SECTION 5.3

HOMOGENEOUS EQUATIONS WITH CONSTANT COEFFICIENTS

This is a purely computational section devoted to the single most widely applicable type ofhigher order differential equations — linear ones with constant coefficients. In Problems 1-20,we write first the characteristic equation and its list of roots, then the corresponding generalsolution of the given differential equation. Explanatory comments are included only when thesolution of the characteristic equation is not routine.

1. ( )( )2 4 2 2 0; 2, 2; ( )r r r r y x− = − + = = − = c1e2x + c2e

-2x

2. ( )22 3 2 3 0; 0, 3 / 2; ( )r r r r r y x− = − = = = c1 + c2e3x/2

3. ( )( )2 3 10 5 2 0; 5, 2; ( )r r r r r y x+ − = + − = = − = c1e2x + c2e

-5x

4. ( )( )22 7 3 2 1 3 0; 1/ 2, 3; ( )r r r r r y x− + = − − = = = c1ex/2 + c2e

3x

5. ( )22 6 9 3 0; 3, 3; ( )r r r r y x+ + = + = = − − = c1e-3x + c2xe-3x

6. ( )2 5 5 0; 5 5 / 2r r r+ + = = − ±

y(x) = e-5x/2[c1exp(x 5 /2) + c2exp(-x 5 /2)]

7. ( )224 12 9 2 3 0; 3 / 2, 3 / 2; ( )r r r r y x− + = − = = − − = c1e3x/2 + c2xe3x/2

8. ( )2 6 13 0; 6 16 / 2 3 2 ; ( )r r r i y x− + = = ± − = ± = e3x(c1cos 2x + c2sin 2x)

9. ( )2 8 25 0; 8 36 / 2 4 3 ; ( )r r r i y x+ + = = − ± − = − ± = e-4x(c1cos 3x + c2sin 3x)

10. ( )4 3 35 3 5 3 0; 0, 0, 0, 3 / 5; ( )r r r r r y x+ = + = = − = c1 + c2x + c3x2 + c4e

-3x/5

11. ( )24 3 2 28 16 4 0; 0, 0, 4, 4; ( )r r r r r r y x− + = − = = = c1 + c2x + c3e4x + c4xe4x

12. ( )34 3 23 3 1 0; 0,1,1,1; ( )r r r r r r r y x− + − = − = = = c1 + c2ex + c3xex + c4x

2ex

13. ( )23 29 12 4 3 2 0; 0, 2 / 3, 2 / 3r r r r r r+ + = + = = − −y(x) = c1 + c2e

-2x/3 + c3xe-2x/3

14. ( )( )4 2 2 24 3 1 1 4 0; 1,1, 2r r r r r i+ − = − + = = − ±y(x) = c1e

x + c2e-x + c3cos 2x + c4sin 2x

15. ( )24 2 2 2 24 8 16 4 ( 2) ( 2) 0; 2, 2, 2, 2r r r r r r− + = − = − + = = − −y(x) = c1e

2x + c2xe2x + c3e-2x + c4xe-2x

16. ( )24 2 218 81 9 0; 3 , 3r r r r i i+ + = + = = ± ±y(x) = (c1 + c2x)cos 3x + (c3 + c4x)sin 3x

17. 4 2 2 26 11 4 (2 1)(3 4) 0; / 2, 2 / 3,r r r r r i i+ + = + + = = ± ±y(x) = c1cos(x / 2 ) + c2sin(x/ 2 ) + c3cos(2x/ 3 ) + c4sin(2x/ 3 )

18. ( )( )4 2 216 4 4 0; 2, 2, 2r r r r i− = − + = = − ±y(x) = c1e

2x + c2e-2x + c3cos 2x + c4sin 2x

19. ( ) ( ) ( )( )23 2 2 21 1 1 1 1 0; 1, 1, 1;r r r r r r r r r+ − − = − + − = − + = = − −y(x) = c1e

x + c2e-x + c3xe-x

20. r4 + 2r3 + 3r2 + 2r + 1 = (r2 + r + 1)2 = 0; ( ) ( )1 3 / 2, 1 3 / 2i i− ± − ±

y = e-x/2(c1 + c2x)cos(x 3 /2) + e-x/2(c3 + c4x)sin(x 3 /2)

21. Imposition of the initial conditions (0) 7, (0) 11y y′= = on the general solution3

1 2( ) x xy x c e c e= + yields the two equations 1 2 1 27, 3 11c c c c+ = + = with solution

1 25, 2.c c= = Hence the desired particular solution is y(x) = 5ex + 2e3x.

22. Imposition of the initial conditions (0) 3, (0) 4y y′= = on the general solution

( ) ( )/ 31 2( ) cos / 3 sin / 3xy x e c x c x− = + yields the two equations

1 1 23, / 3 / 3 4c c c= − + = with solution 1 23, 5 3.c c= = Hence the desired particular

solution is ( ) ( )/ 3( ) 3cos / 3 5 3 sin / 3 .xy x e x x− = +

23. Imposition of the initial conditions (0) 3, (0) 1y y′= = on the general solution

( )31 2( ) cos 4 sin 4xy x e c x c x= + yields the two equations 1 1 23, 3 4 1c c c= + = with

solution 1 23, 2.c c= = − Hence the desired particular solution isy(x) = e3x(3 cos 4x - 2 sin 4x).

24. Imposition of the initial conditions (0) 1, (0) 1, (0) 3y y y′ ′′= = − = on the general

solution 2 / 21 2 3( ) x xy x c c e c e−= + + yields the three equations

1 2 3 2 3 2 31, / 2 2 1, / 4 4 3c c c c c c c+ + = − + = − + =

with solution 1 2 37 / 2, 1/ 2, 4.c c c= − = = Hence the desired particular solution is

y(x) = (-7 + e2x + 8e-x/2)/2.

25. Imposition of the initial conditions (0) 1, (0) 0, (0) 1y y y′ ′′= − = = on the general

solution 2 /31 2 3( ) xy x c c x c e−= + + yields the three equations

1 3 2 3 31, 2 / 3 0, 4 / 9 1c c c c c+ = − − = =

with solution 1 2 313 / 4, 3 / 2, 9 / 4.c c c= − = = Hence the desired particular solution is

y(x) = (-13 + 6x + 9e-2x/3)/4.

26. Imposition of the initial conditions (0) 1, (0) 1, (0) 3y y y′ ′′= = − = on the general

solution 5 51 2 3( ) x xy x c c e c x e− −= + + yields the three equations

1 2 2 3 2 33, 5 4, 25 10 5c c c c c c+ = − + = − =

with solution 1 2 324 / 5, 9 / 5, 5.c c c= = − = − Hence the desired particular solution is

y(x) = (24 - 9e-5x - 25xe-5x)/5.

27. First we spot the root r = 1. Then long division of the polynomial 3 23 4r r+ −by r - 1 yields the quadratic factor 2 24 4 ( 2)r r r+ + = + with rootsr = -2, -2. Hence the general solution is y(x) = c1e

x + c2e-2x + c3xe-2x.

28. First we spot the root r = 2. Then long division of the polynomial 2r3 - r2 - 5r - 2by the factor r - 2 yields the quadratic factor 2r2 + 3r + 1 = (2r + 1)(r + 1) with rootsr = -1, -1/2. Hence the general solution is y(x) = c1e

2x + c2e-x + c3e-

x/2.

29. First we spot the root r = –3. Then long division of the polynomial 3 27r + by

r + 3 yields the quadratic factor 2 3 9r r− + with roots ( )3 1 3 / 2.r i= ± Hence the

general solution is y(x) = c1e-3x + e3x/2[c2cos(3x 3 /2) + c3 sin(3x 3 /2)].

30. First we spot the root r = -1. Then long division of the polynomial

r4 - r3 + r2 - 3r - 6

by r + 1 yields the cubic factor r3 - 2r2 + 3r - 6. Next we spot the root r = 2, and

another long division yields the quadratic factor r2 + 3 with roots r = 3i± . Hence

the general solution is y(x) = c1e-x + c2e

2x + c3cos x 3 + c4sin x 3 .

31. The characteristic equation r3 + 3r2 + 4r - 8 = 0 has the evident root r = 1, and longdivision then yields the quadratic factor r2 + 4r + 8 = (r + 2)2 + 4 corresponding to thecomplex conjugate roots -2 ± 2 i. Hence the general solution is

y(x) = c1ex + e-2x(c2cos 2x + c3sin 2x).

32. The characteristic equation r4 + r3 - 3r2 - 5r - 2 = 0 has root r = 2 that is readilyfound by trial and error, and long division then yields the factorization

(r - 2)(r + 1)3 = 0.

Thus we obtain the general solution y(x) = c1e2x + (c2 + c3x + c4x

2)e-x.

33. Knowing that y = e3x is one solution, we divide the characteristic polynomialr3 + 3r2 - 54 by r - 3 and get the quadratic factor

r2 + 6r + 18 = (r + 3)2 + 9.

Hence the general solution is y(x) = c1e3x + e-3x(c2cos 3x + c3sin 3x).

34. Knowing that y = e2x/3 is one solution, we divide the characteristic polynomial3r3 - 2r2 + 12r - 8 by 3r - 2 and get the quadratic factor r2 + 4. Hence the generalsolution is

y(x) = c1e2x/3 + c2cos 2x + c3sin 2x.

35. The fact that y = cos 2x is one solution tells us that r2 + 4 is a factor of thecharacteristic polynomial

6r4 + 5r3 + 25r2 + 20r + 4.

Then long division yields the quadratic factor 26 5 1 (3 1)(2 1)r r r r+ + = + + with roots1/ 2, 1/ 3.r = − − Hence the general solution is

y(x) = c1e-x/2 + c2e-

x/3 + c3cos 2x + c4sin 2x

36. The fact that y = e-xsin x is one solution tells us that (r + 1)2 + 1 = r2 + 2r + 2is a factor of the characteristic polynomial

9r3 + 11r2 + 4r - 14.

Then long division yields the linear factor 9r - 7. Hence the general solution is

y(x) = c1e7x/9 + e-x(c2cos x + c3sin x).

37. The characteristic equation is r r r r4 3 3 1 0− = − =( ) , so the general solution isy x A Bx Cx Dex( ) = + + +2

. Imposition of the given initial conditions yields theequations

18, 12, 2 13, 7A D B D C D D+ = + = + = =

with solution 11, 5, 3, 7.A B C D= = = = Hence the desired particular solution is

y x x x ex( ) = + + +11 5 3 72.

38. Given that r = 5 is one characteristic root, we divide (r – 5) into the characteristicpolynomial 3 25 100 500r r r− + − and get the remaining factor 2 100r + . Thus thegeneral solution is

y x Ae B x C xx( ) cos sin= + +5 10 10 .

Imposition of the given initial conditions yields the equations

0, 5 10 10, 25 100 250A B A C A B+ = + = − =

with solution 2, 2, 0.A B C= = − = Hence the desired particular solution is

y x e xx( ) cos= −2 2 105.

39. 3 3 2( 2) 6 12 8r r r r− = − + − , so the differential equation is

′′′ − ′′ + ′ − =y y y y6 12 8 0 .

40. 2 3 2( 2)( 4) 2 4 8r r r r r− + = − + − , so the differential equation is

′′′ − ′′ + ′ − =y y y y2 4 8 0 .

41. 2 2 4( 4)( 4) 16r r r+ − = − , so the differential equation is (4) 16 0y y− = .

42. 2 3 6 4 2( 4) 12 48 64r r r r+ = + + + , so the differential equation is

y y y y( ) ( )6 412 48 64 0+ + ′′ + = .

44. (a) x = i, -2i (b) x = -i, 3i

45. The characteristic polynomial is the quadratic polynomial of Problem 44(b). Hence thegeneral solution is

31 2 1 2( ) (cos sin ) (cos3 sin 3 ).ix ixy x c e c e c x i x c x i x−= + = − + +

46. The characteristic polynomial is 2 6 ( 2 )( 3 )r ir r i r i− + = + − so the general solution is

3 21 2 1 2( ) (cos3 sin 3 ) (cos 2 sin 2 ).ix ixy x c e c e c x i x c x i x−= + = + + −

47. The characteristic roots are 2 2 3 (1 3)r i i= ± − + = ± + so the general solution is

( ) ( )(1 3) (1 3 )1 2 1 2( ) cos 3 sin 3 cos 3 sin 3i x i x x xy x c e c e c e x i x c e x i x+ − + −= + = + + −

48. The general solution is y(x) = Aex + Beαx + Ceßx where α = (-1 + i 3 )/2 and

β = (-1 - i 3 )/2. Imposition of the given initial conditions yields the equations

A B C

A B C

A B C

+ + =+ + =

+ + =

1

0

02 2

α β

α β

that we solve for A = B = C = 1/3. Thus the desired particular solution is given

by ( )( 1 3) / 2 ( 1 3 ) / 213( ) ,x i x xy x e e e− + − −= + + which (using Euler's relation) reduces to the

given real-valued solution.

49. The general solution is y = Ae2x + Be-x + C cos x + D sin x. Imposition of the giveninitial conditions yields the equations

A B C

A B D

A B C

A B D

+ + =− + =+ − =− − =

02 0

4 0

8 30

that we solve for A = 2, B = -5, C = 3, and D = -9. Thus

y(x) = 2e2x - 5e-x + 3 cos x - 9 sin x.

50. If 0x > then the differential equation is 0y y′′ + = with general solutioncos sin .y A x B x= + . But if 0x < it is 0y y′′ − = with general solution

cosh sin .y C x D x= + To satisfy the initial conditions 1 1(0) 1, (0) 0y y′= = we choose

1 and 0.A C B D= = = = But to satisfy the initial conditions 2 2(0) 0, (0) 1y y′= = wechoose 0 and 1.A C B D= = = = The corresponding solutions are defined by

1 2

cos if 0, sin if 0,( ) ( )

cosh if 0; sinh if 0.

x x x xy x y x

x x x x

≥ ≥ = = ≤ ≤