5.1 Review of Power Series
C H A P T E R
3
Linear 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 these methods are 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. Another reason to study second order linear 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. As an example, we discuss the oscillations of some basic mechanical and electrical systems at the end of the chapter.
3.1 Homogeneous Equations with Constant Coefficients
A second order ordinary differential equation has the form
d2 y = f t,y, dy ,
(1)
dt2
dt
where f is some given function. Usually, we will denote the independent variable by t
since time is often the independent variable in physical problems, but sometimes we
129
Chapter 3. Second Order Linear Equations will use x instead. We will use y, or occasionally some other letter, to designate the dependent variable. Equation (1) is said to be linear if the function f has the form
f
t, y, dy = g(t) − p(t)dy − q(t)y, (2)
dt
dt
that is, if f is linear in y and y. In Eq. (2) g, p, and q are specified functions of the independent variable t but do not depend on y. In this case we usually rewrite as y + p(t)y + q(t)y = g(t), (3)
where the primes denote differentiation with respect to t. Instead of Eq. (3), we often see the equation P(t)y + Q(t)y + R(t)y = G(t).
(4)
Of course, if P(t) = 0, we can divide Eq. (4) by P(t) and thereby obtain Eq. (3) with p(t) = Q(t) ,q(t) = R(t) ,g(t) = G(t) .
(5) P(t)
P(t)
P(t) In discussing Eq. (3) and in trying to solve it, we will restrict ourselves to intervals in which p, q, and g are continuous functions.1 If Eq. (1) is not of the form (3) or (4), then it is called 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. In addition, there are two special types of second order nonlinear equations that can be solved by a change of variables that reduces them to first order equations. This procedure is outlined in Problems 28 through 43.
An initial value problem consists of a differential equation such as Eq. (1), (3), or (4) together with a pair of initial conditions y(t ) = y ,y(t ) = y ,
(6)
0
0
0
0
where y and y are given numbers. Observe that the initial conditions for a second
0
0
order equation prescribe not only a particular point (t , y ) through which the graph of
0
0
the solution must pass, but also the slope y of the graph at that point. It is reasonable
0
to expect that two initial conditions are needed for a second order 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 equation is said to be homogeneous if the term g(t) in Eq. (3), or the term G(t) in Eq. (4), is zero for all t. Otherwise, the equation is called As a result, the term g(t), or G(t), is sometimes called the nonhomogeneous term. We begin our discussion with homogeneous equations, which we will write in the form P(t)y + Q(t)y + R(t)y = 0.
(7)
1There 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.
3.1Homogeneous Equations with Constant Coefficients Later, in Sections 3.6 and 3.7, 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 func tions P, Q, and R are constants. In this case, Eq. (7) becomes ay + by + cy = 0,
(8)
where a, b, and c are given constants. It turns out that Eq. (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 Eq. (7) if the coefficients are not constants, and a treatment of that case is deferred until Chapter 5.
Before taking up Eq. (8), let us first gain some experience by looking at a simple, but typical, example. Consider the equation y − y = 0, (9)
which is just Eq. (8) with a = 1, b = 0, and c = −1. In words, Eq. (9) says that we seek a function with the property that the second derivative of the function is the same as the function itself. A little thought will probably produce at least one wellknown function from calculus with this property, namely, y (t) = et , the exponential
1
function. A little more thought may also produce a second function, y (t) = e−t . Some
2
further experimentation reveals that constant multiples of these two solutions are also solutions. For example, the functions 2et and 5e−t also satisfy Eq. (9), as you can verify by calculating their second derivatives. In the same way, the functions c y (t) = c et 1 1
1
and c y (t) = c e−t satisfy the differential equation (9) for all values of the constants 2 2
2
c and c . Next, it is of paramount importance to notice that any sum of solutions of
1
2
Eq. (9) is also a solution. In particular, since c y (t) and c y (t) are solutions of Eq. (9), 1 1
2 2 so is the function y = c y (t) + c y (t) = c et + c e−t
(10)
1 1
2 2
1
2
for any values of c and c . Again, this can be verified by calculating the second
1
2
derivative y from Eq. (10). We have y = c et − c e−t and y = c et + c e−t ; thus
1
2
1
2
y is the same as y, and Eq. (9) is satisfied.
Let us summarize what we have done so far in this example. Once we notice that the functions y (t) = et and y (t) = e−t are solutions of Eq. (9), it follows that the general
1
2
linear combination (10) of these functions is also a solution. Since the coefficients c1 and c in Eq. (10) are arbitrary, this expression represents a doubly infinite family of
2
solutions of the differential equation (9).
It is now possible to consider how to pick out a particular member of this infinite family of solutions that also satisfies a given set of initial conditions. For example, suppose that we want the solution of Eq. (9) that also satisfies the initial conditions y(0) = 2,y(0) = −1.
(11)
In other words, we seek the solution that passes through the point (0, 2) and at that point has the slope −1. First, we set t = 0 and y = 2 in Eq. (10); this gives the equation c + c = 2.
(12)
1
2
Chapter 3. Second Order Linear Equations Next, we differentiate Eq. (10) with the result that y = c et − c e−t .
1
2
Then, setting t = 0 and y = −1, we obtain c − c = −1.
(13)
1
2
By solving Eqs. (12) and (13) simultaneously for c and c we find that
1
2
c = 1 ,c = 3 .
(14)
1
2
2
2
Finally, inserting these values in Eq. (10), we obtain y = 1 et + 3 e−t , (15)
2
2
the solution of the initial value problem consisting of the differential equation (9) and the initial conditions (11).
We now return to the more general equation (8), ay + by + cy = 0, which has arbitrary (real) constant coefficients. Based on our experience with let us also seek exponential solutions of Eq. (8). Thus we suppose that y = ert , where r is a parameter to be determined. Then it follows that y = rert and y = r2ert . By substituting these expressions for y, y, and y in Eq. (8), we obtain (ar2 + br + c)ert = 0, or, since ert = 0,ar 2 + br + c = 0.
(16)
Equation (16) is called the the differential equation (8).
Its significance lies in the fact that if r is a root of the polynomial equation (16), then y = ert is a solution of the differential equation (8). Since Eq. (16) is a quadratic equation with real coefficients, it has two roots, which may be real and different, real but repeated, or complex conjugates. We consider the first case here, and the latter two cases in Sections 3.4 and 3.5.
Assuming that the roots of the characteristic equation (16) are real and different, let them be denoted by r and r , where r = r . Then y (t) = er t
t
1
and y (t) = er2 are
1
2
1
2
1
2
two solutions of Eq. (8). Just as in the preceding example, it now follows that y = c y (t) + c y (t) = c er t
t
1
+ c er2
(17)
1 1
2 2
1
2
is also a solution of Eq. (8). To verify that this is so, we can differentiate the expression in Eq. (17); hence y = c r er t
t
1
+ c r er2
(18)
1 1
2 2
and y = c r2er t
t
1
+ c r2er2 .
(19)
1 1
2 2
Substituting these expressions for y, y, and y in Eq. (8) and rearranging terms, we obtain ay + by + cy = c (ar2 + br + c)er t
t
1
+ c (ar2 + br + c)er2 .
(20)
1
1
1
2
2
2
3.1Homogeneous Equations with Constant Coefficients
The quantity in each of the parentheses on the right side of Eq. (20) is zero because r1 and r are roots of Eq. (16); therefore, y as given by Eq. (17) is indeed a solution of
2
Eq. (8), as we wished to verify.
Now suppose that we want to find the particular member of the family of solutions (17) that satisfies the initial conditions (6), y(t ) = y ,y(t ) = y .
0
0
0
0
By substituting t = t and y = y in Eq. (17), we obtain
0
0
c er t
t 1 0 + c er2 0 = y .
(21)
1
2
0
Similarly, setting t = t and y = y in Eq. (18) gives
0
0
c r er t
t 1 0 + c r er2 0 = y .
(22)
1 1
2 2
0
On solving Eqs. (21) and (22) simultaneously for c and c , we find that
1
2
y − y ry r − yc = 0 0 2 e−r t 0 1
0
t
1 0 ,c = e−r2 0 .
(23)
1
r − r
2
r − r
1
2
1
2
Thus, no matter what initial conditions are assigned, that is, regardless of the values of t , y , and y in Eqs. (6), it is always possible to determine c and c so that the
0
0
0
1
2
initial conditions are satisfied; moreover, there is only one possible choice of c and c
1
2
for each set of initial conditions. With the values of c and c given by Eq. (23), the
1
2
expression (17) is the solution of the initial value problem ay + by + cy = 0,y(t ) = y ,y(t ) = y .
(24)
0
0
0
0
It is possible to show, on the basis of the fundamental theorem cited in the next section, that all solutions of Eq. (8) are included in the expression (17), at least for the case in which the roots of Eq. (16) are real and different. Therefore, we call Eq. (17) the general solution of Eq. (8). The fact that any possible initial conditions can be satisfied by the proper choice of the constants in Eq. (17) makes more plausible the idea that this expression does include all solutions of Eq. (8).
Find the general solution of E X A M P L E
1
y + 5y + 6y = 0.
(25)
We assume that y = ert , and it then follows that r must be a root of the characteristic equation r 2 + 5r + 6 = (r + 2)(r + 3) = 0.
Thus the possible values of r are r = −2 and r = −3; the general solution of
1
2
Eq. (25) is y = c e−2t + c e−3t .
(26)
1
2
Find the solution of the initial value problem E X A M P L E
2
y + 5y + 6y = 0,y(0) = 2,y(0) = 3.
(27)
Chapter 3. Second Order Linear Equations
The general solution of the differential equation was found in Example 1 and is given by Eq. (26). To satisfy the first initial condition we set t = 0 and y = 2 in thus c and c must satisfy
1
2
c + c = 2.
(28)
1
2
To use the second initial condition we must first differentiate Eq. (26). This gives y = −2c e−2t − 3c e−3t . Then, setting t = 0 and y = 3, we obtain
1
2
−2c − 3c = 3.
(29)
1
2
By solving Eqs. (28) and (29) we find that c = 9 and c = −7. Using these values in
1
2
the expression (26), we obtain the solution y = 9e−2t − 7e−3t
(30)
of the initial value problem (27). The graph of the solution is shown in Figure 3.1.1.
y
2
y = 9e–2t – 7e–3t
1
0.5
1
1.5
2
t
FIGURE 3.1.1
Solution of y + 5y + 6y = 0, y(0) = 2, y(0) = 3.
Find the solution of the initial value problem E X A M P L E
3
4y − 8y + 3y = 0,y(0) = 2,y(0) = 1 .
(31)
2
If y = ert , then the characteristic equation is 4r 2 − 8r + 3 = 0 and its roots are r = 3/2 and r = 1/2. Therefore the general solution of the differential equation is y = c e3t/2 + c et/2.
(32)
1
2
Applying the initial conditions, we obtain the following two equations for c and c :
1
2
c + c = 2, 3 c + 1c = 1 .
1
2
2 1
2 2
2
The solution of these equations is c = − 1 , c = 5 , and the solution of the initial value
1
2
2
2
problem (31) is y = − 1 e3t/2 + 5 et/2.
(33)
2
2
Figure 3.1.2 shows the graph of the solution.
3.1Homogeneous Equations with Constant Coefficients
y
2
y = – 1 e3t/2 + 5 et/2
2
2
1
0.5
1
1.5
2
t
–1
FIGURE 3.1.2
Solution of 4y − 8y + 3y = 0, y(0) = 2, y(0) = 0.5.
The solution (30) of the initial value problem (27) initially increases (because its initial E X A M P L E slope is positive) but eventually approaches zero (because both terms involve negative
4
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.
One can estimate the coordinates of the maximum point 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 (30), y = 9e−2t − 7e−3t , with respect to t we obtain y = −18e−2t + 21e−3t .
(34)
Setting y equal to zero and multiplying by e3t , we find that the critical value t satisfies
cet = 7/6; hence t = ln(7/6) ∼ = 0.15415.
(35)
c
The corresponding maximum value y is given by
M
y
= 9e−2t ∼
c − 7e−3tc = 108 = 2.20408.
(36) M
49
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 26 you are asked to determine how the coordinates of the maximum point depend on the initial slope.
Returning to the equation ay + by + cy = 0 with arbitrary coefficients, recall that when r = r , its general solution (17) is the sum of two exponential functions.
1
2
Therefore the solution has a relatively simple geometrical behavior: as t increases, the magnitude of the solution either tends to zero (when both exponents are negative) or else grows rapidly (when at least one exponent is positive). These two cases are illustrated by the solutions of Examples 2 and 3, which are shown in and 3.1.2, respectively. 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.
Chapter 3. Second Order Linear Equations
PROBLEMS
In each of Problems 1 through 8 find the general solution of the given differential equation.
1
2
1
2
䉴 19. Find the solution of the initial value problem
4
4
20. Find the solution of the initial value problem
2
䉴 25. Consider the initial value problem
(a) Solve the initial value problem.
0
0
the solution in this case.
䉴 26. Consider the initial value problem (see Example 4)
(a) Solve the initial value problem.
3.2Fundamental Solutions of Linear Homogeneous Equations
m
m
m
m
m
Equations with the Dependent Variable Missing.
For a second order differential equation
Equations with the Independent Variable Missing.
If a second order differential equation
3.2 Fundamental Solutions of Linear Homogeneous Equations
In the preceding section we showed how to solve some differential equations of the form ay + by + cy = 0, where a, b, and c 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 Chapter 3. Second Order Linear Equations turn, this understanding will assist us in finding the solutions of other problems that we will encounter later.
In developing the theory of linear differential equations, it is helpful to introduce a differential operator notation. Let p and q be continuous functions on an open interval I , that is, for α < t < β. The cases α = −∞, or β = ∞, or both, are included. Then, for any function φ that is twice differentiable on I , we define the differential operator L
by the equation L[φ] = φ + pφ + qφ.
(1)
Note that L[φ] is a function on I . The value of L[φ] at a point t is L[φ](t) = φ(t) + p(t)φ(t) + q(t)φ(t).
For example, if p(t) = t2, q(t) = 1 + t, and φ(t) = sin 3t, then L[φ](t) = (sin 3t) + t2(sin 3t) + (1 + t) sin 3t = −9 sin 3t + 3t2 cos 3t + (1 + t) sin 3t.
The operator L is often written as L = D2 + p D + q, where D is the derivative operator.
In this section we study the second order linear homogeneous equation L[φ](t) = 0.
Since it is customary to use the symbol y to denote φ(t), we will usually write this equation in the form L[y] = y + p(t)y + q(t)y = 0.
(2)
With Eq. (2) we associate a set of initial conditions y(t ) = y ,y(t ) = y ,
(3)
0
0
0
0
where t is any point in the interval I , and y and y are given real numbers. We
0
0
0
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
Consider the initial value problem y + p(t)y + q(t)y = g(t),y(t ) = y ,y(t ) = y ,
(4)
0
0
0
0
where p, q, and g are continuous on an open interval I . Then there is exactly one solution y = φ(t) of this problem, and the solution exists throughout the interval I .
We emphasize that the theorem says three things:
1.
The initial value problem has a solution; in other words, a solution exists.
2.
The initial value problem has only one solution; that is, the solution is unique.
3.2Fundamental Solutions of Linear Homogeneous Equations3.
The solution φ is defined throughout the interval I where the coefficients are continuous and is at least twice differentiable there.
For some problems some of these assertions are easy to prove. For example, we found in Section 3.1 that the initial value problem y − y = 0,y(0) = 2,y(0) = −1 (5)
has the solution y = 1 et + 3 e−t .
(6)
2
2
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 (−∞, ∞) 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 Eq. (6). Nevertheless, Theorem 3.2.1 states that this solution is indeed the only solution of the initial value problem (5).
However, 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 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 We will, however, accept Theorem 3.2.1 as true and make use of it whenever necessary.
Find the longest interval in which the solution of the initial value problem E X A M P L E
1
(t2 − 3t)y + ty − (t + 3)y = 0,y(1) = 2,y(1) = 1 is certain to exist.
If the given differential equation is written in the form of Eq. (4), then p(t) = 1/(t − 3), q(t) = −(t + 3)/t (t − 3), and g(t) = 0. The only points of discontinuity of the coefficients are t = 0 and t = 3. Therefore, the longest open interval, containing the initial point t = 1, in which all the coefficients are continuous is 0 < t < 3. Thus, this is the longest interval in which Theorem 3.2.1 guarantees that the solution exists.
Find the unique solution of the initial value problem E X A M P L E
2
y + p(t)y + q(t)y = 0,y(t ) = 0,y(t ) = 0, 0
0
where p and q are continuous in an open interval I containing t .
0
The function y = φ(t) = 0 for all t in I 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.
2A proof of Theorem 3.2.1 may be found, for example, in Chapter 6, Section 8 of the book by Coddington listed in the references.
Chapter 3. Second Order Linear Equations
Let us now assume that y and y are two solutions of Eq. (2); in other words,
1
2
L[y ] = y + py + qy = 0, (7)
1
1
1
1
and similarly for y . Then, just as in the examples in Section 3.1, we can generate
2
more solutions by forming linear combinations of y and y . We state this result as a
1
2
theorem.
Theorem 3.2.2 (Principle of Superposition)
If y and y are two solutions of the differential
1
2
equation (2), L[y] = y + p(t)y + q(t)y = 0, then the linear combination c y + c y is also a solution for any values of the 1 1
2 2 constants c and c .
1
2
A special case of Theorem 3.2.2 occurs if either c or c is zero. Then we conclude
1
2
that any multiple of a solution of Eq. (2) is also a solution.
To prove Theorem 3.2.2 we need only substitute y = c y (t) + c y (t)
(8)
1 1
2 2 for y in Eq. (2). The result is L[c y + c y ] = [c y + c y ] + p[c y + c y ] + q[c y + c y ] 1 1
2 2
1 1
2 2
1 1
2 2
1 1
2 2 = c y + c y + c py + c py + c qy + c qy 1 1
2 2
1
1
2
2
1
1
2
2
= c [y + py + qy ] + c [y + py + qy ]
1
1
1
1
2
2
2
2
= c L[y ] + c L[y ].
1
1
2
2
Since L[y ] = 0 and L[y ] = 0, it follows that L[c y + c y ] = 0 also. Therefore,
1
2
1 1
2 2 regardless of the values of c and c , y as given by Eq. (8) does satisfy the differential
1
2
equation (2) and the proof of Theorem 3.2.2 is complete.
Theorem 3.2.2 states that, beginning with only two solutions of Eq. (2), we can construct a doubly infinite family of solutions by means of Eq. (8). The next question is whether all solutions of Eq. (2) are included in Eq. (8), or whether there may be other solutions of a different form. We begin to address this question by examining whether the constants c and c in Eq. (8) can be chosen so as to satisfy the initial conditions
1
2 (3). These initial conditions require c and c to satisfy the equations
1
2
c y (t ) + c y (t ) = y , 1 1
0
2 2
0
0
(9)
c y (t ) + c y (t ) = y .
1 1
0
2 2
0
0
Upon solving Eqs. (9) for c and c , we find that
1
2
y y (t ) − y y (t ) −y y (t ) + y y (t )c = 0 2
0
0 2
0
,
c = 0 1
0
0 1
0
,
(10)
1
y (t )y (t ) − y (t )y (t )
2
y (t )y (t ) − y (t )y (t )
1
0
2
0
1
0
2
0
1
0
2
0
1
0
2
0
3.2Fundamental Solutions of Linear Homogeneous Equations or, in terms of determinants,
yy (t ) y (t )y
0
2
0
1 0
0
yy (t ) y(t )y
c =
0
2
0
,
c =
1
0
0
.
(11)
1
2
y (t )y (t ) y (t )y (t ) 1 0
2
0
1 0
2
0
y (t )y (t ) y(t )y (t )
1
0
2
0
1
0
2
0
With these values for c and c the expression (8) satisfies the initial conditions (3) as
1
2
well as the differential equation (2).
In order for the expressions for c and c in Eqs. (10) or (11) to make sense, it is
1
2
necessary that the denominators be nonzero. For both c and c the denominator is the
1
2
same, namely, the determinant
y (t )y (t )
W = 1 0
2
0
(t )y (t ) − y (t )y (t ).
(12)
y (t )y (t ) = y1 0 2 0
1
0
2
0
1
0
2
0
The determinant W is called the Wronskian3 or simply the of the solutions y and y . Sometimes we use the more extended notation W (y , y )(t )
1
2
1
2
0
to stand for the expression on the right side of Eq. (12), thereby emphasizing that the Wronskian depends on the functions y and y , and that it is evaluated at the point t .
1
2
0
The preceding argument suffices to establish the following result.
Theorem 3.2.3 Suppose that y and y are two solutions of Eq. (2),
1
2
L[y] = y + p(t)y + q(t)y = 0, and that the Wronskian W = y y − y y 1 2
1 2 is not zero at the point t where the initial conditions (3),
0
y(t ) = y ,y(t ) = y ,
0
0
0
0
are assigned. Then there is a choice of the constants c , c for which y = c y (t) +
1
2
1 1 c y (t) satisfies the differential equation (2) and the initial conditions (3).
2 2
In Example 1 of Section 3.1 we found that y (t) = e−2t and y (t) = e−3t are solutions
1
2
E X A M P L E of the differential equation
3
y + 5y + 6y = 0.
Find the Wronskian of y and y .
1
2
The Wronskian of these two functions is
e−2t
e−3t
W =
−2e−2t
−3e−3t = −e−5t.
3 Wronskian determinants are named for Jo´sef Maria Hoe¨ne´-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.
Chapter 3. Second Order Linear Equations Since W is nonzero for all values of t, the functions y and y can be used to construct
1
2
solutions of the given differential equation, together with initial conditions prescribed at any value of t. One such initial value problem was solved in Example 2 of Section 3.1.
The next theorem justifies the term “general solution” that we introduced in Section 3.1 for the linear combination c y + c y .
1 1
2 2
Theorem 3.2.4
If y and y are two solutions of the differential equation (2),
1
2
L[y] = y + p(t)y + q(t)y = 0, and if there is a point t where the Wronskian of y and y is nonzero, then the family
0
1
2
of solutions y = c y (t) + c y (t) 1 1
2 2 with arbitrary coefficients c and c includes every solution of Eq. (2).
1
2
Let φ be any solution of Eq. (2). To prove the theorem we must show that φ is included in the linear combination c y + c y ; that is, for some choice of the constants c and 1 1
2 2
1
c , the linear combination is equal to φ. Let t be a point where the Wronskian of y
2
0
1
and y is nonzero. Then evaluate φ and φ at this point and call these values y and y ,
2
0
0
respectively; thus y = φ(t ),y = φ(t ).
0
0
0
0
Next, consider the initial value problem y + p(t)y + q(t)y = 0,y(t ) = y ,y(t ) = y .
(13)
0
0
0
0
The function φ is certainly a solution of this initial value problem. On the other hand, since W (y ,y )(t ) is nonzero, it is possible (by Theorem 3.2.3) to choose c and c so
1
2
0
1
2
that y = c y (t) + c y (t) is also a solution of the initial value problem (13). In fact, 1 1
2 2 the proper values of c and c are given by Eqs. (10) or (11). The uniqueness part of
1
2
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 and c ,
1
2
φ(t) = c y (t) + c y (t), 1 1
2 2 and therefore φ is included in the family of functions of c y + c y . Finally, since 1 1
2 2 φ is an arbitrary solution of Eq. (2), it follows that every solution of this equation is included in this family. This completes the proof of Theorem 3.2.4.
Theorem 3.2.4 states that, as long as the Wronskian of y and y is not everywhere
1
2
zero, the linear combination c y + c y contains all solutions of Eq. (2). It is therefore 1 1
2 2 natural (and we have already done this in the preceding section) to call the expression y = c y (t) + c y (t) 1 1
2 2 with arbitrary constant coefficients the Eq. (2). The solutions y1 and y , with a nonzero Wronskian, are said to form a fundamental set of solutions of
2
Eq. (2).
3.2Fundamental Solutions of Linear Homogeneous Equations
To 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. Alternatively, the following example includes all those mentioned in Section 3.1, as well as many other problems of a similar type.
Suppose that y (t) = er t
t
1
and y (t) = er2 are two solutions of an equation of the form
1
2
E X A M P L E (2). Show that they form a fundamental set of solutions if r = r .
1
2
4
We calculate the Wronskian of y and y :
1
2
er tt
1
er2
W =
− r ) exp[(r + r )t].
r er t
t
1
r er2 = (r2
1
1
2
1
2
Since the exponential function is never zero, and since r − r = 0 by the statement of
2
1
the problem, it follows that W is nonzero for every value of t. Consequently, y and y
1
2
form a fundamental set of solutions.
Show that y (t) = t1/2 and y (t) = t−1 form a fundamental set of solutions of
1
2
E X A M P L E
5
2t2 y + 3t y − y = 0,t > 0.
(14)
We will show in Section 5.5 how to solve Eq. (14); see also Problem 38 in Section 3.4.
However, at this stage we can verify by direct substitution that y and y are solutions
1
2
of the differential equation. Since y (t) = 1 t−1/2 and y(t) = − 1 t−3/2, we have
1
2
1
4
2t2(− 1 t−3/2) + 3t ( 1 t−1/2) − t1/2 = (− 1 + 3 − 1)t1/2 = 0.
4
2
2
2
Similarly, y (t) = −t−2 and y(t) = 2t−3, so
2
2
2t2(2t−3) + 3t (−t−2) − t−1 = (4 − 3 − 1)t−1 = 0.
Next we calculate the Wronskian W of y and y :
1
2
t1/2
t−1 W =
1
t−3/2.
(15)
t−1/2
−t−2 = −32
2
Since W = 0 for t > 0, we conclude that y and y form a fundamental set of solutions
1
2
there.
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 may arise as to whether or not a differential equation of the form (2) always has a fundamental set of solutions. The following theorem provides an affirmative answer to this question.
Chapter 3. Second Order Linear Equations
Theorem 3.2.5
Consider the differential equation (2), L[y] = y + p(t)y + q(t)y = 0, whose coefficients p and q are continuous on some open interval I . Choose some point t in I . Let y be the solution of Eq. (2) that also satisfies the initial conditions
0
1
y(t ) = 1,y(t ) = 0, 0
0
and let y be the solution of Eq. (2) that satisfies the initial conditions
2
y(t ) = 0,y(t ) = 1.
0
0
Then y and y form a fundamental set of solutions of Eq. (2).
1
2
First observe that the existence of the functions y and y is assured by the existence
1
2
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
y (t )y (t )
1
0
W (y ,y )(t ) =
1
0
2
0
1
2
0
y(t )y (t ) = 0 1 = 1.
1
0
2
0
Since their Wronskian is not zero at the point t , the functions y and y do form a
0
1
2
fundamental set of solutions, thus completing the proof of Theorem 3.2.5.
Note that the 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 solve the specified initial value problems so as to find the functions y and y indicated in the theorem. Nevertheless, it may be
1
2
reassuring to know that a fundamental set of solutions always exists.
Find the fundamental set of solutions specified by Theorem 3.2.5 for the differential E X A M P L E
equation
6
y − y = 0, (16)
using the initial point t = 0.
0
In Section 3.1 we noted that two solutions of Eq. (16) are y (t) = et and y (t) =
1
2
e−t . The Wronskian of these solutions is W (y ,y )(t) = −2 = 0, so they form a
1
2
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 = 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 (t) the solution of
3
Eq. (16) that satisfies the initial conditions y(0) = 1,y(0) = 0.
(17)
The general solution of Eq. (16) is y = c et + c e−t ,
(18)
1
2
and the initial conditions (17) are satisfied if c = 1/2 and c = 1/2. Thus
1
2
y (t) = 1 et + 1 e−t = cosh t.
3
2
2
3.2Fundamental Solutions of Linear Homogeneous Equations
Similarly, if y (t) satisfies the initial conditions
4
y(0) = 0,y(0) = 1, (19)
then y (t) = 1 et − 1 e−t = sinh t.
4
2
2
Since the Wronskian of y and y is
3
4
W (y ,y )(t) = cosh2 t − sinh2 t = 1, 3
4
these functions also form a fundamental set of solutions, as stated by Therefore, the general solution of Eq. (16) can be written as y = k cosh t + k sinh t,
(20)
1
2
as well as in the form (18). We have used k and k for the arbitrary constants in
1
2
Eq. (20) because they are not the same as the constants c and c in Eq. (18). One
1
2
purpose of this example is to make clear that a given differential equation has more than one fundamental set of solutions; indeed, it has infinitely many. As a rule, you should choose the set that is most convenient.
We can summarize the discussion in this section as follows. To find the general solution of the differential equation y + p(t)y + q(t)y = 0,α < t < β, we must first find two functions y and y that satisfy the differential equation in
1
2
α < t < β. Then we must make sure that there is a point in the interval where the Wronskian W of y and y is nonzero. Under these circumstances y and y form a
1
2
1
2
fundamental set of solutions and the general solution is y = c y (t) + c y (t), 1 1
2 2 where c and c are arbitrary constants. If initial conditions are prescribed at a point in
1
2
α < t < β where W = 0, then c and c can be chosen so as to satisfy these conditions.
1
2
PROBLEMS
In each of Problems 1 through 6 find the Wronskian of the given pair of functions.
xex
Chapter 3. Second Order Linear Equations
1
2
1
2
1
2
1
2
1
2
Explain why this result does not contradict Theorem 3.2.2.
0
0
1
2
differential equation. Do they constitute a fundamental set of solutions?
1
2
1
2
1
2
1
2
If a second order linear homogeneous equation is not exact, it can
3
Linear 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 these methods are 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. Another reason to study second order linear 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. As an example, we discuss the oscillations of some basic mechanical and electrical systems at the end of the chapter.
3.1 Homogeneous Equations with Constant Coefficients
A second order ordinary differential equation has the form
d2 y = f t,y, dy ,
(1)
dt2
dt
where f is some given function. Usually, we will denote the independent variable by t
since time is often the independent variable in physical problems, but sometimes we
129
Chapter 3. Second Order Linear Equations will use x instead. We will use y, or occasionally some other letter, to designate the dependent variable. Equation (1) is said to be linear if the function f has the form
f
t, y, dy = g(t) − p(t)dy − q(t)y, (2)
dt
dt
that is, if f is linear in y and y. In Eq. (2) g, p, and q are specified functions of the independent variable t but do not depend on y. In this case we usually rewrite as y + p(t)y + q(t)y = g(t), (3)
where the primes denote differentiation with respect to t. Instead of Eq. (3), we often see the equation P(t)y + Q(t)y + R(t)y = G(t).
(4)
Of course, if P(t) = 0, we can divide Eq. (4) by P(t) and thereby obtain Eq. (3) with p(t) = Q(t) ,q(t) = R(t) ,g(t) = G(t) .
(5) P(t)
P(t)
P(t) In discussing Eq. (3) and in trying to solve it, we will restrict ourselves to intervals in which p, q, and g are continuous functions.1 If Eq. (1) is not of the form (3) or (4), then it is called 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. In addition, there are two special types of second order nonlinear equations that can be solved by a change of variables that reduces them to first order equations. This procedure is outlined in Problems 28 through 43.
An initial value problem consists of a differential equation such as Eq. (1), (3), or (4) together with a pair of initial conditions y(t ) = y ,y(t ) = y ,
(6)
0
0
0
0
where y and y are given numbers. Observe that the initial conditions for a second
0
0
order equation prescribe not only a particular point (t , y ) through which the graph of
0
0
the solution must pass, but also the slope y of the graph at that point. It is reasonable
0
to expect that two initial conditions are needed for a second order 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 equation is said to be homogeneous if the term g(t) in Eq. (3), or the term G(t) in Eq. (4), is zero for all t. Otherwise, the equation is called As a result, the term g(t), or G(t), is sometimes called the nonhomogeneous term. We begin our discussion with homogeneous equations, which we will write in the form P(t)y + Q(t)y + R(t)y = 0.
(7)
1There 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.
3.1Homogeneous Equations with Constant Coefficients Later, in Sections 3.6 and 3.7, 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 func tions P, Q, and R are constants. In this case, Eq. (7) becomes ay + by + cy = 0,
(8)
where a, b, and c are given constants. It turns out that Eq. (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 Eq. (7) if the coefficients are not constants, and a treatment of that case is deferred until Chapter 5.
Before taking up Eq. (8), let us first gain some experience by looking at a simple, but typical, example. Consider the equation y − y = 0, (9)
which is just Eq. (8) with a = 1, b = 0, and c = −1. In words, Eq. (9) says that we seek a function with the property that the second derivative of the function is the same as the function itself. A little thought will probably produce at least one wellknown function from calculus with this property, namely, y (t) = et , the exponential
1
function. A little more thought may also produce a second function, y (t) = e−t . Some
2
further experimentation reveals that constant multiples of these two solutions are also solutions. For example, the functions 2et and 5e−t also satisfy Eq. (9), as you can verify by calculating their second derivatives. In the same way, the functions c y (t) = c et 1 1
1
and c y (t) = c e−t satisfy the differential equation (9) for all values of the constants 2 2
2
c and c . Next, it is of paramount importance to notice that any sum of solutions of
1
2
Eq. (9) is also a solution. In particular, since c y (t) and c y (t) are solutions of Eq. (9), 1 1
2 2 so is the function y = c y (t) + c y (t) = c et + c e−t
(10)
1 1
2 2
1
2
for any values of c and c . Again, this can be verified by calculating the second
1
2
derivative y from Eq. (10). We have y = c et − c e−t and y = c et + c e−t ; thus
1
2
1
2
y is the same as y, and Eq. (9) is satisfied.
Let us summarize what we have done so far in this example. Once we notice that the functions y (t) = et and y (t) = e−t are solutions of Eq. (9), it follows that the general
1
2
linear combination (10) of these functions is also a solution. Since the coefficients c1 and c in Eq. (10) are arbitrary, this expression represents a doubly infinite family of
2
solutions of the differential equation (9).
It is now possible to consider how to pick out a particular member of this infinite family of solutions that also satisfies a given set of initial conditions. For example, suppose that we want the solution of Eq. (9) that also satisfies the initial conditions y(0) = 2,y(0) = −1.
(11)
In other words, we seek the solution that passes through the point (0, 2) and at that point has the slope −1. First, we set t = 0 and y = 2 in Eq. (10); this gives the equation c + c = 2.
(12)
1
2
Chapter 3. Second Order Linear Equations Next, we differentiate Eq. (10) with the result that y = c et − c e−t .
1
2
Then, setting t = 0 and y = −1, we obtain c − c = −1.
(13)
1
2
By solving Eqs. (12) and (13) simultaneously for c and c we find that
1
2
c = 1 ,c = 3 .
(14)
1
2
2
2
Finally, inserting these values in Eq. (10), we obtain y = 1 et + 3 e−t , (15)
2
2
the solution of the initial value problem consisting of the differential equation (9) and the initial conditions (11).
We now return to the more general equation (8), ay + by + cy = 0, which has arbitrary (real) constant coefficients. Based on our experience with let us also seek exponential solutions of Eq. (8). Thus we suppose that y = ert , where r is a parameter to be determined. Then it follows that y = rert and y = r2ert . By substituting these expressions for y, y, and y in Eq. (8), we obtain (ar2 + br + c)ert = 0, or, since ert = 0,ar 2 + br + c = 0.
(16)
Equation (16) is called the the differential equation (8).
Its significance lies in the fact that if r is a root of the polynomial equation (16), then y = ert is a solution of the differential equation (8). Since Eq. (16) is a quadratic equation with real coefficients, it has two roots, which may be real and different, real but repeated, or complex conjugates. We consider the first case here, and the latter two cases in Sections 3.4 and 3.5.
Assuming that the roots of the characteristic equation (16) are real and different, let them be denoted by r and r , where r = r . Then y (t) = er t
t
1
and y (t) = er2 are
1
2
1
2
1
2
two solutions of Eq. (8). Just as in the preceding example, it now follows that y = c y (t) + c y (t) = c er t
t
1
+ c er2
(17)
1 1
2 2
1
2
is also a solution of Eq. (8). To verify that this is so, we can differentiate the expression in Eq. (17); hence y = c r er t
t
1
+ c r er2
(18)
1 1
2 2
and y = c r2er t
t
1
+ c r2er2 .
(19)
1 1
2 2
Substituting these expressions for y, y, and y in Eq. (8) and rearranging terms, we obtain ay + by + cy = c (ar2 + br + c)er t
t
1
+ c (ar2 + br + c)er2 .
(20)
1
1
1
2
2
2
3.1Homogeneous Equations with Constant Coefficients
The quantity in each of the parentheses on the right side of Eq. (20) is zero because r1 and r are roots of Eq. (16); therefore, y as given by Eq. (17) is indeed a solution of
2
Eq. (8), as we wished to verify.
Now suppose that we want to find the particular member of the family of solutions (17) that satisfies the initial conditions (6), y(t ) = y ,y(t ) = y .
0
0
0
0
By substituting t = t and y = y in Eq. (17), we obtain
0
0
c er t
t 1 0 + c er2 0 = y .
(21)
1
2
0
Similarly, setting t = t and y = y in Eq. (18) gives
0
0
c r er t
t 1 0 + c r er2 0 = y .
(22)
1 1
2 2
0
On solving Eqs. (21) and (22) simultaneously for c and c , we find that
1
2
y − y ry r − yc = 0 0 2 e−r t 0 1
0
t
1 0 ,c = e−r2 0 .
(23)
1
r − r
2
r − r
1
2
1
2
Thus, no matter what initial conditions are assigned, that is, regardless of the values of t , y , and y in Eqs. (6), it is always possible to determine c and c so that the
0
0
0
1
2
initial conditions are satisfied; moreover, there is only one possible choice of c and c
1
2
for each set of initial conditions. With the values of c and c given by Eq. (23), the
1
2
expression (17) is the solution of the initial value problem ay + by + cy = 0,y(t ) = y ,y(t ) = y .
(24)
0
0
0
0
It is possible to show, on the basis of the fundamental theorem cited in the next section, that all solutions of Eq. (8) are included in the expression (17), at least for the case in which the roots of Eq. (16) are real and different. Therefore, we call Eq. (17) the general solution of Eq. (8). The fact that any possible initial conditions can be satisfied by the proper choice of the constants in Eq. (17) makes more plausible the idea that this expression does include all solutions of Eq. (8).
Find the general solution of E X A M P L E
1
y + 5y + 6y = 0.
(25)
We assume that y = ert , and it then follows that r must be a root of the characteristic equation r 2 + 5r + 6 = (r + 2)(r + 3) = 0.
Thus the possible values of r are r = −2 and r = −3; the general solution of
1
2
Eq. (25) is y = c e−2t + c e−3t .
(26)
1
2
Find the solution of the initial value problem E X A M P L E
2
y + 5y + 6y = 0,y(0) = 2,y(0) = 3.
(27)
Chapter 3. Second Order Linear Equations
The general solution of the differential equation was found in Example 1 and is given by Eq. (26). To satisfy the first initial condition we set t = 0 and y = 2 in thus c and c must satisfy
1
2
c + c = 2.
(28)
1
2
To use the second initial condition we must first differentiate Eq. (26). This gives y = −2c e−2t − 3c e−3t . Then, setting t = 0 and y = 3, we obtain
1
2
−2c − 3c = 3.
(29)
1
2
By solving Eqs. (28) and (29) we find that c = 9 and c = −7. Using these values in
1
2
the expression (26), we obtain the solution y = 9e−2t − 7e−3t
(30)
of the initial value problem (27). The graph of the solution is shown in Figure 3.1.1.
y
2
y = 9e–2t – 7e–3t
1
0.5
1
1.5
2
t
FIGURE 3.1.1
Solution of y + 5y + 6y = 0, y(0) = 2, y(0) = 3.
Find the solution of the initial value problem E X A M P L E
3
4y − 8y + 3y = 0,y(0) = 2,y(0) = 1 .
(31)
2
If y = ert , then the characteristic equation is 4r 2 − 8r + 3 = 0 and its roots are r = 3/2 and r = 1/2. Therefore the general solution of the differential equation is y = c e3t/2 + c et/2.
(32)
1
2
Applying the initial conditions, we obtain the following two equations for c and c :
1
2
c + c = 2, 3 c + 1c = 1 .
1
2
2 1
2 2
2
The solution of these equations is c = − 1 , c = 5 , and the solution of the initial value
1
2
2
2
problem (31) is y = − 1 e3t/2 + 5 et/2.
(33)
2
2
Figure 3.1.2 shows the graph of the solution.
3.1Homogeneous Equations with Constant Coefficients
y
2
y = – 1 e3t/2 + 5 et/2
2
2
1
0.5
1
1.5
2
t
–1
FIGURE 3.1.2
Solution of 4y − 8y + 3y = 0, y(0) = 2, y(0) = 0.5.
The solution (30) of the initial value problem (27) initially increases (because its initial E X A M P L E slope is positive) but eventually approaches zero (because both terms involve negative
4
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.
One can estimate the coordinates of the maximum point 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 (30), y = 9e−2t − 7e−3t , with respect to t we obtain y = −18e−2t + 21e−3t .
(34)
Setting y equal to zero and multiplying by e3t , we find that the critical value t satisfies
cet = 7/6; hence t = ln(7/6) ∼ = 0.15415.
(35)
c
The corresponding maximum value y is given by
M
y
= 9e−2t ∼
c − 7e−3tc = 108 = 2.20408.
(36) M
49
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 26 you are asked to determine how the coordinates of the maximum point depend on the initial slope.
Returning to the equation ay + by + cy = 0 with arbitrary coefficients, recall that when r = r , its general solution (17) is the sum of two exponential functions.
1
2
Therefore the solution has a relatively simple geometrical behavior: as t increases, the magnitude of the solution either tends to zero (when both exponents are negative) or else grows rapidly (when at least one exponent is positive). These two cases are illustrated by the solutions of Examples 2 and 3, which are shown in and 3.1.2, respectively. 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.
Chapter 3. Second Order Linear Equations
PROBLEMS
In each of Problems 1 through 8 find the general solution of the given differential equation.
1
2
1
2
䉴 19. Find the solution of the initial value problem
4
4
20. Find the solution of the initial value problem
2
䉴 25. Consider the initial value problem
(a) Solve the initial value problem.
0
0
the solution in this case.
䉴 26. Consider the initial value problem (see Example 4)
(a) Solve the initial value problem.
3.2Fundamental Solutions of Linear Homogeneous Equations
m
m
m
m
m
Equations with the Dependent Variable Missing.
For a second order differential equation
Equations with the Independent Variable Missing.
If a second order differential equation
3.2 Fundamental Solutions of Linear Homogeneous Equations
In the preceding section we showed how to solve some differential equations of the form ay + by + cy = 0, where a, b, and c 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 Chapter 3. Second Order Linear Equations turn, this understanding will assist us in finding the solutions of other problems that we will encounter later.
In developing the theory of linear differential equations, it is helpful to introduce a differential operator notation. Let p and q be continuous functions on an open interval I , that is, for α < t < β. The cases α = −∞, or β = ∞, or both, are included. Then, for any function φ that is twice differentiable on I , we define the differential operator L
by the equation L[φ] = φ + pφ + qφ.
(1)
Note that L[φ] is a function on I . The value of L[φ] at a point t is L[φ](t) = φ(t) + p(t)φ(t) + q(t)φ(t).
For example, if p(t) = t2, q(t) = 1 + t, and φ(t) = sin 3t, then L[φ](t) = (sin 3t) + t2(sin 3t) + (1 + t) sin 3t = −9 sin 3t + 3t2 cos 3t + (1 + t) sin 3t.
The operator L is often written as L = D2 + p D + q, where D is the derivative operator.
In this section we study the second order linear homogeneous equation L[φ](t) = 0.
Since it is customary to use the symbol y to denote φ(t), we will usually write this equation in the form L[y] = y + p(t)y + q(t)y = 0.
(2)
With Eq. (2) we associate a set of initial conditions y(t ) = y ,y(t ) = y ,
(3)
0
0
0
0
where t is any point in the interval I , and y and y are given real numbers. We
0
0
0
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
Consider the initial value problem y + p(t)y + q(t)y = g(t),y(t ) = y ,y(t ) = y ,
(4)
0
0
0
0
where p, q, and g are continuous on an open interval I . Then there is exactly one solution y = φ(t) of this problem, and the solution exists throughout the interval I .
We emphasize that the theorem says three things:
1.
The initial value problem has a solution; in other words, a solution exists.
2.
The initial value problem has only one solution; that is, the solution is unique.
3.2Fundamental Solutions of Linear Homogeneous Equations3.
The solution φ is defined throughout the interval I where the coefficients are continuous and is at least twice differentiable there.
For some problems some of these assertions are easy to prove. For example, we found in Section 3.1 that the initial value problem y − y = 0,y(0) = 2,y(0) = −1 (5)
has the solution y = 1 et + 3 e−t .
(6)
2
2
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 (−∞, ∞) 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 Eq. (6). Nevertheless, Theorem 3.2.1 states that this solution is indeed the only solution of the initial value problem (5).
However, 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 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 We will, however, accept Theorem 3.2.1 as true and make use of it whenever necessary.
Find the longest interval in which the solution of the initial value problem E X A M P L E
1
(t2 − 3t)y + ty − (t + 3)y = 0,y(1) = 2,y(1) = 1 is certain to exist.
If the given differential equation is written in the form of Eq. (4), then p(t) = 1/(t − 3), q(t) = −(t + 3)/t (t − 3), and g(t) = 0. The only points of discontinuity of the coefficients are t = 0 and t = 3. Therefore, the longest open interval, containing the initial point t = 1, in which all the coefficients are continuous is 0 < t < 3. Thus, this is the longest interval in which Theorem 3.2.1 guarantees that the solution exists.
Find the unique solution of the initial value problem E X A M P L E
2
y + p(t)y + q(t)y = 0,y(t ) = 0,y(t ) = 0, 0
0
where p and q are continuous in an open interval I containing t .
0
The function y = φ(t) = 0 for all t in I 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.
2A proof of Theorem 3.2.1 may be found, for example, in Chapter 6, Section 8 of the book by Coddington listed in the references.
Chapter 3. Second Order Linear Equations
Let us now assume that y and y are two solutions of Eq. (2); in other words,
1
2
L[y ] = y + py + qy = 0, (7)
1
1
1
1
and similarly for y . Then, just as in the examples in Section 3.1, we can generate
2
more solutions by forming linear combinations of y and y . We state this result as a
1
2
theorem.
Theorem 3.2.2 (Principle of Superposition)
If y and y are two solutions of the differential
1
2
equation (2), L[y] = y + p(t)y + q(t)y = 0, then the linear combination c y + c y is also a solution for any values of the 1 1
2 2 constants c and c .
1
2
A special case of Theorem 3.2.2 occurs if either c or c is zero. Then we conclude
1
2
that any multiple of a solution of Eq. (2) is also a solution.
To prove Theorem 3.2.2 we need only substitute y = c y (t) + c y (t)
(8)
1 1
2 2 for y in Eq. (2). The result is L[c y + c y ] = [c y + c y ] + p[c y + c y ] + q[c y + c y ] 1 1
2 2
1 1
2 2
1 1
2 2
1 1
2 2 = c y + c y + c py + c py + c qy + c qy 1 1
2 2
1
1
2
2
1
1
2
2
= c [y + py + qy ] + c [y + py + qy ]
1
1
1
1
2
2
2
2
= c L[y ] + c L[y ].
1
1
2
2
Since L[y ] = 0 and L[y ] = 0, it follows that L[c y + c y ] = 0 also. Therefore,
1
2
1 1
2 2 regardless of the values of c and c , y as given by Eq. (8) does satisfy the differential
1
2
equation (2) and the proof of Theorem 3.2.2 is complete.
Theorem 3.2.2 states that, beginning with only two solutions of Eq. (2), we can construct a doubly infinite family of solutions by means of Eq. (8). The next question is whether all solutions of Eq. (2) are included in Eq. (8), or whether there may be other solutions of a different form. We begin to address this question by examining whether the constants c and c in Eq. (8) can be chosen so as to satisfy the initial conditions
1
2 (3). These initial conditions require c and c to satisfy the equations
1
2
c y (t ) + c y (t ) = y , 1 1
0
2 2
0
0
(9)
c y (t ) + c y (t ) = y .
1 1
0
2 2
0
0
Upon solving Eqs. (9) for c and c , we find that
1
2
y y (t ) − y y (t ) −y y (t ) + y y (t )c = 0 2
0
0 2
0
,
c = 0 1
0
0 1
0
,
(10)
1
y (t )y (t ) − y (t )y (t )
2
y (t )y (t ) − y (t )y (t )
1
0
2
0
1
0
2
0
1
0
2
0
1
0
2
0
3.2Fundamental Solutions of Linear Homogeneous Equations or, in terms of determinants,
yy (t ) y (t )y
0
2
0
1 0
0
yy (t ) y(t )y
c =
0
2
0
,
c =
1
0
0
.
(11)
1
2
y (t )y (t ) y (t )y (t ) 1 0
2
0
1 0
2
0
y (t )y (t ) y(t )y (t )
1
0
2
0
1
0
2
0
With these values for c and c the expression (8) satisfies the initial conditions (3) as
1
2
well as the differential equation (2).
In order for the expressions for c and c in Eqs. (10) or (11) to make sense, it is
1
2
necessary that the denominators be nonzero. For both c and c the denominator is the
1
2
same, namely, the determinant
y (t )y (t )
W = 1 0
2
0
(t )y (t ) − y (t )y (t ).
(12)
y (t )y (t ) = y1 0 2 0
1
0
2
0
1
0
2
0
The determinant W is called the Wronskian3 or simply the of the solutions y and y . Sometimes we use the more extended notation W (y , y )(t )
1
2
1
2
0
to stand for the expression on the right side of Eq. (12), thereby emphasizing that the Wronskian depends on the functions y and y , and that it is evaluated at the point t .
1
2
0
The preceding argument suffices to establish the following result.
Theorem 3.2.3 Suppose that y and y are two solutions of Eq. (2),
1
2
L[y] = y + p(t)y + q(t)y = 0, and that the Wronskian W = y y − y y 1 2
1 2 is not zero at the point t where the initial conditions (3),
0
y(t ) = y ,y(t ) = y ,
0
0
0
0
are assigned. Then there is a choice of the constants c , c for which y = c y (t) +
1
2
1 1 c y (t) satisfies the differential equation (2) and the initial conditions (3).
2 2
In Example 1 of Section 3.1 we found that y (t) = e−2t and y (t) = e−3t are solutions
1
2
E X A M P L E of the differential equation
3
y + 5y + 6y = 0.
Find the Wronskian of y and y .
1
2
The Wronskian of these two functions is
e−2t
e−3t
W =
−2e−2t
−3e−3t = −e−5t.
3 Wronskian determinants are named for Jo´sef Maria Hoe¨ne´-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.
Chapter 3. Second Order Linear Equations Since W is nonzero for all values of t, the functions y and y can be used to construct
1
2
solutions of the given differential equation, together with initial conditions prescribed at any value of t. One such initial value problem was solved in Example 2 of Section 3.1.
The next theorem justifies the term “general solution” that we introduced in Section 3.1 for the linear combination c y + c y .
1 1
2 2
Theorem 3.2.4
If y and y are two solutions of the differential equation (2),
1
2
L[y] = y + p(t)y + q(t)y = 0, and if there is a point t where the Wronskian of y and y is nonzero, then the family
0
1
2
of solutions y = c y (t) + c y (t) 1 1
2 2 with arbitrary coefficients c and c includes every solution of Eq. (2).
1
2
Let φ be any solution of Eq. (2). To prove the theorem we must show that φ is included in the linear combination c y + c y ; that is, for some choice of the constants c and 1 1
2 2
1
c , the linear combination is equal to φ. Let t be a point where the Wronskian of y
2
0
1
and y is nonzero. Then evaluate φ and φ at this point and call these values y and y ,
2
0
0
respectively; thus y = φ(t ),y = φ(t ).
0
0
0
0
Next, consider the initial value problem y + p(t)y + q(t)y = 0,y(t ) = y ,y(t ) = y .
(13)
0
0
0
0
The function φ is certainly a solution of this initial value problem. On the other hand, since W (y ,y )(t ) is nonzero, it is possible (by Theorem 3.2.3) to choose c and c so
1
2
0
1
2
that y = c y (t) + c y (t) is also a solution of the initial value problem (13). In fact, 1 1
2 2 the proper values of c and c are given by Eqs. (10) or (11). The uniqueness part of
1
2
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 and c ,
1
2
φ(t) = c y (t) + c y (t), 1 1
2 2 and therefore φ is included in the family of functions of c y + c y . Finally, since 1 1
2 2 φ is an arbitrary solution of Eq. (2), it follows that every solution of this equation is included in this family. This completes the proof of Theorem 3.2.4.
Theorem 3.2.4 states that, as long as the Wronskian of y and y is not everywhere
1
2
zero, the linear combination c y + c y contains all solutions of Eq. (2). It is therefore 1 1
2 2 natural (and we have already done this in the preceding section) to call the expression y = c y (t) + c y (t) 1 1
2 2 with arbitrary constant coefficients the Eq. (2). The solutions y1 and y , with a nonzero Wronskian, are said to form a fundamental set of solutions of
2
Eq. (2).
3.2Fundamental Solutions of Linear Homogeneous Equations
To 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. Alternatively, the following example includes all those mentioned in Section 3.1, as well as many other problems of a similar type.
Suppose that y (t) = er t
t
1
and y (t) = er2 are two solutions of an equation of the form
1
2
E X A M P L E (2). Show that they form a fundamental set of solutions if r = r .
1
2
4
We calculate the Wronskian of y and y :
1
2
er tt
1
er2
W =
− r ) exp[(r + r )t].
r er t
t
1
r er2 = (r2
1
1
2
1
2
Since the exponential function is never zero, and since r − r = 0 by the statement of
2
1
the problem, it follows that W is nonzero for every value of t. Consequently, y and y
1
2
form a fundamental set of solutions.
Show that y (t) = t1/2 and y (t) = t−1 form a fundamental set of solutions of
1
2
E X A M P L E
5
2t2 y + 3t y − y = 0,t > 0.
(14)
We will show in Section 5.5 how to solve Eq. (14); see also Problem 38 in Section 3.4.
However, at this stage we can verify by direct substitution that y and y are solutions
1
2
of the differential equation. Since y (t) = 1 t−1/2 and y(t) = − 1 t−3/2, we have
1
2
1
4
2t2(− 1 t−3/2) + 3t ( 1 t−1/2) − t1/2 = (− 1 + 3 − 1)t1/2 = 0.
4
2
2
2
Similarly, y (t) = −t−2 and y(t) = 2t−3, so
2
2
2t2(2t−3) + 3t (−t−2) − t−1 = (4 − 3 − 1)t−1 = 0.
Next we calculate the Wronskian W of y and y :
1
2
t1/2
t−1 W =
1
t−3/2.
(15)
t−1/2
−t−2 = −32
2
Since W = 0 for t > 0, we conclude that y and y form a fundamental set of solutions
1
2
there.
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 may arise as to whether or not a differential equation of the form (2) always has a fundamental set of solutions. The following theorem provides an affirmative answer to this question.
Chapter 3. Second Order Linear Equations
Theorem 3.2.5
Consider the differential equation (2), L[y] = y + p(t)y + q(t)y = 0, whose coefficients p and q are continuous on some open interval I . Choose some point t in I . Let y be the solution of Eq. (2) that also satisfies the initial conditions
0
1
y(t ) = 1,y(t ) = 0, 0
0
and let y be the solution of Eq. (2) that satisfies the initial conditions
2
y(t ) = 0,y(t ) = 1.
0
0
Then y and y form a fundamental set of solutions of Eq. (2).
1
2
First observe that the existence of the functions y and y is assured by the existence
1
2
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
y (t )y (t )
1
0
W (y ,y )(t ) =
1
0
2
0
1
2
0
y(t )y (t ) = 0 1 = 1.
1
0
2
0
Since their Wronskian is not zero at the point t , the functions y and y do form a
0
1
2
fundamental set of solutions, thus completing the proof of Theorem 3.2.5.
Note that the 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 solve the specified initial value problems so as to find the functions y and y indicated in the theorem. Nevertheless, it may be
1
2
reassuring to know that a fundamental set of solutions always exists.
Find the fundamental set of solutions specified by Theorem 3.2.5 for the differential E X A M P L E
equation
6
y − y = 0, (16)
using the initial point t = 0.
0
In Section 3.1 we noted that two solutions of Eq. (16) are y (t) = et and y (t) =
1
2
e−t . The Wronskian of these solutions is W (y ,y )(t) = −2 = 0, so they form a
1
2
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 = 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 (t) the solution of
3
Eq. (16) that satisfies the initial conditions y(0) = 1,y(0) = 0.
(17)
The general solution of Eq. (16) is y = c et + c e−t ,
(18)
1
2
and the initial conditions (17) are satisfied if c = 1/2 and c = 1/2. Thus
1
2
y (t) = 1 et + 1 e−t = cosh t.
3
2
2
3.2Fundamental Solutions of Linear Homogeneous Equations
Similarly, if y (t) satisfies the initial conditions
4
y(0) = 0,y(0) = 1, (19)
then y (t) = 1 et − 1 e−t = sinh t.
4
2
2
Since the Wronskian of y and y is
3
4
W (y ,y )(t) = cosh2 t − sinh2 t = 1, 3
4
these functions also form a fundamental set of solutions, as stated by Therefore, the general solution of Eq. (16) can be written as y = k cosh t + k sinh t,
(20)
1
2
as well as in the form (18). We have used k and k for the arbitrary constants in
1
2
Eq. (20) because they are not the same as the constants c and c in Eq. (18). One
1
2
purpose of this example is to make clear that a given differential equation has more than one fundamental set of solutions; indeed, it has infinitely many. As a rule, you should choose the set that is most convenient.
We can summarize the discussion in this section as follows. To find the general solution of the differential equation y + p(t)y + q(t)y = 0,α < t < β, we must first find two functions y and y that satisfy the differential equation in
1
2
α < t < β. Then we must make sure that there is a point in the interval where the Wronskian W of y and y is nonzero. Under these circumstances y and y form a
1
2
1
2
fundamental set of solutions and the general solution is y = c y (t) + c y (t), 1 1
2 2 where c and c are arbitrary constants. If initial conditions are prescribed at a point in
1
2
α < t < β where W = 0, then c and c can be chosen so as to satisfy these conditions.
1
2
PROBLEMS
In each of Problems 1 through 6 find the Wronskian of the given pair of functions.
xex
Chapter 3. Second Order Linear Equations
1
2
1
2
1
2
1
2
1
2
Explain why this result does not contradict Theorem 3.2.2.
0
0
1
2
differential equation. Do they constitute a fundamental set of solutions?
1
2
1
2
1
2
1
2
If a second order linear homogeneous equation is not exact, it can