## SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS

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

Px

d 2y dy  Rxy  Gx 2  Qx 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 Gx  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

Px

d 2y dy  Rxy  0  Qx dx 2 dx

If Gx  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 y1x and y2x are both solutions of the linear homogeneous equation (2) and c1 and c2 are any constants, then the function

yx  c1 y1x  c2 y2x is also a solution of Equation 2. Proof Since y1 and y2 are solutions of Equation 2, we have

Pxy1  Qxy1  Rxy1  0 and

Pxy2  Qxy2  Rxy2  0

Therefore, using the basic rules for differentiation, we have Pxy  Qxy  Rxy  Pxc1 y1  c2 y2  Qxc1 y1  c2 y2  Rxc1 y1  c2 y2  Pxc1 y1  c2 y2  Qxc1 y1  c2 y2   Rxc1 y1  c2 y2  c1Pxy1  Qxy1  Rxy1  c2 Pxy2  Qxy2  Rxy2  c10  c20  0

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 tx  5x 2 are linearly dependent, but f x  e x and tx  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 Px is never 0, then the general solution is given by

yx  c1 y1x  c2 y2x 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  ce 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 tx  e3 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  2r  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 e3x

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 (1s13 ) x6  c2 e (1s13 ) x6 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  by2  cy2  a2re rx  r 2xe rx   be rx  rxe rx   cxe rx  2ar  be rx  ar 2  br  cxe rx  0e rx   0xe 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

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

2r  32  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  e3x2 and tx  xe3x2 in Example 3 and some other members of the family of solutions. Notice that all of them approach 0 as x l . ■ ■

y  c1 e3x2  c2 xe3x2 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,   b2a,   s4ac  b 22a.] 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 ix  C2 e ix  C1 e  xcos  x  i sin  x  C2 e  xcos  x  i sin  x  e  x C1  C2  cos  x  iC1  C2  sin  x  e  xc1 cos  x  c2 sin  x where c1  C1  C2 , c2  iC1  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  xc1 cos  x  c2 sin  x

■ ■ Figure 3 shows the graphs of the solutions in Example 4, f x  e 3 x cos 2x and tx  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  s16   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 3xc1 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

yx 0   y0

yx 0   y1

where y0 and y1 are given constants. If P, Q, R, and G are continuous on an interval and Px  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.

y0  1

y0  0

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

tion is yx  c1 e 2x  c2 e3x

20

Differentiating this solution, we get yx  2c1 e 2x  3c2 e3x To satisfy the initial conditions we require that _2

2

0

FIGURE 4

12

y0  c1  c2  1

13

y0  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 e3x 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 sinx   where tan  23

y  y  0

y0  2

y0  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

yx  c1 cos x  c2 sin x

5

yx  c1 sin x  c2 cos x

Since the initial conditions become _2π

y0  c1  2

y0  c2  3

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

yx  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 yx 0   y0

yx 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

y0  1

y1  3

SOLUTION The auxiliary equation is

r 2  2r  1  0

or

r  12  0

whose only root is r  1. Therefore, the general solution is yx  c1 ex  c2 xex

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.

y0  c1  1

5

y1  c1 e1  c2 e1  3 The first condition gives c1  1, so the second condition becomes

_1

5

e1  c2 e1  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  ex  3e  1xex 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  xc1 cos  x  c2 sin  x

r1, r2 real and distinct r1  r2  r r1, r2 complex:   i

20. 2y  5y  3y  0,

S

21. y  16y  0,

Solve the differential equation.

y  2

y0  2,

y0  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 y0 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,

y0  3,

y0  1,

19. 4 y  4 y  y  0,

y0  4

y0  3

y0  1,

y0  1.5

y1  2

y0  1,

y3  0

y  5

y0  2,

31. y  4y  13y  0,

y  2

y0  1, y0  1,

y1  0 y 2  1

y0  2,

32. 9y  18y  10 y  0,

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

y  4

y0  1,

30. y  6y  9y  0,

y0  0, ■

y  1 ■

33. Let L be a nonzero real number.

d y dy 2  5y  0 dx 2 dx ■

y1  1 ■

2

y0  3,

29. y  6y  25y  0, ■

y1  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  y0 dt 2 dt

25. 4 y  y  0,

2

13.

25–32

d 2y dy 6  4y  0 dt 2 dt

12.

y4  4

y  0,

23. y  2y  2y  0,

10. 9y  4y  0

y0  4

22. y  2y  5y  0,

2. y  4y  8y  0

11.

y4  3,

1. y  6y  8y  0

9. 4y  y  0

y0  1,

(a) Show that the boundary-value problem y  y  0, y0  0, yL  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 yx is a solution

of the differential equation ay  by  cy  0, show that lim x l yx  0.

SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS ■ 7

17. y  2e3x2  ex

1. y  c1 e

4x

 c2 e

2x

5. y  c1 e  c2 xe x

x4

9. y  c1  c2 e

x

4x

3. y  e

c1 cos 5x  c2 sin 5x

7. y  c1 cosx2  c2 sinx2 11. y  c1 e (1s2 )t  c2 e (1s2 )t

13. y  et2 [c1 cos(s3t2)  c2 sin(s3t2)] g 15. 40

_0.2

19. y  e x/2  2xe x2 21. y  3 cos 4x  sin 4x 23. y  ex2 cos x  3 sin x 25. y  3 cos( 2 x)  4 sin( 2 x) 1

1

e x3 e 2x  e 1 1  e3

f

27. y 

1

29. No solution

3

31. y  e2x 2 cos 3x  e sin 3x _40

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

33. (b)   n 2 2L2, n a positive integer; y  C sinn xL

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 .

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