so that the expression in equation (19) is a solution of the nonhomogeneous equation (16) rather than the homogeneous equation (18). Thus we differentiate equation (19), obtaining
As in Example 1, we now set the terms involving u_(1)^(')(t)u_{1}^{\prime}(t) and u_(2)^(')(t)u_{2}^{\prime}(t) in equation (20) equal to zero; that is, we require that
Now we substitute for y,y^(')y, y^{\prime}, and y^('')y^{\prime \prime} in equation (16) from equations (19), (22), and (23), respectively. After rearranging the terms in the resulting equation, we find that
Each of the expressions in parentheses in the first two lines of equation (24) is zero because both y_(1)y_{1} and y_(2)y_{2} are solutions of the homogeneous equation (18). Therefore, equation (24) reduces to
Equations (21) and (25) form a system of two linear algebraic equations for the derivatives u_(1)^(')(t)u_{1}^{\prime}(t) and u_(2)^(')(t)u_{2}^{\prime}(t) of the unknown functions. They correspond exactly to equations (6) and (9) in Example 1.
Solving the system of equations (21), (25), we obtain
where W[y_(1),y_(2)]W\left[y_{1}, y_{2}\right] is the Wronskian of y_(1)y_{1} and y_(2)y_{2}. Note that division by W[y_(1),y_(2)]W\left[y_{1}, y_{2}\right] is permissible since y_(1)y_{1} and y_(2)y_{2} are a fundamental set of solutions, and therefore their Wronskian is nonzero. By integrating equations (26), we find the desired functions u_(1)(t)u_{1}(t) and u_(2)(t)u_{2}(t), namely,
{:(27)u_(1)(t)=-int(y_(2)(t)g(t))/(W[y_(1),y_(2)](t))dt+c_(1)","quadu_(2)(t)=int(y_(1)(t)g(t))/(W[y_(1),y_(2)](t))dt+c_(2):}\begin{equation*}
u_{1}(t)=-\int \frac{y_{2}(t) g(t)}{W\left[y_{1}, y_{2}\right](t)} d t+c_{1}, \quad u_{2}(t)=\int \frac{y_{1}(t) g(t)}{W\left[y_{1}, y_{2}\right](t)} d t+c_{2} \tag{27}
\end{equation*}
If the integrals in equations (27) can be evaluated in terms of elementary functions, then we substitute the results in equation (19), thereby obtaining the general solution of equation (16). More generally, the solution can always be expressed in terms of integrals, as stated in the following theorem.
Theorem 3.6.1
Consider the nonhomogeneous second-order linear differential equation
If the functions p,qp, q, and gg are continuous on an open interval II, and if the functions y_(1)y_{1} and y_(2)y_{2} form a fundamental set of solutions of the corresponding homogeneous equation
By examining the expression (30) and reviewing the process by which we derived it, we can see that there may be two major difficulties in carrying out the method of variation of parameters. As we have mentioned earlier, one is the determination of functions y_(1)(t)y_{1}(t) and y_(2)(t)y_{2}(t) that form a fundamental set of solutions of the homogeneous equation (29) when the coefficients in that equation are not constants. The other possible difficulty lies in the evaluation of the integrals appearing in equation (30). This depends entirely on the nature of the functions y_(1),y_(2)y_{1}, y_{2}, and gg. In using equation (30), be sure that the differential equation is exactly in the form (28); otherwise, the nonhomogeneous term g(t)g(t) will not be correctly identified.
A major advantage of the method of variation of parameters is that equation (30) provides an expression for the particular solution Y(t)Y(t) in terms of an arbitrary forcing function g(t)g(t). This expression is a good starting point if you wish to investigate the effect of variations in the forcing function, or if you wish to analyze the response of a system to a number of different forcing functions. (See Problems 18 to 22.)
Problems
In each of Problems 1 through 3, use the method of variation of parameters to find a particular solution of the given differential equation. Then check your answer by using the method of undetermined coefficients.
In each of Problems 4 through 9 , find the general solution of the given differential equation. In Problems 9,g9, g is an arbitrary continuous function.
4. y^('')+y=tan t,quad0 < t < pi//2y^{\prime \prime}+y=\tan t, \quad 0<t<\pi / 2
5. y^('')+9y=9sec^(2)(3t),quad0 < t < pi//6y^{\prime \prime}+9 y=9 \sec ^{2}(3 t), \quad 0<t<\pi / 6
6. y^('')+4y^(')+4y=t^(-2)e^(-2t),quad t > 0y^{\prime \prime}+4 y^{\prime}+4 y=t^{-2} e^{-2 t}, \quad t>0
7. 4y^('')+y=2sec(t//2),quad-pi < t < pi4 y^{\prime \prime}+y=2 \sec (t / 2), \quad-\pi<t<\pi
8. y^('')-2y^(')+y=e^(t)//(1+t^(2))y^{\prime \prime}-2 y^{\prime}+y=e^{t} /\left(1+t^{2}\right)
9. y^('')-5y^(')+6y=g(t)y^{\prime \prime}-5 y^{\prime}+6 y=g(t)
In each of Problems 10 through 15, verify that the given functions y_(1)y_{1} and y_(2)y_{2} satisfy the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous equation. In Problems 14 and 15,g15, g is an arbitrary continuous function.
10. t^(2)y^('')-2y=3t^(2)-1,quad t > 0;quady_(1)(t)=t^(2),quady_(2)(t)=t^(-1)t^{2} y^{\prime \prime}-2 y=3 t^{2}-1, \quad t>0 ; \quad y_{1}(t)=t^{2}, \quad y_{2}(t)=t^{-1}
11. t^(2)y^('')-t(t+2)y^(')+(t+2)y=2t^(3),t > 0t^{2} y^{\prime \prime}-t(t+2) y^{\prime}+(t+2) y=2 t^{3}, t>0; y_(1)(t)=t,quady_(2)(t)=te^(t)y_{1}(t)=t, \quad y_{2}(t)=t e^{t}
12. ty^('')-(1+t)y^(')+y=t^(2)e^(2t),t > 0;quady_(1)(t)=1+t,quady_(2)(t)=e^(t)t y^{\prime \prime}-(1+t) y^{\prime}+y=t^{2} e^{2 t}, t>0 ; \quad y_{1}(t)=1+t, \quad y_{2}(t)=e^{t}
13. x^(2)y^('')-3xy^(')+4y=x^(2)ln x,x > 0;quady_(1)(x)=x^(2)x^{2} y^{\prime \prime}-3 x y^{\prime}+4 y=x^{2} \ln x, x>0 ; \quad y_{1}(x)=x^{2}, y_(2)(x)=x^(2)ln xy_{2}(x)=x^{2} \ln x
14. x^(2)y^('')+xy^(')+(x^(2)-(1)/(4))y=3x^(3//2)sin x,x > 0x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-\frac{1}{4}\right) y=3 x^{3 / 2} \sin x, x>0; y_(1)(x)=x^(-1//2)sin x,quady_(2)(x)=x^(-1//2)cos xy_{1}(x)=x^{-1 / 2} \sin x, \quad y_{2}(x)=x^{-1 / 2} \cos x
15. x^(2)y^('')+xy^(')+(x^(2)-0.25)y=g(x),x > 0x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-0.25\right) y=g(x), x>0; y_(1)(x)=x^(-1//2)sin x,quady_(2)(x)=x^(-1//2)cos xy_{1}(x)=x^{-1 / 2} \sin x, \quad y_{2}(x)=x^{-1 / 2} \cos x
16. By choosing the lower limit of integration in equation (30) in the text as the initial point t_(0)t_{0}, show that Y(t)Y(t) becomes
Y(t)=int_(t_(0))^(t)(y_(1)(s)y_(2)(t)-y_(1)(t)y_(2)(s))/(y_(1)(s)y_(2)^(')(s)-y_(1)^(')(s)y_(2)(s))g(s)dsY(t)=\int_{t_{0}}^{t} \frac{y_{1}(s) y_{2}(t)-y_{1}(t) y_{2}(s)}{y_{1}(s) y_{2}^{\prime}(s)-y_{1}^{\prime}(s) y_{2}(s)} g(s) d s
Show that Y(t)Y(t) is a solution of the initial value problem
respectively. In other words, the nonhomogeneities in the differential equation and in the initial conditions can be dealt with separately. Observe that uu is easy to find if a fundamental set of solutions of L[u]=0L[u]=0 is known. And, as shown in Problem 16, the function vv is given by equation (30).
18. a. Use the result of Problem 16 to show that the solution of the initial value problem
where L[y]=(D-a)(D-b)yL[y]=(D-a)(D-b) y for real numbers aa and bb with a!=ba \neq b. Note that L[y]=y^('')-(a+b)y^(')+abyL[y]=y^{\prime \prime}-(a+b) y^{\prime}+a b y.
20. Use the result of Problem 16 to find the solution of the initial value problem
where L[y]=(D-(lambda+i mu))(D-(lambda-i mu))yL[y]=(D-(\lambda+i \mu))(D-(\lambda-i \mu)) y; that is, L[y]=y^('')-2lambday^(')+(lambda^(2)+mu^(2))yL[y]=y^{\prime \prime}-2 \lambda y^{\prime}+\left(\lambda^{2}+\mu^{2}\right) y. Note that the roots of the characteristic equation are lambda+-i mu\lambda \pm i \mu.
21. Use the result of Problem 16 to find the solution of the initial value problem
where L[y]=(D-a)^(2)yL[y]=(D-a)^{2} y, that is, L[y]=y^('')-2ay^(')+a^(2)yL[y]=y^{\prime \prime}-2 a y^{\prime}+a^{2} y, and aa is any real number.
22. By combining the results of Problems 19 through 21 , show that the solution of the initial value problem
where bb and cc are constants, can be written in the form
{:(37)y=phi(t)=int_(t_(0))^(t)K(t-s)g(s)ds:}\begin{equation*}
y=\phi(t)=\int_{t_{0}}^{t} K(t-s) g(s) d s \tag{37}
\end{equation*}
where the function KK depends only on the solutions y_(1)y_{1} and y_(2)y_{2} of the corresponding homogeneous equation and is independent of the nonhomogeneous term. Once KK is determined, all nonhomogeneous problems involving the same differential operator LL are reduced to the evaluation of an integral. Note also that although KK depends on both tt and ss, only the combination t-st-s appears, so KK is actually a function of a single variable. When we think of g(t)g(t) as the input to the problem and of phi(t)\phi(t) as the output, it follows from equation (37) that the output depends on the input over the entire interval from the
initial point t_(0)t_{0} to the current value tt. The integral in equation (37) is called the convolution of KK and gg, and KK is referred to as the kernel.
23. The method of reduction of order (Section 3.4) can also be used for the nonhomogeneous equation
provided one solution y_(1)y_{1} of the corresponding homogeneous equation is known. Let y=v(t)y_(1)(t)y=v(t) y_{1}(t) and show that yy satisfies equation (38) if vv is a solution of
Equation (39) is a first-order linear differential equation for v^(')v^{\prime}. By solving equation (39) for v^(')v^{\prime}, integrating the result to find vv, and then multiplying by y_(1)(t)y_{1}(t), you can find the general solution of equation (38). This method simultaneously finds both the second homogeneous solution and a particular solution.
In each of Problems 24 through 26, use the method outlined in Problem 23 to solve the given differential equation.
24. t^(2)y^('')-2ty^(')+2y=4t^(2),quad t > 0;quady_(1)(t)=tt^{2} y^{\prime \prime}-2 t y^{\prime}+2 y=4 t^{2}, \quad t>0 ; \quad y_{1}(t)=t
25. t^(2)y^('')+7ty^(')+5y=t,quad t > 0;quady_(1)(t)=t^(-1)t^{2} y^{\prime \prime}+7 t y^{\prime}+5 y=t, \quad t>0 ; \quad y_{1}(t)=t^{-1}
26. ty^('')-(1+t)y^(')+y=t^(2)e^(2t),t > 0;quady_(1)(t)=1+t quadt y^{\prime \prime}-(1+t) y^{\prime}+y=t^{2} e^{2 t}, t>0 ; \quad y_{1}(t)=1+t \quad (see Problem 12)
3.7 Mechanical and Electrical Vibrations
One of the reasons why second-order linear differential equations with constant coefficients are worth studying is that they serve as mathematical models of many important physical processes. Two important areas of application are the fields of mechanical and electrical oscillations. For example, the motion of a mass on a vibrating spring, the torsional oscillations of a shaft with a flywheel, the flow of electric current in a simple series circuit, and many other physical problems are all described by the solution of an initial value problem of the form
This illustrates a fundamental relationship between mathematics and physics: many physical problems may have mathematically equivalent models. Thus, once we know how to solve the initial value problem (1), it is only necessary to make appropriate interpretations of the constants a,ba, b, and cc, and of the functions yy and gg, to obtain solutions of different physical problems.
We will study the motion of a mass on a spring in detail because understanding the behavior of this simple system is the first step in the investigation of more complex vibrating systems. Further, the principles involved are common to many problems.
Consider a mass mm hanging at rest on the end of a vertical spring of original length ll, as shown in Figure 3.7.1. The mass causes an elongation LL of the spring in the downward (positive) direction. In this static situation there are two forces acting at the point where the mass is attached to the spring; see Figure 3.7.2. The gravitational force, or weight of the mass, acts downward and has magnitude w=mgw=m g, where gg is the acceleration due to gravity. There is also a force F_(s)F_{s}, due to the spring, that acts upward. If we assume that the elongation LL of the spring is small, the spring force is very nearly proportional to LL; this is known as Hooke's ^(10){ }^{10} law. Thus we write F_(s)=-kLF_{s}=-k L, where the constant of proportionality kk is called the
spring constant, and the minus sign is due to the fact that the spring force acts in the upward (negative) direction. Since the mass is in equilibrium, the two forces balance each other, which means that
$$
For a given weight w=mgw=m g, you can measure LL and then use equation (2) to determine kk. Note that kk has the units of force per unit length.
FIGURE 3.7.1 A spring-mass system.
FIGURE 3.7.2 Force diagram for a spring-mass system.
In the corresponding dynamic problem, we are interested in studying the motion of the mass when it is acted on by an external force or is initially displaced. Let u(t)u(t), measured positive in the downward direction, denote the displacement of the mass from its equilibrium position at time tt; see Figure 3.7.1. Then u(t)u(t) is related to the forces acting on the mass through Newton's law of motion
{:(3)mu^('')(t)=f(t)",":}\begin{equation*}
m u^{\prime \prime}(t)=f(t), \tag{3}
\end{equation*}
where u^('')u^{\prime \prime} is the acceleration of the mass and ff is the net force acting on the mass. Observe that both uu and ff are functions of time. In this dynamic problem there are now four separate forces that must be considered.
The weight w=mgw=m g of the mass always acts downward.
The spring force F_(s)F_{s} is assumed to be proportional to the total elongation L+uL+u of the spring and always acts to restore the spring to its natural position. If L+u > 0L+u>0, then the spring is extended, and the spring force is directed upward. In this case
On the other hand, if L+u < 0L+u<0, then the spring is compressed a distance |L+u||L+u|, and the spring force, which is now directed downward, is given by F_(s)=k|L+u|F_{s}=k|L+u|. However, when L+u < 0L+u<0, it follows that |L+u|=-(L+u)|L+u|=-(L+u), so F_(s)F_{s} is again given by equation (4). Thus, regardless of the position of the mass, the force exerted by the spring is always expressed by equation (4).
3. The damping or resistive force F_(d)F_{d} always acts in the direction opposite to the direction of motion of the mass. This force may arise from several sources: resistance from the air or other medium in which the mass moves, internal energy dissipation due to the extension or compression of the spring, friction between the mass and the guides (if any) that constrain its motion to one dimension, or a mechanical device (dashpot) that imparts a resistive force to the mass. In any case, we assume that the resistive force is proportional to the speed |du//dt||d u / d t| of the mass; this is usually referred to as viscous damping. If du//dt > 0d u / d t>0, then uu is increasing, so the mass is moving downward. Then F_(d)F_{d} is directed
upward and is given by
where gamma\gamma is a positive constant of proportionality known as the damping constant. On the other hand, if du//dt < 0d u / d t<0, then uu is decreasing, the mass is moving upward, and F_(d)F_{d} is directed downward. In this case, F_(d)=gamma|u^(')(t)|F_{d}=\gamma\left|u^{\prime}(t)\right|; since |u^(')(t)|=-u^(')(t)\left|u^{\prime}(t)\right|=-u^{\prime}(t), it follows that F_(d)(t)F_{d}(t) is again given by equation (5). Thus, regardless of the direction of motion of the mass, the damping force is always expressed by equation (5).
The damping force may be rather complicated, and the assumption that it is modeled adequately by equation (5) may be open to question. Some dashpots do behave as equation (5) states, and if the other sources of dissipation are small, it may be possible to neglect them altogether or to adjust the damping constant gamma\gamma to approximate them. An important benefit of the assumption (5) is that it leads to a linear (rather than a nonlinear) differential equation. In turn, this means that a thorough analysis of the system is straightforward, as we will show in this section and in Section 3.8.
4. An applied external force F(t)F(t) is directed downward or upward as F(t)F(t) is positive or negative. This could be a force due to the motion of the mount to which the spring is attached, or it could be a force applied directly to the mass. Often the external force is periodic.
Taking account of these forces, we can now rewrite Newton's law (3) as
Since mg-kL=0m g-k L=0 by equation (2), it follows that the equation of motion of the mass is
{:(7)mu^('')(t)+gammau^(')(t)+ku(t)=F(t):}\begin{equation*}
m u^{\prime \prime}(t)+\gamma u^{\prime}(t)+k u(t)=F(t) \tag{7}
\end{equation*}
where the constants m,gammam, \gamma, and kk are positive. Note that equation (7) has the same form as equation (1), that is, it is a nonhomogeneous second-order linear differential equation with constant coefficients.
It is important to understand that equation (7) is only an approximate equation for the displacement u(t)u(t). In particular, both equations (4) and (5) should be viewed as approximations for the spring force and the damping force, respectively. In our derivation we have also neglected the mass of the spring in comparison with the mass of the attached body.
The complete formulation of the vibration problem requires that we specify two initial conditions, namely, the initial position u_(0)u_{0} and the initial velocity v_(0)v_{0} of the mass:
It follows from Theorem 3.2.1 that these conditions give a mathematical problem that has a unique solution for any values of the constants u_(0)u_{0} and v_(0)v_{0}. This is consistent with our physical intuition that if the mass is set in motion with a given initial displacement and velocity, then its position will be determined uniquely at all future times. The position of the mass is given (approximately) by the solution of the second-order linear differential equation (7) subject to the prescribed initial conditions (8).
EXAMPLE 1
A mass weighing 4 lb stretches a spring 2 in . Suppose that the mass is given an additional 6-in displacement in the positive direction and then released. The mass is in medium that exerts a viscous resistance of 6 lb when the mass has a velocity of 3ft//s3 \mathrm{ft} / \mathrm{s}. Under the assumptions discussed in this section, formulate the initial value problem that governs the motion of the mass.
Solution:
The required initial value problem consists of the differential equation (7) and initial conditions (8), so our task is to determine the various constants that appear in these equations. The first step is to
choose the units of measurement. Based on the statement of the problem, it is natural to use the English rather than the metric system of units. The only time unit mentioned is the second, so we will measure tt in seconds. On the other hand, both the foot and the inch appear in the statement as units of length. It is immaterial which one we use, but having made a choice, we must be consistent. To be definite, let us measure the displacement uu in feet.
Since nothing is said in the statement of the problem about an external force, we assume that F(t)=0F(t)=0. To determine mm, note that
The damping coefficient gamma\gamma is determined from the statement that gammau^(')\gamma u^{\prime} is equal to 6 lb when u^(')u^{\prime} is 3ft//s3 \mathrm{ft} / \mathrm{s}. Therefore,
The second initial condition is implied by the word "released" in the statement of the problem, which we interpret to mean that the mass is set in motion with no initial velocity.
Undamped Free Vibrations. If there is no external force, then F(t)=0F(t)=0 in equation (7). Let us also suppose that there is no damping so that gamma=0\gamma=0; this is an idealized configuration of the system, seldom (if ever) completely attainable in practice. However, if the actual damping is very small, then the assumption of no damping may yield satisfactory results over short to moderate time intervals. In this case the equation of motion (7) reduces to
{:(11)mu^('')+ku=0:}\begin{equation*}
m u^{\prime \prime}+k u=0 \tag{11}
\end{equation*}
The characteristic equation for equation (11) is
mr^(2)+k=0m r^{2}+k=0
and its roots are r=+-isqrt(k//m)r= \pm i \sqrt{k / m}. Thus the general solution of equation (11) is
{:(15)u=R cos delta cos(omega_(0)t)+R sin delta sin(omega_(0)t):}\begin{equation*}
u=R \cos \delta \cos \left(\omega_{0} t\right)+R \sin \delta \sin \left(\omega_{0} t\right) \tag{15}
\end{equation*}
By comparing equation (15) with equation (12), we find that A,B,RA, B, R, and delta\delta are related by the equations
{:(16)A=R cos delta","quad B=R sin delta:}\begin{equation*}
A=R \cos \delta, \quad B=R \sin \delta \tag{16}
\end{equation*}
Thus
{:(17)R=sqrt(A^(2)+B^(2))","quad tan delta=(B)/(A):}\begin{equation*}
R=\sqrt{A^{2}+B^{2}}, \quad \tan \delta=\frac{B}{A} \tag{17}
\end{equation*}
In calculating delta\delta, we must take care to choose the correct quadrant; this can be done by checking the signs of cos delta\cos \delta and sin delta\sin \delta in equations (16).
The graph of equation (14), or the equivalent equation (12), for a typical set of initial conditions is shown in Figure 3.7.3. The graph is a displaced cosine wave that describes a periodic, or simple harmonic, motion of the mass. The period of the motion is
The circular frequency omega_(0)=sqrt(k//m)\omega_{0}=\sqrt{k / m}, measured in radians per unit time, is called the natural frequency of the vibration. The maximum displacement RR of the mass from equilibrium is the amplitude of the motion. The dimensionless parameter delta\delta is called the phase, or phase angle, and measures the displacement of the wave from its normal position corresponding to delta=0\delta=0.
FIGURE 3.7.3 Simple harmonic motion; u=R cos(omega_(0)t-delta)u=R \cos \left(\omega_{0} t-\delta\right).
Note that the horizontal axis is labeled as omega_(0)t\omega_{0} t.
Note that the motion described by equation (14) has a constant amplitude that does not diminish with time. This reflects the fact that, in the absence of damping, there is no way for the system to dissipate the energy imparted to it by the initial displacement and velocity. Further, for a given mass mm and spring constant kk, the system always vibrates at the same frequency omega_(0)\omega_{0}, regardless of the initial conditions. However, the initial conditions do help to determine the amplitude of the motion. Finally, observe from equation (18) that the period TT increases as the mass mm increases, so larger masses vibrate more slowly. On the other hand, the period TT decreases as the spring constant kk increases, which means that stiffer springs cause the system to vibrate more rapidly.
EXAMPLE 2
Suppose that a mass weighing 10 lb stretches a spring 2 in . If the mass is displaced an additional 2 in and is then set in motion with an initial upward velocity of 1ft//s1 \mathrm{ft} / \mathrm{s}, determine the position of the mass at any later time. Also determine the period, amplitude, and phase of the motion.
Solution:
The spring constant is k=10lb//2in=60lb//ftk=10 \mathrm{lb} / 2 \mathrm{in}=60 \mathrm{lb} / \mathrm{ft}, and the mass is m=w//g=10//32lb*s^(2)//ftm=w / g=10 / 32 \mathrm{lb} \cdot \mathrm{s}^{2} / \mathrm{ft}. Hence the equation of motion reduces to
The natural frequency is omega_(0)=8sqrt3~=13.856rad//s\omega_{0}=8 \sqrt{3} \cong 13.856 \mathrm{rad} / \mathrm{s}, so the period is T=2pi//omega_(0)~=0.453sT=2 \pi / \omega_{0} \cong 0.453 \mathrm{~s}. The amplitude RR and phase delta\delta are found from equations (17). We have
R^(2)=(1)/(36)+(1)/(192)=(19)/(576)," so "R~=0.182ftR^{2}=\frac{1}{36}+\frac{1}{192}=\frac{19}{576}, \text { so } R \cong 0.182 \mathrm{ft}
The second of equations (17) yields tan delta=-sqrt3//4\tan \delta=-\sqrt{3} / 4. There are two solutions of this equation, one in the second quadrant and one in the fourth. In the present problem, cos delta > 0\cos \delta>0 and sin delta < 0\sin \delta<0, so delta\delta is in the fourth quadrant. In fact,
The graph of the solution (20) is shown in Figure 3.7.4.
FIGURE 3.7.4 An undamped free vibration: u^('')+192 u=0,u(0)=1//6,u^(')(0)=-1u^{\prime \prime}+192 u=0, u(0)=1 / 6, u^{\prime}(0)=-1.
Note that the scale for the horizontal axis is omega_(0)t\omega_{0} t.
Damped Free Vibrations. When the effects of damping are included, the differential equation governing the motion of the mass is
{:(21)mu^('')+gammau^(')+ku=0:}\begin{equation*}
m u^{\prime \prime}+\gamma u^{\prime}+k u=0 \tag{21}
\end{equation*}
We are especially interested in examining the effect of variations in the damping coefficient gamma\gamma for given values of the mass mm and spring constant kk. The corresponding characteristic equation is
mr^(2)+gamma r+k=0m r^{2}+\gamma r+k=0
and its roots are
{:(22)r_(1)","r_(2)=(-gamma+-sqrt(gamma^(2)-4km))/(2m)=(gamma)/(2m)(-1+-sqrt(1-(4km)/(gamma^(2)))):}\begin{equation*}
r_{1}, r_{2}=\frac{-\gamma \pm \sqrt{\gamma^{2}-4 k m}}{2 m}=\frac{\gamma}{2 m}\left(-1 \pm \sqrt{1-\frac{4 k m}{\gamma^{2}}}\right) \tag{22}
\end{equation*}
Depending on the sign of gamma^(2)-4km\gamma^{2}-4 k m, the solution uu has one of the following forms:
Since m,gammam, \gamma, and kk are positive, gamma^(2)-4km\gamma^{2}-4 k m is always less than gamma^(2)\gamma^{2}. Hence, if gamma^(2)-4km >= 0\gamma^{2}-4 k m \geq 0, then the values of r_(1)r_{1} and r_(2)r_{2} given by equation (22) are negative. If gamma^(2)-4km < 0\gamma^{2}-4 k m<0, then the values of r_(1)r_{1} and r_(2)r_{2} are complex, but with negative real part. Thus, in all cases, the solution uu tends to zero as t rarr oot \rightarrow \infty; this occurs regardless of the values of the arbitrary constants AA and BB-that is, regardless of the initial conditions. This confirms our intuitive expectation, namely, that damping gradually dissipates the energy initially imparted to the system, and consequently the motion dies out with increasing time.
The most interesting case is the third one, which occurs when the damping is small. If we let A=R cos deltaA=R \cos \delta and B=R sin deltaB=R \sin \delta in equation (25), then we obtain
The displacement uu lies between the curves u=+-Re^(-gamma t//(2m))u= \pm R e^{-\gamma t /(2 m)}; hence it resembles a cosine wave whose amplitude decreases as tt increases. A typical example is sketched in Figure 3.7.5. The motion is called a damped oscillation or a damped vibration. The amplitude factor RR depends on m,gamma,km, \gamma, k, and the initial conditions.
FIGURE 3.7.5 Damped vibration; u=Re^(-gamma t//2m)cos(mu t-delta)u=R e^{-\gamma t / 2 m} \cos (\mu t-\delta).
Note that the scale for the horizontal axis is mu t\mu t.
Although the motion is not periodic, the parameter mu\mu determines the frequency with which the mass oscillates back and forth; consequently, mu\mu is called the quasi-frequency. By comparing mu\mu with the frequency omega_(0)\omega_{0} of undamped motion, we find that
{:(27)(mu)/(omega_(0))=((4km-gamma^(2))^(1//2)//(2m))/(sqrt(k//m))=(1-(gamma^(2))/(4km))^(1//2)~=1-(gamma^(2))/(8km):}\begin{equation*}
\frac{\mu}{\omega_{0}}=\frac{\left(4 k m-\gamma^{2}\right)^{1 / 2} /(2 m)}{\sqrt{k / m}}=\left(1-\frac{\gamma^{2}}{4 k m}\right)^{1 / 2} \cong 1-\frac{\gamma^{2}}{8 k m} \tag{27}
\end{equation*}
The last approximation is valid when gamma^(2)//4km\gamma^{2} / 4 \mathrm{~km} is small; we refer to this situation as "small damping." Thus the effect of small damping is to reduce slightly the frequency of the oscillation. By analogy with equation (18), the quantity T_(d)=2pi//muT_{d}=2 \pi / \mu is called the quasi-period of the motion. It is the time between successive maxima or successive minima of the position of the mass, or between successive passages of the mass through its equilibrium position while going in the same direction. The relation between T_(d)T_{d} and TT is given by
{:(28)(T_(d))/(T)=(omega_(0))/(mu)=(1-(gamma^(2))/(4km))^(-1//2)~=1+(gamma^(2))/(8km):}\begin{equation*}
\frac{T_{d}}{T}=\frac{\omega_{0}}{\mu}=\left(1-\frac{\gamma^{2}}{4 k m}\right)^{-1 / 2} \cong 1+\frac{\gamma^{2}}{8 k m} \tag{28}
\end{equation*}
where again the last approximation is valid when gamma^(2)//4km\gamma^{2} / 4 \mathrm{~km} is small. Thus small damping increases the quasi-period.
Equations (27) and (28) reinforce the significance of the dimensionless ratio gamma^(2)//(4km)\gamma^{2} /(4 \mathrm{~km}). It is not the magnitude of gamma\gamma alone that determines whether damping is large or small, but the magnitude of gamma^(2)\gamma^{2} compared to 4 km . When gamma^(2)//(4km)\gamma^{2} /(4 \mathrm{~km}) is small, then damping has a small effect on the quasi-frequency and quasi-period of the motion. On the other hand, if we want to study the detailed motion of the mass for all time, then we can never neglect the damping force, no matter how small.
As gamma^(2)//(4km)\gamma^{2} /(4 \mathrm{~km}) increases, the quasi-frequency mu\mu decreases and the quasi-period T_(d)T_{d} increases. In fact, mu rarr0\mu \rightarrow 0 and T_(d)rarr ooT_{d} \rightarrow \infty as gamma rarr2sqrtkm\gamma \rightarrow 2 \sqrt{\mathrm{~km}}. As indicated by equations (23), (24), and (25), the nature of the solution changes as gamma\gamma passes through the value 2sqrtkm2 \sqrt{\mathrm{~km}}. The motion with gamma=2sqrt(km)\gamma=2 \sqrt{k m} is said to be critically damped. For larger values of gamma\gamma, gamma > 2sqrt(km)\gamma>2 \sqrt{k m}, the motion is said to be overdamped. In these cases, given by equations (24) and (23), respectively, the mass may pass through its equilibrium position at most once (see Figure 3.7.6) and then creep back to it. The mass does not oscillate about the equilibrium, as it does for small gamma\gamma. Two typical examples of critically damped motion are shown in Figure 3.7.6, and the situation is discussed further in Problems 15 and 16.
FIGURE 3.7.6 Critically damped motions: u^('')+u^(')+0.25 u=0u^{\prime \prime}+u^{\prime}+0.25 u=0; u=(A+Bt)e^(-t//2)u=(A+B t) e^{-t / 2}. The solid blue curve is the solution satisfying u(0)=1//2,u^(')(0)=7//4u(0)=1 / 2, u^{\prime}(0)=7 / 4; the dashed green curve satisfies u(0)=1//2,u^(')(0)=-7//4u(0)=1 / 2, u^{\prime}(0)=-7 / 4.
EXAMPLE 3
The motion of a certain spring-mass system is governed by the differential equation
where uu is measured in feet and tt in seconds. If u(0)=2u(0)=2 and u^(')(0)=0u^{\prime}(0)=0, determine the position of the mass at any time. Find the quasi-frequency and the quasi-period, as well as the time at which the mass first passes through its equilibrium position. Also find the time tau\tau such that |u(t)| < 0.1|u(t)|<0.1 for all t > taut>\tau.
FIGURE 3.7.7 Vibration with small damping (solid blue curve) and with no damping (dashed green curve). Both motions have the same initial conditions: u(0)=2,u^(')(0)=0u(0)=2, u^{\prime}(0)=0.
where delta\delta is in the first quadrant with tan delta=1//sqrt255\tan \delta=1 / \sqrt{255}, so delta~=0.06254\delta \cong 0.06254. The displacement of the mass as a function of time is shown in Figure 3.7.7. For purposes of comparison, we also show the motion if the damping term is neglected.
The quasi-frequency is mu=sqrt255//16~=0.998\mu=\sqrt{255} / 16 \cong 0.998, and the quasi-period is T_(d)=2pi//mu~=6.295sT_{d}=2 \pi / \mu \cong 6.295 \mathrm{~s}. These values differ only slightly from the corresponding values ( 1 and 2pi2 \pi, respectively) for the undamped oscillation. This is evident also from the graphs in Figure 3.7.7, which rise and fall almost together. The damping coefficient is small in this example: only one-sixteenth of the critical value, in fact. Nevertheless, the amplitude of the oscillation is reduced rather rapidly.
FIGURE 3.7.8 Solution of Example 3 for 40 <= t <= 6040 \leq t \leq 60; determination of the time tau\tau after which |u(t)| < 0.1|u(t)|<0.1.
Figure 3.7.8 shows the graph of the solution for 40 <= t <= 6040 \leq t \leq 60, together with the graphs of u=+-0.1u= \pm 0.1. From the graph it appears that tau\tau is about 47.5 , and by a more precise calculation we find that tau~=47.5149s\tau \cong 47.5149 \mathrm{~s}.
To find the time at which the mass first passes through its equilibrium position, we refer to equation (30) and set sqrt255t//16-delta\sqrt{255} t / 16-\delta equal to pi//2\pi / 2, the smallest positive zero of the cosine function. Then, by solving for tt, we obtain
FIGURE 3.7.9 A simple electric circuit.
Electric Circuits. A second example of the occurrence of second-order linear differential equations with constant coefficients is their use as a model of the flow of electric current in the simple series circuit shown in Figure 3.7.9. The current II, measured in amperes (A), is a function of time tt. The resistance RR in ohms (Omega)(\Omega), the capacitance CC in farads (F)(\mathrm{F}), and the inductance LL in henrys (H)(\mathrm{H}) are all positive and are assumed to be known constants. The impressed voltage EE in volts (V)(\mathrm{V}) is a given function of time. Another physical quantity that enters the discussion is the total charge QQ in coulombs (C) on the capacitor at time tt. The relation between charge QQ and current II is
The flow of current in the circuit is governed by Kirchhoff's ^(11){ }^{11} second law: In a closed circuit the impressed voltage is equal to the sum of the voltage drops in the rest of the circuit.
According to the elementary laws of electricity, we know that
The voltage drop across the resistor is RIR I.
The voltage drop across the capacitor is (Q)/(C)\frac{Q}{C}.
The voltage drop across the inductor is L(dI)/(dt)L \frac{d I}{d t}.
Hence, by Kirchhoff's law,
{:(32)L(dI)/(dt)+RI+(1)/(C)Q=E(t):}\begin{equation*}
L \frac{d I}{d t}+R I+\frac{1}{C} Q=E(t) \tag{32}
\end{equation*}
The units for voltage, resistance, current, charge, capacitance, inductance, and time are all related:
Thus to know the charge at any time it is sufficient to know the charge on the capacitor and the current in the circuit at some initial time t_(0)t_{0}.
Alternatively, we can obtain a differential equation for the current II by differentiating equation (33) with respect to tt, and then substituting for dQ//dtd Q / d t from equation (31). The result is
{:(35)LI^('')+RI^(')+(1)/(C)I=E^(')(t):}\begin{equation*}
L I^{\prime \prime}+R I^{\prime}+\frac{1}{C} I=E^{\prime}(t) \tag{35}
\end{equation*}
Hence I_(0)^(')I_{0}^{\prime} is also determined by the initial charge and current, which are physically measurable quantities.
The most important conclusion from this discussion is that the flow of current in the circuit is described by an initial value problem of precisely the same form as the one that describes the motion of a spring-mass system. This is a good example of the unifying role of mathematics: once you know how to solve second-order linear equations with constant coefficients, you can interpret the results in terms of mechanical vibrations, electric circuits, or any other physical situation that leads to the same problem.
Problems
In each of Problems 1 and 2 , determine omega_(0),R\omega_{0}, R, and delta\delta so as to write the given expression in the form u=R cos(omega_(0)t-delta)u=R \cos \left(\omega_{0} t-\delta\right).
A mass of 100 g stretches a spring 5 cm . If the mass is set in motion from its equilibrium position with a downward velocity of 10cm//s10 \mathrm{~cm} / \mathrm{s}, and if there is no damping, determine the position uu of the mass at any time tt. When does the mass first return to its equilibrium position?
A mass weighing 3 lb stretches a spring 3 in . If the mass is pushed upward, contracting the spring a distance of 1 in and then set in motion with a downward velocity of 2ft//s2 \mathrm{ft} / \mathrm{s}, and if there is no damping, find the position uu of the mass at any time tt. Determine the frequency, period, amplitude, and phase of the motion.
(G) 5. A mass of 20 g stretches a spring 5 cm . Suppose that the mass is also attached to a viscous damper with a damping constant of 400dyn*s//cm400 \mathrm{dyn} \cdot \mathrm{s} / \mathrm{cm}. If the mass is pulled down an additional 2 cm and then released, find its position uu at any time tt. Plot uu versus tt. Determine the quasi-frequency and the quasi-period. Determine the ratio of the quasiperiod to the period of the corresponding undamped motion. Also find the time tau\tau such that |u(t)| < 0.05cm|u(t)|<0.05 \mathrm{~cm} for all t > taut>\tau.
A spring is stretched 10 cm by a force of 3 N . A mass of 2 kg is hung from the spring and is also attached to a viscous damper that exerts a force of 3 N when the velocity of the mass is 5m//s5 \mathrm{~m} / \mathrm{s}. If the mass is pulled down 5 cm below its equilibrium position and given an initial downward velocity of 10cm//s10 \mathrm{~cm} / \mathrm{s}, determine its position uu at any time tt. Find the quasi-frequency mu\mu and the ratio of mu\mu to the natural frequency of the corresponding undamped motion.
A series circuit has a capacitor of 10^(-5)F10^{-5} \mathrm{~F}, a resistor of 3xx10^(2)Omega3 \times 10^{2} \Omega, and an inductor of 0.2 H . The initial charge on the capacitor is 10^(-6)C10^{-6} \mathrm{C} and there is no initial current. Find the charge QQ on the capacitor at any time tt.
A vibrating system satisfies the equation u^('')+gammau^(')+u=0u^{\prime \prime}+\gamma u^{\prime}+u=0. Find the value of the damping coefficient gamma\gamma for which the quasi-period of the damped motion is 50%50 \% greater than the period of the corresponding undamped motion.
Show that the period of motion of an undamped vibration of a mass hanging from a vertical spring is 2pisqrt(L//g)2 \pi \sqrt{L / g}, where LL is the elongation of the spring due to the mass, and gg is the acceleration due to gravity.
Show that the solution of the initial value problem
can be expressed as the sum u=v+wu=v+w, where vv satisfies the initial conditions v(t_(0))=u_(0),v^(')(t_(0))=0,wv\left(t_{0}\right)=u_{0}, v^{\prime}\left(t_{0}\right)=0, w satisfies the initial conditions w(t_(0))=0,w^(')(t_(0))=u_(0)^(')w\left(t_{0}\right)=0, w^{\prime}\left(t_{0}\right)=u_{0}^{\prime}, and both vv and ww satisfy the same differential equation as uu. This is another instance of superposing solutions of simpler problems to obtain the solution of a more general problem.
11. a. Show that A cos(omega_(0)t)+B sin(omega_(0)t)A \cos \left(\omega_{0} t\right)+B \sin \left(\omega_{0} t\right) can be written in the form r sin(omega_(0)t-theta)r \sin \left(\omega_{0} t-\theta\right). Determine rr and theta\theta in terms of AA and BB.
b. If R cos(omega_(0)t-delta)=r sin(omega_(0)t-theta)R \cos \left(\omega_{0} t-\delta\right)=r \sin \left(\omega_{0} t-\theta\right), determine the relationship among R,r,deltaR, r, \delta, and theta\theta.
12. If a series circuit has a capacitor of C=0.8 xx10^(-6)FC=0.8 \times 10^{-6} \mathrm{~F} and an inductor of L=0.2HL=0.2 \mathrm{H}, find the resistance RR so that the circuit is critically damped.
13. Assume that the system described by the differential equation mu^('')+gammau^(')+ku=0m u^{\prime \prime}+\gamma u^{\prime}+k u=0 is either critically damped or overdamped. Show that the mass can pass through the equilibrium position at most once, regardless of the initial conditions.
Hint: Determine all possible values of tt for which u=0u=0.
14. Assume that the system described by the differential equation mu^('')+gammau^(')+ku=0m u^{\prime \prime}+\gamma u^{\prime}+k u=0 is critically damped and that the initial conditions are u(0)=u_(0),u^(')(0)=v_(0)u(0)=u_{0}, u^{\prime}(0)=v_{0}. If v_(0)=0v_{0}=0, show that u rarr0u \rightarrow 0 as t rarr oot \rightarrow \infty but that uu is never zero. If u_(0)u_{0} is positive, determine a condition on v_(0)v_{0} that will ensure that the mass passes through its equilibrium position after it is released.
15. Logarithmic Decrement. a. For the damped oscillation described by equation (26), show that the time between successive maxima is T_(d)=2pi//muT_{d}=2 \pi / \mu.
b. Show that the ratio of the displacements at two successive maxima is given by exp(gammaT_(d)//(2m))\exp \left(\gamma T_{d} /(2 m)\right). Observe that this ratio does not depend on which pair of maxima is chosen. The natural logarithm of this ratio is called the logarithmic decrement and is denoted by Delta\Delta.
c. Show that Delta=pi gamma//(m mu)\Delta=\pi \gamma /(m \mu). Since m,mum, \mu, and Delta\Delta are quantities that can be measured easily for a mechanical system, this result provides a convenient and practical method for determining the damping constant of the system, which is more difficult to measure directly. In particular, for the motion of a vibrating mass in a viscous fluid, the damping constant depends on the viscosity of the fluid; for simple geometric shapes the form of this dependence is known, and the preceding relation allows the experimental determination of the viscosity. This is one of the most accurate ways of determining the viscosity of a gas at high pressure.
16. Referring to Problem 15, find the logarithmic decrement of the system in Problem 5.
17. The position of a certain spring-mass system satisfies the initial value problem
If the period and amplitude of the resulting motion are observed to be pi\pi and 3, respectively, determine the values of kk and vv.
18. Consider the initial value problem
Assume that gamma^(2) < 4km\gamma^{2}<4 \mathrm{~km}.
a. Solve the initial value problem.
b. Write the solution in the form u(t)=Re^(-gamma t//(2m))cos(mu t-delta)u(t)=R e^{-\gamma t /(2 m)} \cos (\mu t-\delta).
Determine RR in terms of m,gamma,k,u_(0)m, \gamma, k, u_{0}, and v_(0)v_{0}.
c. Investigate the dependence of RR on the damping coefficient gamma\gamma for fixed values of the other parameters.
19. A cubic block of side ll and mass density rho\rho per unit volume is floating in a fluid of mass density rho_(0)\rho_{0} per unit volume, where rho_(0) > rho\rho_{0}>\rho. If the block is slightly depressed and then released, it oscillates in the vertical direction. Assuming that the viscous damping of the fluid and air can be neglected, derive the differential equation of motion and determine the period of the motion.
Hint: Use Archimedes ^(12){ }^{12} principle: an object that is completely or partially submerged in a fluid is acted on by an upward (buoyant) force equal to the weight of the displaced fluid.
20. The position of a certain undamped spring-mass system satisfies the initial value problem
a. Find the solution of this initial value problem.
G b. Plot uu versus tt and u^(')u^{\prime} versus tt on the same axes.
G c. Plot u^(')u^{\prime} versus uu; that is, plot u(t)u(t) and u^(')(t)u^{\prime}(t) parametrically with tt as the parameter. This plot is known as a phase plot, and the uu^(')u u^{\prime}-plane is called the phase plane. Observe that a closed curve in the phase plane corresponds to a periodic solution u(t)u(t). What is the direction of motion on the phase plot as tt increases?
21. The position of a certain spring-mass system satisfies the initial value problem
a. Find the solution of this initial value problem.
G b\mathbf{b}. Plot uu versus tt and u^(')u^{\prime} versus tt on the same axes.
(G) Plot u^(')u^{\prime} versus uu in the phase plane (see Problem 20). Identify several corresponding points on the curves in parts bb and c. What is the direction of motion on the phase plot as tt increases?
22. In the absence of damping, the motion of a spring-mass system satisfies the initial value problem
a. Show that the kinetic energy initially imparted to the mass is mb^(2)//2m b^{2} / 2 and that the potential energy initially stored in the spring is ka^(2)//2k a^{2} / 2, so initially the total energy in the system is (ka^(2)+mb^(2))//2\left(k a^{2}+m b^{2}\right) / 2.
b. Solve the given initial value problem.
c. Using the solution in part b, determine the total energy in the system at any time tt. Your result should confirm the principle of conservation of energy for this system.
23. Suppose that a mass mm slides without friction on a horizontal surface. The mass is attached to a spring with spring constant kk, as shown in Figure 3.7.10, and is also subject to viscous air resistance with coefficient gamma\gamma. Show that the displacement u(t)u(t) of the mass from its equilibrium position satisfies equation (21). How does the derivation of the equation of motion in this case differ from the derivation given in the text?
FIGURE 3.7.10 A spring-mass system.
24. In the spring-mass system of Problem 23 , suppose that the spring force is not given by Hooke's law but instead satisfies the relation
where k > 0k>0 and epsilon\epsilon is small but may be of either sign. The spring is called a hardening spring if epsilon > 0\epsilon>0 and a softening spring if epsilon < 0\epsilon<0. Why are these terms appropriate?
a. Show that the displacement u(t)u(t) of the mass from its equilibrium position satisfies the differential equation
In the remainder of this problem, assume that m=1,k=1m=1, k=1, and gamma=0\gamma=0.
b. Find u(t)u(t) when epsilon=0\epsilon=0 and also determine the amplitude and period of the motion.
(G) Let epsilon=0.1\epsilon=0.1. Plot a numerical approximation to the solution. Does the motion appear to be periodic? Estimate the amplitude and period.
(G) d. Repeat part c for epsilon=0.2\epsilon=0.2 and epsilon=0.3\epsilon=0.3.
(G) e. Plot your estimated values of the amplitude AA and the period TT versus epsilon\epsilon. Describe the way in which AA and TT, respectively, depend on epsilon\epsilon.
G f. Repeat parts c,d\mathrm{c}, \mathrm{d}, and e for negative values of epsilon\epsilon.
3.8 Forced Periodic Vibrations
We will now investigate the situation in which a periodic external force is applied to a springmass system. The behavior of this simple system models that of many oscillatory systems with an external force due, for example, to a motor attached to the system. We will first consider the case in which damping is present and will look later at the idealized special case in which there is assumed to be no damping.
Forced Vibrations with Damping. The algebraic calculations can be fairly complicated in this kind of problem, so we will begin with a relatively simple example.
EXAMPLE 1
Suppose that the motion of a certain spring-mass system satisfies the differential equation
{:(1)u^('')+u^(')+(5)/(4)u=3cos t:}\begin{equation*}
u^{\prime \prime}+u^{\prime}+\frac{5}{4} u=3 \cos t \tag{1}
\end{equation*}
Find the solution of this initial value problem and describe the behavior of the solution for large tt.
Solution:
The homogeneous equation corresponding to equation (1) has the characteristic equation r^(2)+r+(5)/(4)=0r^{2}+r+\frac{5}{4}=0 with roots r=-(1)/(2)+-ir=-\frac{1}{2} \pm i. Thus a general solution u_(c)(t)u_{c}(t) of this homogeneous equation is
A particular solution of equation (1) has the form U(t)=A cos t+B sin tU(t)=A \cos t+B \sin t, where AA and BB are found by substituting U(t)U(t) for uu in equation (1). We have U^(')(t)=-A sin t+B cos tU^{\prime}(t)=-A \sin t+B \cos t and U^('')(t)=-A cos t-B sin tU^{\prime \prime}(t)=-A \cos t-B \sin t. Thus, from equation (1) we obtain
so c_(1)=(22)/(17)c_{1}=\frac{22}{17} and c_(2)=(14)/(17)c_{2}=\frac{14}{17}. Thus we finally arrive at the solution of the given initial value problem (1), (2), namely,
The graph of the solution (6) is shown by the green curve in Figure 3.8.1.
It is important to note that the solution consists of two distinct parts. The first two terms on the right-hand side of equation (6) contain the exponential factor e^(-t//2)e^{-t / 2}; as a result, they rapidly approach zero. It is customary to call these terms the transient solution. The remaining terms in equation (6) involve only sines and cosines, so they represent an oscillation that continues indefinitely. We refer to them as the steady-state solution. The dotted red and dashed blue curves in Figure 3.8.1 show the transient and the steady-state parts of the solution, respectively. The transient part comes from the solution of the homogeneous equation corresponding to equation (1) and is needed to satisfy the initial conditions. The steady-state solution is the particular solution of the full nonhomogeneous equation. After a fairly short time, the transient solution is vanishingly small and the full solution is essentially indistinguishable from the steady state.
FIGURE 3.8.1 Solution of the initial value problem (1), (2): u^('')+u^(')+5u//4=3cos t,u(0)=2,u^(')(0)=3u^{\prime \prime}+u^{\prime}+5 u / 4=3 \cos t, u(0)=2, u^{\prime}(0)=3. The full solution (solid green) is the sum of the transient solution (dotted red) and steady-state solution (dashed blue).
The equation of motion of a general spring-mass system subject to an external force F(t)F(t) is equation (7) in Section 3.7:
{:(7)mu^('')(t)+gammau^(')(t)+ku(t)=F(t):}\begin{equation*}
m u^{\prime \prime}(t)+\gamma u^{\prime}(t)+k u(t)=F(t) \tag{7}
\end{equation*}
where m,gammam, \gamma, and kk are the mass, damping coefficient, and spring constant of the spring-mass system. Suppose now that the external force is given by F_(0)cos(omega t)F_{0} \cos (\omega t), where F_(0)F_{0} and omega\omega are positive constants representing the amplitude and frequency, respectively, of the force. Then equation (7) becomes
The first two terms on the right-hand side of equation (9) are the general solution u_(c)(t)u_{c}(t) of the homogeneous equation corresponding to equation (8), and the latter two terms are a particular solution U(t)U(t) of the full nonhomogeneous equation. The coefficients AA and BB can be found, as usual, by substituting these terms into the differential equation (8), while the arbitrary constants c_(1)c_{1} and c_(2)c_{2} are available to satisfy initial conditions, if any are prescribed. The solutions u_(1)(t)u_{1}(t)
and u_(2)(t)u_{2}(t) of the homogeneous equation depend on the roots r_(1)r_{1} and r_(2)r_{2} of the characteristic equation mr^(2)+gamma r+k=0m r^{2}+\gamma r+k=0. Since m,gammam, \gamma, and kk are all positive, it follows that r_(1)r_{1} and r_(2)r_{2} either are real and negative or are complex conjugates with a negative real part. In either case, both u_(1)(t)u_{1}(t) and u_(2)(t)u_{2}(t) approach zero as t rarr oot \rightarrow \infty. Since u_(c)(t)u_{c}(t) dies out as tt increases, it is called the transient solution. In many applications, it is of little importance and (depending on the value of gamma\gamma ) may well be undetectable after only a few seconds.
The remaining terms in equation (9)-namely, U(t)=A cos(omega t)+B sin(omega t)U(t)=A \cos (\omega t)+B \sin (\omega t)-do not die out as tt increases but persist indefinitely, or as long as the external force is applied. They represent a steady oscillation with the same frequency as the external force and are called the steady-state solution or the forced response of the system. The transient solution enables us to satisfy whatever initial conditions may be imposed. With increasing time, the energy put into the system by the initial displacement and velocity is dissipated through the damping force, and the motion then becomes the response of the system to the external force. Without damping, the effect of the initial conditions would persist for all time.
It is convenient to express U(t)U(t) as a single trigonometric term rather than as a sum of two terms. Recall that we did this for other similar expressions in Section 3.7. Thus we write
The amplitude RR and phase delta\delta depend directly on AA and BB and indirectly on the parameters in the differential equation (8). It is possible to show, by straightforward but somewhat lengthy algebraic computations, that
{:(11)R=(F_(0))/(Delta)","quad cos delta=(m(omega_(0)^(2)-omega^(2)))/(Delta)","quad" and "sin delta=(gamma omega)/(Delta):}\begin{equation*}
R=\frac{F_{0}}{\Delta}, \quad \cos \delta=\frac{m\left(\omega_{0}^{2}-\omega^{2}\right)}{\Delta}, \quad \text { and } \sin \delta=\frac{\gamma \omega}{\Delta} \tag{11}
\end{equation*}
where
{:(12)Delta=sqrt(m^(2)(omega_(0)^(2)-omega^(2))^(2)+gamma^(2)omega^(2))" and "omega_(0)^(2)=(k)/(m):}\begin{equation*}
\Delta=\sqrt{m^{2}\left(\omega_{0}^{2}-\omega^{2}\right)^{2}+\gamma^{2} \omega^{2}} \text { and } \omega_{0}^{2}=\frac{k}{m} \tag{12}
\end{equation*}
Recall that omega_(0)\omega_{0} is the natural frequency of the unforced system in the absence of damping.
We now investigate how the amplitude RR of the steady-state oscillation depends on the frequency omega\omega of the external force. Substituting from equation (12) into the expression for RR in equation (11) and executing some algebraic manipulations, we find that
{:(13)(Rk)/(F_(0))=((1-((omega)/(omega_(0)))^(2))^(2)+Gamma((omega)/(omega_(0)))^(2))^(-1//2)" where "Gamma=(gamma^(2))/(mk):}\begin{equation*}
\frac{R k}{F_{0}}=\left(\left(1-\left(\frac{\omega}{\omega_{0}}\right)^{2}\right)^{2}+\Gamma\left(\frac{\omega}{\omega_{0}}\right)^{2}\right)^{-1 / 2} \text { where } \Gamma=\frac{\gamma^{2}}{m k} \tag{13}
\end{equation*}
Observe that the quantity Rk//F_(0)R k / F_{0} is the ratio of the amplitude RR of the forced response to F_(0)//kF_{0} / k, the static displacement of the spring produced by a force F_(0)F_{0}.
For low frequency excitation-that is, as omega rarr0\omega \rightarrow 0-it follows from equation (13) that Rk//F_(0)rarr1R k / F_{0} \rightarrow 1 or R rarrF_(0)//kR \rightarrow F_{0} / k. At the other extreme, for very high frequency excitation, equation (13) implies that R rarr0R \rightarrow 0 as omega rarr oo\omega \rightarrow \infty. At an intermediate value of omega\omega the amplitude may have a maximum. To find this maximum point, we can differentiate RR with respect to omega\omega and set the result equal to zero. In this way we find that the maximum amplitude occurs when omega=omega_("max ")\omega=\omega_{\text {max }}, where
{:(14)omega_(max)^(2)=omega_(0)^(2)-(gamma^(2))/(2m^(2))=omega_(0)^(2)(1-(gamma^(2))/(2mk)):}\begin{equation*}
\omega_{\max }^{2}=\omega_{0}^{2}-\frac{\gamma^{2}}{2 m^{2}}=\omega_{0}^{2}\left(1-\frac{\gamma^{2}}{2 m k}\right) \tag{14}
\end{equation*}
Note that omega_(max) < omega_(0)\omega_{\max }<\omega_{0} and that omega_(max)\omega_{\max } is close to omega_(0)\omega_{0} when gamma\gamma is small. The maximum value of RR is
{:(15)R_(max)=(F_(0))/(gammaomega_(0)sqrt(1-(gamma^(2)//4mk)))~=(F_(0))/(gammaomega_(0))(1+(gamma^(2))/(8mk)):}\begin{equation*}
R_{\max }=\frac{F_{0}}{\gamma \omega_{0} \sqrt{1-\left(\gamma^{2} / 4 m k\right)}} \cong \frac{F_{0}}{\gamma \omega_{0}}\left(1+\frac{\gamma^{2}}{8 m k}\right) \tag{15}
\end{equation*}
where the last expression is an approximation that is valid when gamma\gamma is small (see Problem 5). If (gamma^(2))/(mk) > 2\frac{\gamma^{2}}{m k}>2, then omega_(max)\omega_{\max } as given by equation (14) is imaginary; in this case the maximum value of RR occurs for omega=0\omega=0, and RR is a monotone decreasing function of omega\omega. Recall that critical damping occurs when (gamma^(2))/(mk)=4\frac{\gamma^{2}}{m k}=4.
For small gamma\gamma it follows from equation (15) that R_(max)~=(F_(0))/(gammaomega_(0))R_{\max } \cong \frac{F_{0}}{\gamma \omega_{0}}. Thus, for lightly damped systems, the amplitude RR of the forced response when omega\omega is near omega_(0)\omega_{0} is quite large even for relatively small external forces, and the smaller the value of gamma\gamma, the more pronounced is this effect. This phenomenon is known as resonance, and it is often an important design consideration. Resonance can be either good or bad, depending on the circumstances. It must be taken very seriously in the design of structures, such as buildings and bridges, where it can produce instabilities that might lead to the catastrophic failure of the structure. On the other hand, resonance can be put to good use in the design of instruments, such as seismographs, that are intended to detect weak periodic incoming signals.
FIGURE 3.8.2 Forced vibration with damping: amplitude of steady-state response versus frequency of driving force for several values of the dimensionless damping parameter Gamma=gamma^(2)//mk\Gamma=\gamma^{2} / m k.
Figure 3.8.2 contains some representative graphs of (Rk)/(F_(0))\frac{R k}{F_{0}} versus (omega)/(omega_(0))\frac{\omega}{\omega_{0}} for several values of Gamma=(gamma^(2))/(mk)\Gamma=\frac{\gamma^{2}}{m k}. We refer to Gamma\Gamma as a damping parameter, as the following examples will explain. The graph corresponding to Gamma=0.015625\Gamma=0.015625 is included because this is the value of Gamma\Gamma that occurs in Example 2 below. Note particularly the sharp peak in the curve corresponding to Gamma=0.015625\Gamma=0.015625 near (omega)/(omega_(0))=1\frac{\omega}{\omega_{0}}=1. The limiting case as Gamma rarr0\Gamma \rightarrow 0 is also shown. It follows from equation (13), or from equations (11) and (12), that R rarr(F_(0))/(m|omega_(0)^(2)-omega^(2)|)R \rightarrow \frac{F_{0}}{m\left|\omega_{0}^{2}-\omega^{2}\right|} as gamma rarr0\gamma \rightarrow 0 and hence (Rk)/(F_(0))\frac{R k}{F_{0}} is asymptotic to the vertical line omega=omega_(0)\omega=\omega_{0}, as shown in the figure. As the damping in the system increases, the peak response gradually diminishes.
Figure 3.8.2 also illustrates the usefulness of dimensionless variables. You can easily verify that each of the quantities (Rk)/(F_(0)),(omega)/(omega_(0))\frac{R k}{F_{0}}, \frac{\omega}{\omega_{0}}, and Gamma\Gamma is dimensionless (see Problem 9d). The importance of this observation is that the number of significant parameters in the problem has been reduced to three rather than the five that appear in equation (8). Thus only one family of curves, of which a few are shown in Figure 3.8.2, is needed to describe the response-versusfrequency behavior of all systems governed by equation (8).
The phase angle delta\delta also depends in an interesting way on omega\omega. For omega\omega near zero, it follows from equations (11) and (12) that cos delta~=1\cos \delta \cong 1 and sin delta~=0\sin \delta \cong 0. Thus delta~=0\delta \cong 0, and the response is nearly in phase with the excitation, meaning that they rise and fall together and, in particular, assume their respective maxima nearly together and their respective minima nearly together.
For omega=omega_(0)\omega=\omega_{0} we find that cos delta=0\cos \delta=0 and sin delta=1\sin \delta=1, so delta=pi//2\delta=\pi / 2. In this case the response lags behind the excitation by pi//2\pi / 2; that is, the peaks of the response occur pi//2\pi / 2 later than the peaks of the excitation, and similarly for the valleys. Finally, for omega\omega very large, we have cos delta~=-1\cos \delta \cong-1 and sin delta~=0\sin \delta \cong 0. Thus delta~=pi\delta \cong \pi so that the response is nearly out of phase with the excitation; this means that the response is minimum when the excitation is maximum, and vice versa. Figure 3.8.3 shows the graphs of delta\delta versus omega//omega_(0)\omega / \omega_{0} for several values of Gamma\Gamma. For small damping, the phase transition from near delta=0\delta=0 to near delta=pi\delta=\pi occurs rather abruptly, whereas for larger values of the damping parameter, the transition takes place more gradually.
FIGURE 3.8.3 Forced vibration with damping: phase of steady-state response versus frequency of driving force for several values of the dimensionless damping parameter Gamma=gamma^(2)//mk\Gamma=\gamma^{2} / m k.
Show plots of the solution for different values of the forcing frequency omega\omega, and compare them with corresponding plots of the forcing function.
Solution:
For this system we have omega_(0)=1\omega_{0}=1 and Gamma=1//64=0.015625\Gamma=1 / 64=0.015625. Its unforced motion was discussed in Example 3 of Section 3.7, and Figure 3.7.7 shows the graph of the solution of the unforced problem. Figures 3.8.4, 3.8.5, and 3.8 .6 show the solution of the forced problem (16) for omega=0.3,omega=1\omega=0.3, \omega=1, and omega=2\omega=2, respectively. The graph of the corresponding forcing function is also shown in each figure. In this example the static displacement, F_(0)//kF_{0} / k, is equal to 3 .
Figure 3.8.4 shows the low frequency case, omega//omega_(0)=0.3\omega / \omega_{0}=0.3. After the initial transient response is substantially damped out, the remaining steady-state response is essentially in phase with the excitation, and the amplitude of the response is somewhat larger than the static displacement. To be specific, R~=3.2939R \cong 3.2939 and delta~=0.041185\delta \cong 0.041185.
The resonant case, omega//omega_(0)=1\omega / \omega_{0}=1, is shown in Figure 3.8.5. Here, the amplitude of the steady-state response is eight times the static displacement, and the figure also shows the predicted phase lag of pi//2\pi / 2 relative to the external force.
The case of comparatively high frequency excitation is shown in Figure 3.8.6. Observe that the amplitude of the steady forced response is approximately one-third the static displacement and that the phase difference between the excitation and the response is approximately pi\pi. More precisely, we find that R~=0.99655R \cong 0.99655 and that delta~=3.0585\delta \cong 3.0585.
FIGURE 3.8.4 A forced vibration with damping; the solution (solid blue) of equation (16) with omega=0.3\omega=0.3 : u^('')+(1)/(8)u^(')+u=3cos(0.3 t)u^{\prime \prime}+\frac{1}{8} u^{\prime}+u=3 \cos (0.3 t), u(0)=2,u^(')(0)=0u(0)=2, u^{\prime}(0)=0. The dashed red curve is the external force: F(t)=3cos(0.3 t)F(t)=3 \cos (0.3 t).
FIGURE 3.8.5 A forced vibration with damping; the solution (solid blue) of equation (16) with omega=1:u^('')+(1)/(8)u^(')+u=3cos t,u(0)=2\omega=1: u^{\prime \prime}+\frac{1}{8} u^{\prime}+u=3 \cos t, u(0)=2, u^(')(0)=0u^{\prime}(0)=0. The dashed red curve is the external force: F(t)=3cos tF(t)=3 \cos t.
Forcing function
Solution
FIGURE 3.8.6 A forced vibration with damping; the solution (solid blue) of equation (16) with omega=2\omega=2 : u^('')+(1)/(8)u^(')+u=3cos(2t),u(0)=2u^{\prime \prime}+\frac{1}{8} u^{\prime}+u=3 \cos (2 t), u(0)=2, u^(')(0)=0u^{\prime}(0)=0. The dashed red curve is the external force: F(t)=3cos(2t)F(t)=3 \cos (2 t).
Forced Vibrations Without Damping. We now assume that gamma=0\gamma=0 in equation (8), thereby obtaining the equation of motion of an undamped forced oscillator,
The form of the general solution of equation (17) is different, depending on whether the forcing frequency omega\omega is different from or equal to the natural frequency omega_(0)=sqrt(k//m)\omega_{0}=\sqrt{k / m} of the unforced system. First consider the case omega!=omega_(0)\omega \neq \omega_{0}; then the general solution of equation (17) is
The constants c_(1)c_{1} and c_(2)c_{2} are determined by the initial conditions. The resulting motion is, in general, the sum of two periodic motions of different frequencies ( omega_(0)\omega_{0} and omega\omega ) and different amplitudes as well.
It is particularly interesting to suppose that the mass is initially at rest so that the initial conditions are u(0)=0u(0)=0 and u^(')(0)=0u^{\prime}(0)=0. Then the energy driving the system comes entirely from the external force, with no contribution from the initial conditions. In this case it turns out that the constants c_(1)c_{1} and c_(2)c_{2} in equation (18) are given by
This is the sum of two periodic functions of different periods but the same amplitude. Making use of the trigonometric identities for cos(A+-B)\cos (A \pm B) with A=(1)/(2)(omega_(0)+omega)tA=\frac{1}{2}\left(\omega_{0}+\omega\right) t and B=(1)/(2)(omega_(0)-omega)tB=\frac{1}{2}\left(\omega_{0}-\omega\right) t, we can write equation (20) in the form
If |omega_(0)-omega|\left|\omega_{0}-\omega\right| is small, then omega_(0)+omega\omega_{0}+\omega is much greater than |omega_(0)-omega|\left|\omega_{0}-\omega\right|. Consequently, sin((1)/(2)(omega_(0)+omega)t)\sin \left(\frac{1}{2}\left(\omega_{0}+\omega\right) t\right) is a rapidly oscillating function compared to sin((1)/(2)(omega_(0)-omega)t)\sin \left(\frac{1}{2}\left(\omega_{0}-\omega\right) t\right). Thus the motion is a rapid oscillation with frequency (1)/(2)(omega_(0)+omega)\frac{1}{2}\left(\omega_{0}+\omega\right) but with a slowly varying sinusoidal amplitude
This type of motion, possessing a periodic variation of amplitude, exhibits what is called a beat. For example, such a phenomenon occurs in acoustics when two tuning forks of nearly equal frequency are excited simultaneously. In this case the periodic variation of amplitude is quite apparent to the unaided ear. In electronics, the variation of the amplitude with time is called amplitude modulation.
In this case omega_(0)=1,omega=0.8\omega_{0}=1, \omega=0.8, and F_(0)=(1)/(2)F_{0}=\frac{1}{2}, so from equation (21) the solution of the given problem is
A graph of this solution is shown in Figure 3.8.7. The amplitude variation has a slow frequency of 0.1 and a corresponding slow period of 2pi//0.1=20 pi2 \pi / 0.1=20 \pi. Note that a half-period of 10 pi10 \pi corresponds to a single cycle of increasing and then decreasing amplitude. The displacement of the spring-mass system oscillates with a relatively fast frequency of 0.9 , which is only slightly less than the natural frequency omega_(0)\omega_{0}.
Now imagine that the forcing frequency omega\omega is increased, say, to omega=0.9\omega=0.9. Then the slow frequency is halved to 0.05 , and the corresponding slow half-period is doubled to 20 pi20 \pi. The multiplier 2.7778 also increases substantially, to 5.263 . However, the fast frequency is only marginally increased, to 0.95 . Can you visualize what happens as omega\omega takes on values closer and closer to the natural frequency omega_(0)=1\omega_{0}=1 ?
FIGURE 3.8.7 A beat; the solution (solid blue) of equation (22): u^('')+u=(1)/(2)cos(0.8 t)u^{\prime \prime}+u=\frac{1}{2} \cos (0.8 t), u(0)=0,u^(')(0)=0u(0)=0, u^{\prime}(0)=0 is u=2.778 sin(0.1 t)sin(0.9 t)u=2.778 \sin (0.1 t) \sin (0.9 t). The dashed red curve is the external force F(t)=(1)/(2)cos(0.8 t)F(t)=\frac{1}{2} \cos (0.8 t).
Now let us return to equation (17) and consider the case of resonance, where omega=omega_(0)\omega=\omega_{0}; that is, the frequency of the forcing function is the same as the natural frequency of the system. Then the nonhomogeneous term F_(0)cos(omega t)F_{0} \cos (\omega t) is a solution of the homogeneous equation. In this case the solution of equation (17) is
{:(24)u=c_(1)cos omega_(0)t+c_(2)sin omega_(0)t+(F_(0))/(2momega_(0))t sin(omega_(0)t):}\begin{equation*}
u=c_{1} \cos \omega_{0} t+c_{2} \sin \omega_{0} t+\frac{F_{0}}{2 m \omega_{0}} t \sin \left(\omega_{0} t\right) \tag{24}
\end{equation*}
The general solution of the differential equation is
u=c_(1)cos t+c_(2)sin t+(t)/(4)sin tu=c_{1} \cos t+c_{2} \sin t+\frac{t}{4} \sin t
and the initial conditions require that c_(1)=c_(2)=0c_{1}=c_{2}=0. Thus the solution of the given initial value problem is
{:(26)u=(t)/(4)sin t:}\begin{equation*}
u=\frac{t}{4} \sin t \tag{26}
\end{equation*}
The graph of the solution is shown in Figure 3.8.8.
FIGURE 3.8.8 Resonance; the solution (solid blue) of equation (25):
u^('')+u=(1)/(2)cos t,u(0)=0,u^(')(0)=0" is "u=(t)/(4)sin tu^{\prime \prime}+u=\frac{1}{2} \cos t, u(0)=0, u^{\prime}(0)=0 \text { is } u=\frac{t}{4} \sin t
Because of the term t sin(omega_(0)t)t \sin \left(\omega_{0} t\right), the solution (24) predicts that the motion will become unbounded as t rarr oot \rightarrow \infty regardless of the values of c_(1)c_{1} and c_(2)c_{2}, and Figure 3.8.8 bears this out. Of course, in reality, unbounded oscillations do not occur, because the spring cannot stretch infinitely far. Moreover, as soon as uu becomes large, the mathematical model on which equation (17) is based is no longer valid, since the assumption that the spring force depends linearly on the displacement requires that uu be small. As we have seen, if damping is included in the model, the predicted motion remains bounded; however, the response to the input function F_(0)cos(omega t)F_{0} \cos (\omega t) may be quite large if the damping is small and omega\omega is close to omega_(0)\omega_{0}.
Problems
In each of Problems 1 through 3, write the given expression as a product of two trigonometric functions of different frequencies.
sin(7t)-sin(6t)\sin (7 t)-\sin (6 t)
cos(pi t)+cos(2pi t)\cos (\pi t)+\cos (2 \pi t)
sin(3t)+sin(4t)\sin (3 t)+\sin (4 t)
A mass of 5 kg stretches a spring 10 cm . The mass is acted on by an external force of 10 sin(t//2)N10 \sin (t / 2) \mathrm{N} (newtons) and moves in a medium that imparts a viscous force of 2 N when the speed of the mass is 4cm//s4 \mathrm{~cm} / \mathrm{s}. If the mass is set in motion from its equilibrium position with an initial velocity of 3cm//s3 \mathrm{~cm} / \mathrm{s}, formulate the initial value problem describing the motion of the mass.
a. Find the solution of the initial value problem in Problem 4.
b. Identify the transient and steady-state parts of the solution.
G c. Plot the graph of the steady-state solution.
N d. If the given external force is replaced by a force of 2cos(omega t)2 \cos (\omega t) of frequency omega\omega, find the value of omega\omega for which the amplitude of the forced response is maximum.
(N) 6. A mass that weighs 8 lb stretches a spring 6 in . The system is acted on by an external force of 8sin(8t)lb8 \sin (8 t) \mathrm{lb}. If the mass is pulled down 3 in and then released, determine the position of the mass at any time. Determine the first four times at which the velocity of the mass is zero.
7. A spring is stretched 6 in by a mass that weighs 8 lb . The mass is attached to a dashpot mechanism that has a damping constant of (1)/(4)lb*s//ft\frac{1}{4} \mathrm{lb} \cdot \mathrm{s} / \mathrm{ft} and is acted on by an external force of 4cos(2t)lb4 \cos (2 t) \mathrm{lb}.
a. Determine the steady-state response of this system.
b. If the given mass is replaced by a mass mm, determine the value of mm for which the amplitude of the steady-state response is maximum.
8. A spring-mass system has a spring constant of 3N//m3 \mathrm{~N} / \mathrm{m}. A mass of 2 kg is attached to the spring, and the motion takes place in a viscous fluid that offers a resistance numerically equal to the magnitude of the instantaneous velocity. If the system is driven by an external force of (3cos(3t)-2sin(3t))N(3 \cos (3 t)-2 \sin (3 t)) \mathrm{N}, determine the steady-state response. Express your answer in the form R cos(omega t-delta)R \cos (\omega t-\delta).
9. In this problem we ask you to supply some of the details in the analysis of a forced damped oscillator.
a. Derive equations (10), (11), and (12) for the steady-state solution of equation (8).
b. Derive the expression in equation (13) for Rk//F_(0)R k / F_{0}.
c. Show that omega_("max ")^(2)\omega_{\text {max }}^{2} and R_("max ")R_{\text {max }} are given by equations (14) and (15), respectively.
d. Verify that Rk//F_(0),omega//omega_(0)R k / F_{0}, \omega / \omega_{0}, and Gamma=gamma^(2)//(mk)\Gamma=\gamma^{2} /(m k) are all dimensionless quantities.
10. Find the velocity of the steady-state response given by equation (10). Then show that the velocity is maximum when omega=omega_(0)\omega=\omega_{0}.
11. Find the solution of the initial value problem
Hint: Treat each time interval separately, and match the solutions in the different intervals by requiring uu and u^(')u^{\prime} to be continuous functions of tt.
12. A series circuit has a capacitor of 0.25 xx10^(-6)F0.25 \times 10^{-6} \mathrm{~F}, a resistor of 5xx10^(3)Omega5 \times 10^{3} \Omega, and an inductor of 1 H . The initial charge on the capacitor is zero. If a 12 V battery is connected to the circuit and the circuit is closed at t=0t=0, determine the charge on the capacitor at t=0.001st=0.001 \mathrm{~s}, at t=0.01st=0.01 \mathrm{~s}, and at any time tt. Also determine the limiting charge as t rarr oot \rightarrow \infty.
(N) 13. Consider the forced but undamped system described by the initial value problem
a. Find the solution u(t)u(t) for omega!=1\omega \neq 1.
G b. Plot the solution u(t)u(t) versus tt for omega=0.7,omega=0.8\omega=0.7, \omega=0.8, and omega=0.9\omega=0.9. Describe how the response u(t)u(t) changes as omega\omega varies in this interval. What happens as omega\omega takes on values closer and closer
to 1 ? Note that the natural frequency of the unforced system is omega_(0)=1\omega_{0}=1.
14. Consider the vibrating system described by the initial value problem
a. Find the solution for omega!=1\omega \neq 1.
(G) b. Plot the solution u(t)u(t) versus tt for omega=0.7,omega=0.8\omega=0.7, \omega=0.8, and omega=0.9\omega=0.9. Compare the results with those of Problem 13; that is, describe the effect of the nonzero initial conditions.
(G) 15. For the initial value problem in Problem 13, plot u^(')u^{\prime} versus uu for omega=0.7,omega=0.8\omega=0.7, \omega=0.8, and omega=0.9\omega=0.9. (Recall that such a plot is called a phase plot.) Use a tt interval that is long enough so that the phase plot appears as a closed curve. Mark your curve with arrows to show the direction in which it is traversed as tt increases.
Problems 16 through 18 deal with the initial value problem
In each of these problems:
(G) a. Plot the given forcing function F(t)F(t) versus tt, and also plot the solution u(t)u(t) versus tt on the same set of axes. Use a tt interval that is long enough so the initial transients are substantially eliminated. Observe the relation between the amplitude and phase of the forcing term and the amplitude and phase of the response. Note that omega_(0)=sqrt(k//m)=2\omega_{0}=\sqrt{k / m}=2.
(G) Draw the phase plot of the solution; that is, plot u^(')u^{\prime} versus uu.
16. quad F(t)=3cos(t//4)\quad F(t)=3 \cos (t / 4)
17. F(t)=3cos(2t)F(t)=3 \cos (2 t)
18. F(t)=3cos(6t)F(t)=3 \cos (6 t)
G 19. A spring-mass system with a hardening spring (Problem 24 of Section 3.7) is acted on by a periodic external force. In the absence of damping, suppose that the displacement of the mass satisfies the initial value problem
a. Let omega=1\omega=1 and plot a computer-generated solution of the given problem. Does the system exhibit a beat?
b. Plot the solution for several values of omega\omega between 1//21 / 2 and 2 . Describe how the solution changes as omega\omega increases.
References
Coddington, E. A., An Introduction to Ordinary Differential Equations (Englewood Cliffs, NJ: Prentice-Hall, 1961; New York: Dover, 1989).
There are many books on mechanical vibrations and electric circuits. One that deals with both is
Close, C. M., and Frederick, D. K., Modeling and Analysis of Dynamic Systems (3rd ed.) (New York: Wiley, 2001).
A classic book on mechanical vibrations is
Den Hartog, J. P., Mechanical Vibrations (4th ed.) (New York: McGraw-Hill, 1956; New York: Dover, 1985).
An intermediate-level book is
Thomson, W. T., Theory of Vibrations with Applications (5th ed.) (Englewood Cliffs, NJ: Prentice-Hall, 1997).
An elementary book on electric circuits is
Bobrow, L. S., Elementary Linear Circuit Analysis (New York: Oxford University Press, 1996).
^(1){ }^{1} There is a corresponding treatment of higher-order linear equations in Chapter 4. If you wish, you may read the appropriate parts of Chapter 4 in parallel with Chapter 3.
^(2){ }^{2} A proof of Theorem 3.2.1 can be found, for example, in Chapter 6, Section 8 of the book by Coddington listed in the references at the end of this chapter.
^(4){ }^{4} The result in Theorem 3.2.7 was derived by the Norwegian mathematician Niels Henrik Abel (1802-1829) in 1827 and is known as Abel's formula. Abel also showed that there is no general formula for solving a quintic, or fifth degree, polynomial equation in terms of explicit algebraic operations on the coefficients, thereby resolving a question that had remained unanswered since the sixteenth century. His greatest contributions, however, were in analysis, particularly in the study of elliptic functions. Unfortunately, his work was not widely noticed until after his death. The distinguished French mathematician Legendre called it a "monument more lasting than bronze."
^(5){ }^{5} Recall from calculus that the reordering of terms in the right-hand side of equation (9) is allowed because the series converges absolutely for all -oo < t < oo-\infty<t<\infty.
^(7){ }^{7} Jean d'Alembert (1717-1783), a French mathematician, was a contemporary of Euler and Daniel Bernoulli and is known primarily for his work in mechanics and differential equations. D'Alembert's principle in mechanics and d'Alembert's paradox in hydrodynamics are named for him, and the wave equation first appeared in his paper on vibrating strings in 1747 . In his later years he devoted himself primarily to philosophy and to his duties as science editor of Diderot's Encyclopédie.
^(8){ }^{8} R. S. Luthar, "Another Approach to a Standard Differential Equation," Two Year College Mathematics Journal 10 (1979), pp. 200-201. Also see D. C. Sandell and F. M. Stein, "Factorization of Operators of Second-Order Linear Homogeneous Ordinary Differential Equations," Two Year College Mathematics Journal 8 (1977), pp. 132-141, for a more general discussion of factoring operators.
^(9){ }^{9} An alternate, and more mathematically appealing, derivation of the second condition can be found in Problems 17 to 19 in Section 7.9.
^(10){ }^{10} Robert Hooke (1635-1703) was an English scientist with wide-ranging interests. His most important book, Micrographia, was published in 1665 and described a variety of microscopical observations. Hooke first published his law of elastic behavior in 1676 as ceiiinosssttuv; in 1678 he gave the interpretation ut tensio sic vis, which means, roughly, "as the force so is the displacement."
^(11){ }^{11} Gustav Kirchhoff (1824-1887) was a German physicist and professor at Breslau, Heidelberg, and Berlin. He formulated the basic laws of electric circuits about 1845 while still a student at Albertus University in his native Königsberg. In 1857 he discovered that an electric current in a resistanceless wire travels at the speed of light. He is also famous for fundamental work in electromagnetic absorption and emission and was one of the founders of spectroscopy.
^(12){ }^{12} Archimedes (287-212 BCE) was the foremost of the ancient Greek mathematicians. He lived in Syracuse on the island of Sicily. His most notable discoveries were in geometry, but he also made important contributions to hydrostatics and other branches of mechanics. His method of exhaustion is a precursor of the integral calculus developed by Newton and Leibniz almost two millennia later. He died at the hands of a Roman soldier during the Second Punic War.
