4 SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS so the only root is . By (10) the general solution is CASE III In this case the roots and of the auxiliary equation …

SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS A second-order linear differential equation has the form

1

Px

d 2y dy Rxy Gx 2 Qx dx dx

where P, Q, R, and G are continuous functions. Equations of this type arise in the study of the motion of a spring. In Additional Topics: Applications of Second-Order Differential Equations we will further pursue this application as well as the application to electric circuits. In this section we study the case where Gx 0, for all x, in Equation 1. Such equations are called homogeneous linear equations. Thus, the form of a second-order linear homogeneous differential equation is

2

Px

d 2y dy Rxy 0 Qx dx 2 dx

If Gx 0 for some x, Equation 1 is nonhomogeneous and is discussed in Additional Topics: Nonhomogeneous Linear Equations. Two basic facts enable us to solve homogeneous linear equations. The first of these says that if we know two solutions y1 and y2 of such an equation, then the linear combination y c1 y1 c2 y2 is also a solution. 3 Theorem If y1x and y2x are both solutions of the linear homogeneous equation (2) and c1 and c2 are any constants, then the function

yx c1 y1x c2 y2x is also a solution of Equation 2. Proof Since y1 and y2 are solutions of Equation 2, we have

Pxy1 Qxy1 Rxy1 0 and

Pxy2 Qxy2 Rxy2 0

Therefore, using the basic rules for differentiation, we have Pxy Qxy Rxy Pxc1 y1 c2 y2 Qxc1 y1 c2 y2 Rxc1 y1 c2 y2 Pxc1 y1 c2 y2 Qxc1 y1 c2 y2 Rxc1 y1 c2 y2 c1Pxy1 Qxy1 Rxy1 c2 Pxy2 Qxy2 Rxy2 c10 c20 0

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Thus, y c1 y1 c2 y2 is a solution of Equation 2. The other fact we need is given by the following theorem, which is proved in more advanced courses. It says that the general solution is a linear combination of two linearly independent solutions y1 and y2. This means that neither y1 nor y2 is a constant multiple of the other. For instance, the functions f x x 2 and tx 5x 2 are linearly dependent, but f x e x and tx xe x are linearly independent. 1

2 ■ SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS

4 Theorem If y1 and y2 are linearly independent solutions of Equation 2, and Px is never 0, then the general solution is given by

yx c1 y1x c2 y2x where c1 and c2 are arbitrary constants. Theorem 4 is very useful because it says that if we know two particular linearly independent solutions, then we know every solution. In general, it is not easy to discover particular solutions to a second-order linear equation. But it is always possible to do so if the coefficient functions P, Q, and R are constant functions, that is, if the differential equation has the form ay by cy 0

5

where a, b, and c are constants and a 0. It’s not hard to think of some likely candidates for particular solutions of Equation 5 if we state the equation verbally. We are looking for a function y such that a constant times its second derivative y plus another constant times y plus a third constant times y is equal to 0. We know that the exponential function y e rx (where r is a constant) has the property that its derivative is a constant multiple of itself: y re rx. Furthermore, y r 2e rx. If we substitute these expressions into Equation 5, we see that y e rx is a solution if ar 2e rx bre rx ce rx 0 ar 2 br ce rx 0

or

But e rx is never 0. Thus, y e rx is a solution of Equation 5 if r is a root of the equation ar 2 br c 0

6

Equation 6 is called the auxiliary equation (or characteristic equation) of the differential equation ay by cy 0. Notice that it is an algebraic equation that is obtained from the differential equation by replacing y by r 2, y by r, and y by 1. Sometimes the roots r1 and r 2 of the auxiliary equation can be found by factoring. In other cases they are found by using the quadratic formula: r1

7

b sb 2 4ac 2a

r2

b sb 2 4ac 2a

We distinguish three cases according to the sign of the discriminant b 2 4ac. b2 4ac 0 In this case the roots r1 and r 2 of the auxiliary equation are real and distinct, so y1 e r 1 x and y2 e r 2 x are two linearly independent solutions of Equation 5. (Note that e r 2 x is not a constant multiple of e r 1 x.) Therefore, by Theorem 4, we have the following fact. CASE I

■

If the roots r1 and r 2 of the auxiliary equation ar 2 br c 0 are real and unequal, then the general solution of ay by cy 0 is Thomson Brooks-Cole copyright 2007

8

y c1 e r 1 x c2 e r 2 x

SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS ■ 3

In Figure 1 the graphs of the basic solutions f x e 2 x and tx e3 x of the differential equation in Example 1 are shown in black and red, respectively. Some of the other solutions, linear combinations of f and t , are shown in blue. ■ ■

8

EXAMPLE 1 Solve the equation y y 6y 0. SOLUTION The auxiliary equation is

r 2 r 6 r 2r 3 0 whose roots are r 2, 3. Therefore, by (8) the general solution of the given differential equation is

5f+g

y c1 e 2x c2 e3x

f+5g f+g f

g

_1

g-f

f-g _5

1

We could verify that this is indeed a solution by differentiating and substituting into the differential equation. EXAMPLE 2 Solve 3

FIGURE 1

d 2y dy y 0. dx 2 dx

SOLUTION To solve the auxiliary equation 3r 2 r 1 0 we use the quadratic

formula: r

1 s13 6

Since the roots are real and distinct, the general solution is y c1 e (1s13 ) x6 c2 e (1s13 ) x6 b 2 4 ac 0 In this case r1 r2 ; that is, the roots of the auxiliary equation are real and equal. Let’s denote by r the common value of r1 and r 2. Then, from Equations 7, we have CASE II

9

■

r

b 2a

so

2ar b 0

We know that y1 e rx is one solution of Equation 5. We now verify that y2 xe rx is also a solution: ay2 by2 cy2 a2re rx r 2xe rx be rx rxe rx cxe rx 2ar be rx ar 2 br cxe rx 0e rx 0xe rx 0 The first term is 0 by Equations 9; the second term is 0 because r is a root of the auxiliary equation. Since y1 e rx and y2 xe rx are linearly independent solutions, Theorem 4 provides us with the general solution. If the auxiliary equation ar 2 br c 0 has only one real root r, then the general solution of ay by cy 0 is 10

y c1 e rx c2 xe rx

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EXAMPLE 3 Solve the equation 4y 12y 9y 0. SOLUTION The auxiliary equation 4r 2 12r 9 0 can be factored as

2r 32 0

4 ■ SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS

so the only root is r 32 . By (10) the general solution is

Figure 2 shows the basic solutions f x e3x2 and tx xe3x2 in Example 3 and some other members of the family of solutions. Notice that all of them approach 0 as x l . ■ ■

y c1 e3x2 c2 xe3x2 b 2 4ac 0 In this case the roots r1 and r 2 of the auxiliary equation are complex numbers. (See Additional Topics: Complex Numbers for information about complex numbers.) We can write CASE III

f-g 8 f 5f+g _2

f+g

f+5g

■

r1 i

2 g-f

g

r 2 i

where and are real numbers. [In fact, b2a, s4ac b 22a.] Then, using Euler’s equation

_5

e i cos i sin

FIGURE 2

from Additional Topics: Complex Numbers, we write the solution of the differential equation as y C1 e r 1 x C2 e r 2 x C1 e ix C2 e ix C1 e xcos x i sin x C2 e xcos x i sin x e x C1 C2 cos x iC1 C2 sin x e xc1 cos x c2 sin x where c1 C1 C2 , c2 iC1 C2. This gives all solutions (real or complex) of the differential equation. The solutions are real when the constants c1 and c2 are real. We summarize the discussion as follows. If the roots of the auxiliary equation ar 2 br c 0 are the complex numbers r1 i, r 2 i, then the general solution of ay by cy 0 is 11

y e xc1 cos x c2 sin x

■ ■ Figure 3 shows the graphs of the solutions in Example 4, f x e 3 x cos 2x and tx e 3 x sin 2x, together with some linear combinations. All solutions approach 0 as x l .

3 f+g

g

f-g

SOLUTION The auxiliary equation is r 2 6r 13 0. By the quadratic formula, the

roots are r

6 s36 52 6 s16 3 2i 2 2

By (11) the general solution of the differential equation is

f

_3

EXAMPLE 4 Solve the equation y 6y 13y 0.

2

y e 3xc1 cos 2x c2 sin 2x INITIAL-VALUE AND BOUNDARY-VALUE PROBLEMS

_3

FIGURE 3

An initial-value problem for the second-order Equation 1 or 2 consists of finding a solution y of the differential equation that also satisfies initial conditions of the form

Thomson Brooks-Cole copyright 2007

yx 0 y0

yx 0 y1

where y0 and y1 are given constants. If P, Q, R, and G are continuous on an interval and Px 0 there, then a theorem found in more advanced books guarantees the existence and uniqueness of a solution to this initial-value problem. Examples 5 and 6 illustrate the technique for solving such a problem.

SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS ■ 5

EXAMPLE 5 Solve the initial-value problem

y y 6y 0 ■ ■ Figure 4 shows the graph of the solution of the initial-value problem in Example 5. Compare with Figure 1.

y0 1

y0 0

SOLUTION From Example 1 we know that the general solution of the differential equa-

tion is yx c1 e 2x c2 e3x

20

Differentiating this solution, we get yx 2c1 e 2x 3c2 e3x To satisfy the initial conditions we require that _2

2

0

FIGURE 4

12

y0 c1 c2 1

13

y0 2c1 3c2 0

From (13) we have c2 23 c1 and so (12) gives c1 23 c1 1

c1 35

c2 25

Thus, the required solution of the initial-value problem is y 35 e 2x 25 e3x EXAMPLE 6 Solve the initial-value problem The solution to Example 6 is graphed in Figure 5. It appears to be a shifted sine curve and, indeed, you can verify that another way of writing the solution is ■ ■

y s13 sinx where tan 23

y y 0

y0 2

y0 3

SOLUTION The auxiliary equation is r 2 1 0, or r 2 1, whose roots are i. Thus

0, 1, and since e 0x 1, the general solution is

yx c1 cos x c2 sin x

5

yx c1 sin x c2 cos x

Since the initial conditions become _2π

2π

y0 c1 2

y0 c2 3

Therefore, the solution of the initial-value problem is _5

yx 2 cos x 3 sin x

FIGURE 5

A boundary-value problem for Equation 1 consists of finding a solution y of the differential equation that also satisfies boundary conditions of the form yx 0 y0

yx 1 y1

In contrast with the situation for initial-value problems, a boundary-value problem does not always have a solution. EXAMPLE 7 Solve the boundary-value problem

y 2y y 0

y0 1

y1 3

SOLUTION The auxiliary equation is

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r 2 2r 1 0

or

r 12 0

whose only root is r 1. Therefore, the general solution is yx c1 ex c2 xex

6 ■ SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS

The boundary conditions are satisfied if

■ ■ Figure 6 shows the graph of the solution of the boundary-value problem in Example 7.

y0 c1 1

5

y1 c1 e1 c2 e1 3 The first condition gives c1 1, so the second condition becomes

_1

5

e1 c2 e1 3 Solving this equation for c2 by first multiplying through by e, we get 1 c2 3e

_5

FIGURE 6

c2 3e 1

so

Thus, the solution of the boundary-value problem is y ex 3e 1xex Summary: Solutions of ay by c 0

Roots of ar 2 br c 0

General solution y c1 e r 1 x c2 e r 2 x y c1 e rx c2 xe rx y e xc1 cos x c2 sin x

r1, r2 real and distinct r1 r2 r r1, r2 complex: i

EXERCISES A Click here for answers. 1–13

20. 2y 5y 3y 0,

Click here for solutions.

S

21. y 16y 0,

Solve the differential equation.

y 2

y0 2,

y0 1

3. y 8y 41y 0

4. 2y y y 0

24. y 12y 36y 0,

5. y 2y y 0

6. 3y 5y

7. 4y y 0

8. 16y 24y 9y 0

d 2y dy 2 y0 dt 2 dt

■

■

■

■

■

■

■

■

■

■

■

d 2y dy 2y 0 14. 6 dx 2 dx

17–24

■

■

■

■

■

■

Solve the initial-value problem.

17. 2y 5y 3y 0, 18. y 3y 0,

y0 3,

y0 1,

19. 4 y 4 y y 0,

y0 4

y0 3

y0 1,

y0 1.5

■

■

■

■

■

■

■

■

y1 2

y0 1,

y3 0

y 5

y0 2,

31. y 4y 13y 0,

y 2

y0 1, y0 1,

■

■

■

■

■

y1 0 y 2 1

y0 2,

32. 9y 18y 10 y 0,

d 2y dy 8 16y 0 15. dx 2 dx

■

■

y 4

y0 1,

30. y 6y 9y 0,

■

y0 0, ■

■

y 1 ■

■

■

33. Let L be a nonzero real number.

d y dy 2 5y 0 dx 2 dx ■

y1 1 ■

■

2

■

■

y0 3,

29. y 6y 25y 0, ■

y1 0, ■

Solve the boundary-value problem, if possible.

28. y 100 y 0,

Graph the two basic solutions of the differential equation and several other solutions. What features do the solutions have in common?

■

■

27. y 3y 2y 0,

; 14–16

16.

■

26. y 2y 0,

d y dy y0 dt 2 dt

■

■

25. 4 y y 0,

2

13.

■

25–32

d 2y dy 6 4y 0 dt 2 dt

12.

y4 4

y 0,

23. y 2y 2y 0,

10. 9y 4y 0

y0 4

22. y 2y 5y 0,

2. y 4y 8y 0

11.

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y4 3,

1. y 6y 8y 0

9. 4y y 0

y0 1,

■

(a) Show that the boundary-value problem y y 0, y0 0, yL 0 has only the trivial solution y 0 for the cases 0 and 0. (b) For the case 0, find the values of for which this problem has a nontrivial solution and give the corresponding solution. 34. If a, b, and c are all positive constants and yx is a solution

of the differential equation ay by cy 0, show that lim x l yx 0.

SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS ■ 7

ANSWERS S

17. y 2e3x2 ex

Click here for solutions.

1. y c1 e

4x

c2 e

2x

5. y c1 e c2 xe x

x4

9. y c1 c2 e

x

4x

3. y e

c1 cos 5x c2 sin 5x

7. y c1 cosx2 c2 sinx2 11. y c1 e (1s2 )t c2 e (1s2 )t

13. y et2 [c1 cos(s3t2) c2 sin(s3t2)] g 15. 40

_0.2

19. y e x/2 2xe x2 21. y 3 cos 4x sin 4x 23. y ex2 cos x 3 sin x 25. y 3 cos( 2 x) 4 sin( 2 x) 1

1

e x3 e 2x e 1 1 e3

f

27. y

1

29. No solution

3

31. y e2x 2 cos 3x e sin 3x _40

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All solutions approach 0 as x l and approach as x l .

33. (b) n 2 2L2, n a positive integer; y C sinn xL

8 ■ SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS

SOLUTIONS

1. The auxiliary equation is r2 − 6r + 8 = 0 ⇒ (r − 4)(r − 2) = 0 ⇒ r = 4, r = 2. Then by (8) the general solution is y = c1 e4x + c2 e2x . 3. The auxiliary equation is r2 + 8r + 41 = 0 ⇒ r = −4 ± 5i. Then by (11) the general solution is y = e−4x (c1 cos 5x + c2 sin 5x). 5. The auxiliary equation is r2 − 2r + 1 = (r − 1)2 = 0 ⇒ r = 1. Then by (10), the general solution is y = c1 ex + c2 xex .

7. The auxiliary equation is 4r2 + 1 = 0

r = ± 12 i, so y = c1 cos

⇒

1 x + c2 sin 12 x . 2

9. The auxiliary equation is 4r2 + r = r(4r + 1) = 0 ⇒ r = 0, r = − 14 , so y = c1 + c2 e−x/4 . 11. The auxiliary equation is r2 − 2r − 1 = 0 ⇒ r = 1 ±

√ √ √ 2, so y = c1 e(1+ 2)t + c2 e(1− 2)t .

13. The auxiliary equation is r2 + r + 1 = 0 ⇒ r = − 12 ±

√

3 i, 2

k √ √ l so y = e−t/2 c1 cos 23 t + c2 sin 23 t .

15. r2 − 8r + 16 = (r − 4)2 = 0 so y = c1 e4x + c2 xe4x .

The graphs are all asymptotic to the x-axis as x → −∞, and as x → ∞ the solutions tend to ±∞.

17. 2r2 + 5r + 3 = (2r + 3)(r + 1) = 0, so r = − 32 , r = −1 and the general solution is y = c1 e−3x/2 + c2 e−x . Then y(0) = 3 ⇒ c1 + c2 = 3 and y 0 (0) = −4 ⇒ − 32 c1 − c2 = −4, so c1 = 2 and c2 = 1. Thus the solution to the initial-value problem is y = 2e−3x/2 + e−x . 19. 4r2 − 4r + 1 = (2r − 1)2 = 0 ⇒ r = ⇒ c1 = 1 and y 0 (0) = −1.5 ⇒

1 c 2 1

1 2

and the general solution is y = c1 ex/2 + c2 xex/2 . Then y(0) = 1

+ c2 = −1.5, so c2 = −2 and the solution to the initial-value problem is

y = ex/2 − 2xex/2 . 21. r2 + 16 = 0 ⇒ r = ±4i and the general solution is y = e0x (c1 cos 4x + c2 sin 4x) = c1 cos 4x + c2 sin 4x. Then y π4 = −3 ⇒ −c1 = −3 ⇒ c1 = 3 and y 0 π4 = 4 ⇒ −4c2 = 4 ⇒ c2 = −1, so the solution to the initial-value problem is y = 3 cos 4x − sin 4x.

23. r2 + 2r + 2 = 0 ⇒ r = −1 ± i and the general solution is y = e−x (c1 cos x + c2 sin x). Then 2 = y(0) = c1 and 1 = y 0 (0) = c2 − c1

⇒ c2 = 3 and the solution to the initial-value problem is y = e−x (2 cos x + 3 sin x).

25. 4r2 + 1 = 0 ⇒ r = ± 12 i and the general solution is y = c1 cos 12 x + c2 sin 12 x . Then 3 = y(0) = c1 and −4 = y(π) = c2 , so the solution of the boundary-value problem is y = 3 cos 12 x − 4 sin 12 x .

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27. r2 − 3r + 2 = (r − 2)(r − 1) = 0 ⇒ r = 1, r = 2 and the general solution is y = c1 ex + c2 e2x . Then

1 = y(0) = c1 + c2 and 0 = y(3) = c1 e3 + c2 e6 so c2 = 1/(1 − e3 ) and c1 = e3 /(e3 − 1). The solution of the boundary-value problem is y =

e2x ex+3 + . 3 e −1 1 − e3

SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS ■ 9

29. r2 − 6r + 25 = 0 ⇒ r = 3 ± 4i and the general solution is y = e3x (c1 cos 4x + c2 sin 4x). But 1 = y(0) = c1 and 2 = y(π) = c1 e3π

⇒ c1 = 2/e3π , so there is no solution.

31. r2 + 4r + 13 = 0 ⇒ r = −2 ± 3i and the general solution is y = e−2x (c1 cos 3x + c2 sin 3x). But 2 = y(0) = c1 and 1 = y π2 = e−π (−c2 ), so the solution to the boundary-value problem is y = e−2x (2 cos 3x − eπ sin 3x).

33. (a) Case 1 (λ = 0): y 00 + λy = 0 ⇒ y 00 = 0 which has an auxiliary equation r2 = 0 ⇒ r = 0 ⇒

y = c1 + c2 x where y(0) = 0 and y(L) = 0. Thus, 0 = y(0) = c1 and 0 = y(L) = c2 L ⇒ c1 = c2 = 0.

Thus, y = 0.

√ Case 2 (λ < 0): y 00 + λy = 0 has auxiliary equation r2 = −λ ⇒ r = ± −λ (distinct and real since λ < 0) ⇒

√

y = c1 e √

0 = y(L) = c1 e

−λL √

Multiplying (∗) by e

√

−λx

+ c2 e− √ − −λL

+ c2 e

−λL

−λx

where y(0) = 0 and y(L) = 0. Thus, 0 = y(0) = c1 + c2 (∗) and

(†).

√ √ and subtracting (†) gives c2 e −λL − e− −λL = 0 ⇒ c2 = 0 and thus

c1 = 0 from (∗). Thus, y = 0 for the cases λ = 0 and λ < 0.

√ √ √ λ ⇒ y = c1 cos λ x + c2 sin λ x √ where y(0) = 0 and y(L) = 0. Thus, 0 = y(0) = c1 and 0 = y(L) = c2 sin λL since c1 = 0. Since we √ √ cannot have a trivial solution, c2 6= 0 and thus sin λ L = 0 ⇒ λ L = nπ where n is an integer

(b) y 00 + λy = 0 has an auxiliary equation r2 + λ = 0 ⇒ r = ±i

Thomson Brooks-Cole copyright 2007

⇒

λ = n2 π 2 /L2 and y = c2 sin(nπx/L) where n is an integer.

1

Px

d 2y dy Rxy Gx 2 Qx dx dx

where P, Q, R, and G are continuous functions. Equations of this type arise in the study of the motion of a spring. In Additional Topics: Applications of Second-Order Differential Equations we will further pursue this application as well as the application to electric circuits. In this section we study the case where Gx 0, for all x, in Equation 1. Such equations are called homogeneous linear equations. Thus, the form of a second-order linear homogeneous differential equation is

2

Px

d 2y dy Rxy 0 Qx dx 2 dx

If Gx 0 for some x, Equation 1 is nonhomogeneous and is discussed in Additional Topics: Nonhomogeneous Linear Equations. Two basic facts enable us to solve homogeneous linear equations. The first of these says that if we know two solutions y1 and y2 of such an equation, then the linear combination y c1 y1 c2 y2 is also a solution. 3 Theorem If y1x and y2x are both solutions of the linear homogeneous equation (2) and c1 and c2 are any constants, then the function

yx c1 y1x c2 y2x is also a solution of Equation 2. Proof Since y1 and y2 are solutions of Equation 2, we have

Pxy1 Qxy1 Rxy1 0 and

Pxy2 Qxy2 Rxy2 0

Therefore, using the basic rules for differentiation, we have Pxy Qxy Rxy Pxc1 y1 c2 y2 Qxc1 y1 c2 y2 Rxc1 y1 c2 y2 Pxc1 y1 c2 y2 Qxc1 y1 c2 y2 Rxc1 y1 c2 y2 c1Pxy1 Qxy1 Rxy1 c2 Pxy2 Qxy2 Rxy2 c10 c20 0

Thomson Brooks-Cole copyright 2007

Thus, y c1 y1 c2 y2 is a solution of Equation 2. The other fact we need is given by the following theorem, which is proved in more advanced courses. It says that the general solution is a linear combination of two linearly independent solutions y1 and y2. This means that neither y1 nor y2 is a constant multiple of the other. For instance, the functions f x x 2 and tx 5x 2 are linearly dependent, but f x e x and tx xe x are linearly independent. 1

2 ■ SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS

4 Theorem If y1 and y2 are linearly independent solutions of Equation 2, and Px is never 0, then the general solution is given by

yx c1 y1x c2 y2x where c1 and c2 are arbitrary constants. Theorem 4 is very useful because it says that if we know two particular linearly independent solutions, then we know every solution. In general, it is not easy to discover particular solutions to a second-order linear equation. But it is always possible to do so if the coefficient functions P, Q, and R are constant functions, that is, if the differential equation has the form ay by cy 0

5

where a, b, and c are constants and a 0. It’s not hard to think of some likely candidates for particular solutions of Equation 5 if we state the equation verbally. We are looking for a function y such that a constant times its second derivative y plus another constant times y plus a third constant times y is equal to 0. We know that the exponential function y e rx (where r is a constant) has the property that its derivative is a constant multiple of itself: y re rx. Furthermore, y r 2e rx. If we substitute these expressions into Equation 5, we see that y e rx is a solution if ar 2e rx bre rx ce rx 0 ar 2 br ce rx 0

or

But e rx is never 0. Thus, y e rx is a solution of Equation 5 if r is a root of the equation ar 2 br c 0

6

Equation 6 is called the auxiliary equation (or characteristic equation) of the differential equation ay by cy 0. Notice that it is an algebraic equation that is obtained from the differential equation by replacing y by r 2, y by r, and y by 1. Sometimes the roots r1 and r 2 of the auxiliary equation can be found by factoring. In other cases they are found by using the quadratic formula: r1

7

b sb 2 4ac 2a

r2

b sb 2 4ac 2a

We distinguish three cases according to the sign of the discriminant b 2 4ac. b2 4ac 0 In this case the roots r1 and r 2 of the auxiliary equation are real and distinct, so y1 e r 1 x and y2 e r 2 x are two linearly independent solutions of Equation 5. (Note that e r 2 x is not a constant multiple of e r 1 x.) Therefore, by Theorem 4, we have the following fact. CASE I

■

If the roots r1 and r 2 of the auxiliary equation ar 2 br c 0 are real and unequal, then the general solution of ay by cy 0 is Thomson Brooks-Cole copyright 2007

8

y c1 e r 1 x c2 e r 2 x

SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS ■ 3

In Figure 1 the graphs of the basic solutions f x e 2 x and tx e3 x of the differential equation in Example 1 are shown in black and red, respectively. Some of the other solutions, linear combinations of f and t , are shown in blue. ■ ■

8

EXAMPLE 1 Solve the equation y y 6y 0. SOLUTION The auxiliary equation is

r 2 r 6 r 2r 3 0 whose roots are r 2, 3. Therefore, by (8) the general solution of the given differential equation is

5f+g

y c1 e 2x c2 e3x

f+5g f+g f

g

_1

g-f

f-g _5

1

We could verify that this is indeed a solution by differentiating and substituting into the differential equation. EXAMPLE 2 Solve 3

FIGURE 1

d 2y dy y 0. dx 2 dx

SOLUTION To solve the auxiliary equation 3r 2 r 1 0 we use the quadratic

formula: r

1 s13 6

Since the roots are real and distinct, the general solution is y c1 e (1s13 ) x6 c2 e (1s13 ) x6 b 2 4 ac 0 In this case r1 r2 ; that is, the roots of the auxiliary equation are real and equal. Let’s denote by r the common value of r1 and r 2. Then, from Equations 7, we have CASE II

9

■

r

b 2a

so

2ar b 0

We know that y1 e rx is one solution of Equation 5. We now verify that y2 xe rx is also a solution: ay2 by2 cy2 a2re rx r 2xe rx be rx rxe rx cxe rx 2ar be rx ar 2 br cxe rx 0e rx 0xe rx 0 The first term is 0 by Equations 9; the second term is 0 because r is a root of the auxiliary equation. Since y1 e rx and y2 xe rx are linearly independent solutions, Theorem 4 provides us with the general solution. If the auxiliary equation ar 2 br c 0 has only one real root r, then the general solution of ay by cy 0 is 10

y c1 e rx c2 xe rx

Thomson Brooks-Cole copyright 2007

EXAMPLE 3 Solve the equation 4y 12y 9y 0. SOLUTION The auxiliary equation 4r 2 12r 9 0 can be factored as

2r 32 0

4 ■ SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS

so the only root is r 32 . By (10) the general solution is

Figure 2 shows the basic solutions f x e3x2 and tx xe3x2 in Example 3 and some other members of the family of solutions. Notice that all of them approach 0 as x l . ■ ■

y c1 e3x2 c2 xe3x2 b 2 4ac 0 In this case the roots r1 and r 2 of the auxiliary equation are complex numbers. (See Additional Topics: Complex Numbers for information about complex numbers.) We can write CASE III

f-g 8 f 5f+g _2

f+g

f+5g

■

r1 i

2 g-f

g

r 2 i

where and are real numbers. [In fact, b2a, s4ac b 22a.] Then, using Euler’s equation

_5

e i cos i sin

FIGURE 2

from Additional Topics: Complex Numbers, we write the solution of the differential equation as y C1 e r 1 x C2 e r 2 x C1 e ix C2 e ix C1 e xcos x i sin x C2 e xcos x i sin x e x C1 C2 cos x iC1 C2 sin x e xc1 cos x c2 sin x where c1 C1 C2 , c2 iC1 C2. This gives all solutions (real or complex) of the differential equation. The solutions are real when the constants c1 and c2 are real. We summarize the discussion as follows. If the roots of the auxiliary equation ar 2 br c 0 are the complex numbers r1 i, r 2 i, then the general solution of ay by cy 0 is 11

y e xc1 cos x c2 sin x

■ ■ Figure 3 shows the graphs of the solutions in Example 4, f x e 3 x cos 2x and tx e 3 x sin 2x, together with some linear combinations. All solutions approach 0 as x l .

3 f+g

g

f-g

SOLUTION The auxiliary equation is r 2 6r 13 0. By the quadratic formula, the

roots are r

6 s36 52 6 s16 3 2i 2 2

By (11) the general solution of the differential equation is

f

_3

EXAMPLE 4 Solve the equation y 6y 13y 0.

2

y e 3xc1 cos 2x c2 sin 2x INITIAL-VALUE AND BOUNDARY-VALUE PROBLEMS

_3

FIGURE 3

An initial-value problem for the second-order Equation 1 or 2 consists of finding a solution y of the differential equation that also satisfies initial conditions of the form

Thomson Brooks-Cole copyright 2007

yx 0 y0

yx 0 y1

where y0 and y1 are given constants. If P, Q, R, and G are continuous on an interval and Px 0 there, then a theorem found in more advanced books guarantees the existence and uniqueness of a solution to this initial-value problem. Examples 5 and 6 illustrate the technique for solving such a problem.

SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS ■ 5

EXAMPLE 5 Solve the initial-value problem

y y 6y 0 ■ ■ Figure 4 shows the graph of the solution of the initial-value problem in Example 5. Compare with Figure 1.

y0 1

y0 0

SOLUTION From Example 1 we know that the general solution of the differential equa-

tion is yx c1 e 2x c2 e3x

20

Differentiating this solution, we get yx 2c1 e 2x 3c2 e3x To satisfy the initial conditions we require that _2

2

0

FIGURE 4

12

y0 c1 c2 1

13

y0 2c1 3c2 0

From (13) we have c2 23 c1 and so (12) gives c1 23 c1 1

c1 35

c2 25

Thus, the required solution of the initial-value problem is y 35 e 2x 25 e3x EXAMPLE 6 Solve the initial-value problem The solution to Example 6 is graphed in Figure 5. It appears to be a shifted sine curve and, indeed, you can verify that another way of writing the solution is ■ ■

y s13 sinx where tan 23

y y 0

y0 2

y0 3

SOLUTION The auxiliary equation is r 2 1 0, or r 2 1, whose roots are i. Thus

0, 1, and since e 0x 1, the general solution is

yx c1 cos x c2 sin x

5

yx c1 sin x c2 cos x

Since the initial conditions become _2π

2π

y0 c1 2

y0 c2 3

Therefore, the solution of the initial-value problem is _5

yx 2 cos x 3 sin x

FIGURE 5

A boundary-value problem for Equation 1 consists of finding a solution y of the differential equation that also satisfies boundary conditions of the form yx 0 y0

yx 1 y1

In contrast with the situation for initial-value problems, a boundary-value problem does not always have a solution. EXAMPLE 7 Solve the boundary-value problem

y 2y y 0

y0 1

y1 3

SOLUTION The auxiliary equation is

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r 2 2r 1 0

or

r 12 0

whose only root is r 1. Therefore, the general solution is yx c1 ex c2 xex

6 ■ SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS

The boundary conditions are satisfied if

■ ■ Figure 6 shows the graph of the solution of the boundary-value problem in Example 7.

y0 c1 1

5

y1 c1 e1 c2 e1 3 The first condition gives c1 1, so the second condition becomes

_1

5

e1 c2 e1 3 Solving this equation for c2 by first multiplying through by e, we get 1 c2 3e

_5

FIGURE 6

c2 3e 1

so

Thus, the solution of the boundary-value problem is y ex 3e 1xex Summary: Solutions of ay by c 0

Roots of ar 2 br c 0

General solution y c1 e r 1 x c2 e r 2 x y c1 e rx c2 xe rx y e xc1 cos x c2 sin x

r1, r2 real and distinct r1 r2 r r1, r2 complex: i

EXERCISES A Click here for answers. 1–13

20. 2y 5y 3y 0,

Click here for solutions.

S

21. y 16y 0,

Solve the differential equation.

y 2

y0 2,

y0 1

3. y 8y 41y 0

4. 2y y y 0

24. y 12y 36y 0,

5. y 2y y 0

6. 3y 5y

7. 4y y 0

8. 16y 24y 9y 0

d 2y dy 2 y0 dt 2 dt

■

■

■

■

■

■

■

■

■

■

■

d 2y dy 2y 0 14. 6 dx 2 dx

17–24

■

■

■

■

■

■

Solve the initial-value problem.

17. 2y 5y 3y 0, 18. y 3y 0,

y0 3,

y0 1,

19. 4 y 4 y y 0,

y0 4

y0 3

y0 1,

y0 1.5

■

■

■

■

■

■

■

■

y1 2

y0 1,

y3 0

y 5

y0 2,

31. y 4y 13y 0,

y 2

y0 1, y0 1,

■

■

■

■

■

y1 0 y 2 1

y0 2,

32. 9y 18y 10 y 0,

d 2y dy 8 16y 0 15. dx 2 dx

■

■

y 4

y0 1,

30. y 6y 9y 0,

■

y0 0, ■

■

y 1 ■

■

■

33. Let L be a nonzero real number.

d y dy 2 5y 0 dx 2 dx ■

y1 1 ■

■

2

■

■

y0 3,

29. y 6y 25y 0, ■

y1 0, ■

Solve the boundary-value problem, if possible.

28. y 100 y 0,

Graph the two basic solutions of the differential equation and several other solutions. What features do the solutions have in common?

■

■

27. y 3y 2y 0,

; 14–16

16.

■

26. y 2y 0,

d y dy y0 dt 2 dt

■

■

25. 4 y y 0,

2

13.

■

25–32

d 2y dy 6 4y 0 dt 2 dt

12.

y4 4

y 0,

23. y 2y 2y 0,

10. 9y 4y 0

y0 4

22. y 2y 5y 0,

2. y 4y 8y 0

11.

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y4 3,

1. y 6y 8y 0

9. 4y y 0

y0 1,

■

(a) Show that the boundary-value problem y y 0, y0 0, yL 0 has only the trivial solution y 0 for the cases 0 and 0. (b) For the case 0, find the values of for which this problem has a nontrivial solution and give the corresponding solution. 34. If a, b, and c are all positive constants and yx is a solution

of the differential equation ay by cy 0, show that lim x l yx 0.

SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS ■ 7

ANSWERS S

17. y 2e3x2 ex

Click here for solutions.

1. y c1 e

4x

c2 e

2x

5. y c1 e c2 xe x

x4

9. y c1 c2 e

x

4x

3. y e

c1 cos 5x c2 sin 5x

7. y c1 cosx2 c2 sinx2 11. y c1 e (1s2 )t c2 e (1s2 )t

13. y et2 [c1 cos(s3t2) c2 sin(s3t2)] g 15. 40

_0.2

19. y e x/2 2xe x2 21. y 3 cos 4x sin 4x 23. y ex2 cos x 3 sin x 25. y 3 cos( 2 x) 4 sin( 2 x) 1

1

e x3 e 2x e 1 1 e3

f

27. y

1

29. No solution

3

31. y e2x 2 cos 3x e sin 3x _40

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All solutions approach 0 as x l and approach as x l .

33. (b) n 2 2L2, n a positive integer; y C sinn xL

8 ■ SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS

SOLUTIONS

1. The auxiliary equation is r2 − 6r + 8 = 0 ⇒ (r − 4)(r − 2) = 0 ⇒ r = 4, r = 2. Then by (8) the general solution is y = c1 e4x + c2 e2x . 3. The auxiliary equation is r2 + 8r + 41 = 0 ⇒ r = −4 ± 5i. Then by (11) the general solution is y = e−4x (c1 cos 5x + c2 sin 5x). 5. The auxiliary equation is r2 − 2r + 1 = (r − 1)2 = 0 ⇒ r = 1. Then by (10), the general solution is y = c1 ex + c2 xex .

7. The auxiliary equation is 4r2 + 1 = 0

r = ± 12 i, so y = c1 cos

⇒

1 x + c2 sin 12 x . 2

9. The auxiliary equation is 4r2 + r = r(4r + 1) = 0 ⇒ r = 0, r = − 14 , so y = c1 + c2 e−x/4 . 11. The auxiliary equation is r2 − 2r − 1 = 0 ⇒ r = 1 ±

√ √ √ 2, so y = c1 e(1+ 2)t + c2 e(1− 2)t .

13. The auxiliary equation is r2 + r + 1 = 0 ⇒ r = − 12 ±

√

3 i, 2

k √ √ l so y = e−t/2 c1 cos 23 t + c2 sin 23 t .

15. r2 − 8r + 16 = (r − 4)2 = 0 so y = c1 e4x + c2 xe4x .

The graphs are all asymptotic to the x-axis as x → −∞, and as x → ∞ the solutions tend to ±∞.

17. 2r2 + 5r + 3 = (2r + 3)(r + 1) = 0, so r = − 32 , r = −1 and the general solution is y = c1 e−3x/2 + c2 e−x . Then y(0) = 3 ⇒ c1 + c2 = 3 and y 0 (0) = −4 ⇒ − 32 c1 − c2 = −4, so c1 = 2 and c2 = 1. Thus the solution to the initial-value problem is y = 2e−3x/2 + e−x . 19. 4r2 − 4r + 1 = (2r − 1)2 = 0 ⇒ r = ⇒ c1 = 1 and y 0 (0) = −1.5 ⇒

1 c 2 1

1 2

and the general solution is y = c1 ex/2 + c2 xex/2 . Then y(0) = 1

+ c2 = −1.5, so c2 = −2 and the solution to the initial-value problem is

y = ex/2 − 2xex/2 . 21. r2 + 16 = 0 ⇒ r = ±4i and the general solution is y = e0x (c1 cos 4x + c2 sin 4x) = c1 cos 4x + c2 sin 4x. Then y π4 = −3 ⇒ −c1 = −3 ⇒ c1 = 3 and y 0 π4 = 4 ⇒ −4c2 = 4 ⇒ c2 = −1, so the solution to the initial-value problem is y = 3 cos 4x − sin 4x.

23. r2 + 2r + 2 = 0 ⇒ r = −1 ± i and the general solution is y = e−x (c1 cos x + c2 sin x). Then 2 = y(0) = c1 and 1 = y 0 (0) = c2 − c1

⇒ c2 = 3 and the solution to the initial-value problem is y = e−x (2 cos x + 3 sin x).

25. 4r2 + 1 = 0 ⇒ r = ± 12 i and the general solution is y = c1 cos 12 x + c2 sin 12 x . Then 3 = y(0) = c1 and −4 = y(π) = c2 , so the solution of the boundary-value problem is y = 3 cos 12 x − 4 sin 12 x .

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27. r2 − 3r + 2 = (r − 2)(r − 1) = 0 ⇒ r = 1, r = 2 and the general solution is y = c1 ex + c2 e2x . Then

1 = y(0) = c1 + c2 and 0 = y(3) = c1 e3 + c2 e6 so c2 = 1/(1 − e3 ) and c1 = e3 /(e3 − 1). The solution of the boundary-value problem is y =

e2x ex+3 + . 3 e −1 1 − e3

SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS ■ 9

29. r2 − 6r + 25 = 0 ⇒ r = 3 ± 4i and the general solution is y = e3x (c1 cos 4x + c2 sin 4x). But 1 = y(0) = c1 and 2 = y(π) = c1 e3π

⇒ c1 = 2/e3π , so there is no solution.

31. r2 + 4r + 13 = 0 ⇒ r = −2 ± 3i and the general solution is y = e−2x (c1 cos 3x + c2 sin 3x). But 2 = y(0) = c1 and 1 = y π2 = e−π (−c2 ), so the solution to the boundary-value problem is y = e−2x (2 cos 3x − eπ sin 3x).

33. (a) Case 1 (λ = 0): y 00 + λy = 0 ⇒ y 00 = 0 which has an auxiliary equation r2 = 0 ⇒ r = 0 ⇒

y = c1 + c2 x where y(0) = 0 and y(L) = 0. Thus, 0 = y(0) = c1 and 0 = y(L) = c2 L ⇒ c1 = c2 = 0.

Thus, y = 0.

√ Case 2 (λ < 0): y 00 + λy = 0 has auxiliary equation r2 = −λ ⇒ r = ± −λ (distinct and real since λ < 0) ⇒

√

y = c1 e √

0 = y(L) = c1 e

−λL √

Multiplying (∗) by e

√

−λx

+ c2 e− √ − −λL

+ c2 e

−λL

−λx

where y(0) = 0 and y(L) = 0. Thus, 0 = y(0) = c1 + c2 (∗) and

(†).

√ √ and subtracting (†) gives c2 e −λL − e− −λL = 0 ⇒ c2 = 0 and thus

c1 = 0 from (∗). Thus, y = 0 for the cases λ = 0 and λ < 0.

√ √ √ λ ⇒ y = c1 cos λ x + c2 sin λ x √ where y(0) = 0 and y(L) = 0. Thus, 0 = y(0) = c1 and 0 = y(L) = c2 sin λL since c1 = 0. Since we √ √ cannot have a trivial solution, c2 6= 0 and thus sin λ L = 0 ⇒ λ L = nπ where n is an integer

(b) y 00 + λy = 0 has an auxiliary equation r2 + λ = 0 ⇒ r = ±i

Thomson Brooks-Cole copyright 2007

⇒

λ = n2 π 2 /L2 and y = c2 sin(nπx/L) where n is an integer.