SECTION 5.5

UNDETERMINED COEFFICIENTSAND VARIATION OF PARAMETERS

The method of undetermined coefficients is based on "educated guessing". If we can guesscorrectly the form of a particular solution of a nonhomogeneous linear equation with constantcoefficients, then we can determine the particular solution explicitly by substitution in the givendifferential equation. It is pointed out at the end of Section 5.5 that this simple approach is notalways successful — in which case the method of variation of parameters is available if acomplementary function is known. However, undetermined coefficients does turn out to workwell with a surprisingly large number of the nonhomogeneous linear differential equations thatarise in elementary scientific applications.

In each of Problems 1-20 we give first the form of the trial solution ytrial, then the equations inthe coefficients we get when we substitute ytrial into the differential equation and collect liketerms, and finally the resulting particular solution yp.

1. 3trial ; 25 1;xy Ae A= = yp = (1/25)e3x

2. trial ; 2 4, 2 3;y A Bx A B B= + − − = − = yp = -(5 + 6x)/4

3. trial cos3 sin 3 ; 15 3 0, 3 15 2;y A x B x A B A B= + − − = − = yp = (cos 3x - 5 sin 3x)/39

4. trial ; 9 12 0, 9 3;x xy Ae B x e A B B= + + = = yp = (-4ex + 3xex)/9

5. First we substitute sin2x = (1 - cos 2x)/2 on the right-hand side of the differential

equation. Then:

trial cos 2 sin 2 ; 1/ 2, 3 2 1/ 2, 2 3 0;y A B x C x A B C B C= + + = − + = − − − = yp = (13 + 3 cos 2x - 2 sin 2x)/26

6. 2trial ; 7 4 4 0, 7 8 0, 7 1;y A B x C x A B C B C C= + + + + = + = =

yp = (4 - 56x + 49x2)/343

7. First we substitute sinh x = (ex - e-x)/2 on the right-hand side of the differential

equation. Then:

trial ; 3 1/ 2, 3 1/ 2;x xy Ae B e A B−= + − = − = − yp = (e-x - ex)/6 = -(1/3)sinh x

8. First we note that cosh 2x is part of the complementary function

c 1 2cosh 2 sinh 2 .y c x c x= + . Then:

( )trial cosh 2 sinh 2 ; 4 0, 4 1;y x A x B x A B= + = = yp = (1/4)x sinh 2x

9. First we note that ex is part of the complementary function yc = c1ex + c2e

-3x. Then:

trial ( ) ; 3 0, 4 2 0, 8 1;xy A x B Cx e A B C C= + + − = + = = yp = -(1/3) + (2x2 - x)ex/16.

10. First we note the duplication with the complementary function c 1 2cos3 sin 3 .y c x c x= +Then:

( )trial cos3 sin 3 ; 6 2, 6 3;y x A x B x B A= + = − = yp = (2x sin 3x - 3x cos 3x)/6

11. First we note the duplication with the complementary function

c 1 2 3cos 2 sin 2 .y c x c x c x= + + Then:

( )trial ; 4 1, 8 3;y x A B x A B= + = − = yp = (3x2 - 2x)/8

12. First we note the duplication with the complementary function

c 1 2 3cos sin .y c x c x c x= + + Then:

( )trial cos sin ; 2, 2 0, 2 1;y Ax x B x C x A B C= + + = − = − = − yp = 2x + (1/2)x sin x

13. ( )trial cos sin ; 7 4 0, 4 7 1;xy e A x B x A B A B= + + = − + = yp = ex(7 sin x - 4 cos x)/65

14. First we note the duplication with the complementary function

( ) ( )c 1 2 3 4 .x xy c c x e c c x e−= + + + Then:

( )2trial ; 8 24 0, 24 1;xy x A B x e A B B= + + = = yp = (-3x2ex + x3ex)/24

15. This is something of a trick problem. We cannot solve the characteristic equation5 45 1 0r r+ − = to find the complementary function, but we can see that it contains no

constant term (why?). Hence the trial solution trialy A= leads immediately to the

particular solution yp = -17.

16. ( )2 3trial ;xy A B Cx Dx e= + + +

9 5, 18 6 2 0, 18 12 0, 18 2;A B C D C D D= + + = + = = yp = (45 + e3x - 6xe3x + 9x2e3x)/81

17. First we note the duplication with the complementary function c 1 2cos sin .y c x c x= +Then:

[ ]trial ( ) cos ( )sin ;y x A Bx x C Dx x= + + +2 2 0, 4 1, 2 2 1, 4 0;B C D A D B+ = = − + = − = yp = (x2sin x - x cos x)/4

18. First we note the duplication with the complementary function 2 2

c 1 2 3 4 .x x x xy c e c e c e c e− −= + + + Then:

ytrial = x(Aex) + x(B + Cx) e2x; 6 1, 12 38 0, 24 1;A B C C− = + = = −yp = -(24xex - 19xe2x + 6x2e2x)/144

19. First we note the duplication with the part 1 2c c x+ of the complementary function

(which corresponds to the factor 2r of the characteristic polynomial). Then:

ytrial = x2(A + Bx + Cx2); 4 12 1, 12 48 0, 24 3;A B B C C+ = − + = =yp = (10x2 - 4x3 + x4)/8

20. First we note that the characteristic polynomial 3r r− has the zero r = 1 correspondingto the duplicating part ex of the complementary function. Then:

( )trial ; 7, 3 1;xy A x Be A B= + − = = yp = -7 + (1/3)xex

In Problems 21-30 we list first the complementary function yc, then the initially proposed trialfunction yi, and finally the actual trial function yp in which duplication with thecomplementary function has been eliminated.

21. ( )c 1 2cos sin ;xy e c x c x= + ( )i cos sinxy e A x B x= +

( )p cos sinxy x e A x B x= ⋅ +

22. ( ) ( ) ( )2c 1 2 3 4 5 ;x xy c c x c x c e c e−= + + + + ( ) ( )2

ixy A Bx Cx D e= + + +

( ) ( )3 2p

xy x A Bx Cx x D e= ⋅ + + + ⋅

23. c 1 2cos sin ;y c x c x= + i ( ) cos 2 ( )sin 2y A Bx x C Dx x= + + + [ ]p ( ) cos 2 ( )sin 2y x A Bx x C Dx x= ⋅ + + +

24. 3 4c 1 2 3 ;x xy c c e c e−= + + 3

i ( ) ( ) xy A Bx C Dx e−= + + + 3

p ( ) ( ) xy x A Bx x C Dx e−= ⋅ + + ⋅ +

25. 2c 1 2 ;x xy c e c e− −= + 2

i ( ) ( )x xy A Bx e C Dx e− −= + + + 2

p ( ) ( )x xy x A Bx e x C Dx e− −= ⋅ + + ⋅ +

26. ( )3c 1 2cos 2 sin 2 ;xy e c x c x= + 3 3

i ( ) cos 2 ( ) sin 2x xy A Bx e x C Dx e x= + + +

3 3p ( ) cos 2 ( ) sin 2x xy x A Bx e x C Dx e x = ⋅ + + +

27. ( ) ( )c 1 2 3 3cos sin cos 2 sin 2y c x c x c x c x= + + +

( ) ( )i cos sin cos 2 sin 2y A x B x C x D x= + + +

( ) ( )p cos sin cos 2 sin 2y x A x B x C x D x = ⋅ + + +

28. ( ) ( )c 1 2 3 3cos3 sin 3y c c x c x c x= + + +

( ) ( )2 2i cos3 sin 3y A Bx Cx x D Ex Fx x= + + + + +

( ) ( )2 2p cos3 sin 3y x A Bx Cx x D Ex Fx x = ⋅ + + + + +

29. ( )2 2 2c 1 2 3 4 5

x x xy c c x c x e c e c e−= + + + + ; ( ) 2 2i

x x xy A Bx e C e D e−= + + +

( ) ( ) ( )3 2 2p

x x xy x A Bx e x C e x D e−= ⋅ + + ⋅ + ⋅

30. ( ) ( )c 1 2 3 4x xy c c x e c c x e−= + + +

( ) ( )2 2i p cos siny y A Bx Cx x D Ex Fx x= = + + + + +

In Problems 31-40 we list first the complementary function yc, the trial solution ytr for themethod of undetermined coefficients, and the corresponding general solution yg = yc + yp whereyp results from determining the coefficients in ytr so as to satisfy the given nonhomogeneousdifferential equation. Then we list the linear equations obtained by imposing the given initialconditions, and finally the resulting particular solution y(x).

31. c 1 2 tr g 1 2cos 2 sin 2 ; ; cos 2 sin 2 / 2y c x c x y A Bx y c x c x x= + = + = + + ;

1 21, 2 1/ 2 2; ( ) cos 2 (3 / 4)sin 2 / 2c c y x x x x= + = = + +

32. 2 2c 1 2 tr g 1 2; ; / 6x x x x x xy c e c e y Ae y c e c e e− − − −= + = = + + ;

( )21 2 1 21/ 6 0, 2 1/ 6 3; ( ) 15 16 / 6x x xc c c c y x e e e− −+ + = − − + = = − +

33. c 1 2 trcos3 sin 3 ; cos 2 sin 2y c x c x y A x B x= + = +

g 1 2cos3 sin 3 (1/ 5)sin 2y c x c x x= + +

( )1 21, 3 2 / 5 0; ( ) 15cos3 2sin 3 3sin 2 /15c c y x x x x= + = = − +

34. ( )c 1 2 trcos sin ; cos siny c x c x y x A x B x= + = ⋅ +

g 1 2cos sin (1/ 2) sin 2y c x c x x x= + +

1 21, 1; ( ) cos sin (1/ 2) sinc c y x x x x x= = − = − +

35. ( )c 1 2 trcos sin ;xy e c x c x y A B x= + = +

( )g 1 2cos sin 1 / 2xy e c x c x x= + + +

( )1 1 21 3, 1/ 2 0; ( ) 4cos 5sin / 2 1 / 2xc c c y x e x x x+ = + + = = − + +

36. ( )2 2 2 2c 1 2 3 4 tr;x xy c c x c e c e y x A B x C x−= + + + = ⋅ + +

2 2 2 4g 1 2 3 4 /16 / 48x xy c c x c e c e x x−= + + + − −

1 3 4 2 3 4 3 4 3 41, 2 2 1, 4 4 1/8 1, 8 8 1c c c c c c c c c c+ + = − + = + − = − − + = −

( )2 2 2 4( ) 234 240 9 33 12 4 /192x xy x x e e x x−= + − − − −

37. ( ) ( )2c 1 2 3 tr;x x xy c c e c x e y x A x B Cx e= + + = ⋅ + ⋅ +

2 3g 1 2 3 / 2 / 6x x x xy c c e c x e x x e x e= + + + − +

1 2 2 3 2 30, 1 0, 2 1 1c c c c c c+ = + + = + − =

( )2 3( ) 4 24 18 3 / 6xy x x e x x x= + + − + − +

38. ( )c 1 2 trcos sin ; cos3 sin 3xy e c x c x y A x B x−= + = +

( ) ( )g 1 2cos sin 6cos3 7sin 3 /85xy e c x c x x x−= + − +

1 1 26 /185 2, 21/ 85 0c c c− = − + − =

( ) ( )( ) 176cos 197sin 6cos3 7sin 3 / 85xy x e x x x x− = + − +

39. ( ) ( )2c 1 2 3 tr;x xy c c x c e y x A Bx x Ce− −= + + = ⋅ + + ⋅

2 3g 1 2 3 / 2 / 6x xy c c x c e x x x e− −= + + − + +

1 3 2 3 31, 1 0, 3 1c c c c c+ = − + = − =

( )2 3( ) 18 18 3 / 6 (4 ) xy x x x x x e−= − + − + + +

40. c 1 2 3 4 trcos sin ;x xy c e c e c x c x y A−= + + + =

g 1 2 3 4cos sin 5x xy c e c e c x c x−= + + + −

1 2 3 1 2 4 1 2 3 1 2 45 0, 0, 0, 0c c c c c c c c c c c c+ + − = − + + = + − = − + − =

( )( ) 5 5 10cos 20 / 4x xy x e e x−= + + −

41. The trial solution 2 3 4 5try A Bx Cx Dx Ex Fx= + + + + + leads to the equations

2 2 6 24 0

2 2 6 24 120 0

2 3 12 60 0

2 4 20 0

2 5 0

2 8

A B C D E

B C D E F

C D E F

D E F

E F

F

− − − + =− − − − + =− − − − =− − − =− − =− =

that are readily solve by back-substitution. The resulting particular solution is

y(x) = –255 - 450x + 30x2 + 20x3 + 10x4 - 4x5

42. The characteristic equation 4 3 2 2 0r r r r− − − − = has roots 1, 2,r i= − ± so the

complementary function is 2c 1 2 3 4cos sin .x xy c e c e c x c x−= + + + We find that the

coefficients satisfy the equations

1 2 3

1 2 4

1 2 3

1 2 4

255 0

2 450 0

4 60 0

8 120 0

c c c

c c c

c c c

c c c

+ + − =− + + − =

+ − + =− + − + =

Solution of this system gives finally the particular solution c py y y= + where yp is the

particular solution of Problem 41 and

2c 10 35 210cos 390sin .x xy e e x x−= + + +

43. (a) 3cos3 sin 3 (cos sin )x i x x i x+ = +3 2 2 3cos 3 cos sin 3cos sin sinx i x x x x i x= + − −

When we equate real parts we get the equation

( )( )3 2 3cos 3 cos 1 cos 4cos 3cosx x x x x− − = −

and readily solve for 3 3 14 4cos cos cos3 .x x x= + The formula for 3sin x is derived

similarly by equating imaginary parts in the first equation above.

(b) Upon substituting the trial solution p cos sin cos3 sin 3y A x B x C x D x= + + +in the differential equation 3 1

4 44 cos cos3 ,y y x x′′ + = + we find that A = 1/4, B = 0,C = –1/20, D = 0. The resulting general solution is

y(x) = c1cos 2x + c2sin 2x + (1/4)cos x - (1/20)cos 3x.

44. We use the identity 1 12 2sin sin 3 cos 2 cos 4 ,x x x x= − and hence substitute the trial

solution p cos 2 sin 2 cos 4 sin 4y A x B x C x D x= + + + in the differential equation1 12 2cos 2 cos 4 .y y y x x′′ ′+ + = − We find that A = –3/26, B = 1/13, C = –14/482,

D = 2/141. The resulting general solution is

y(x) = e-x/2(c1 cos x 3 /2 + c2 sin x 3 /2)

+ (-3 cos 2x + 2 sin 2x)/26 + (-15 cos 4x + 4 sin 4x)/482.

45. We substitute

sin4x = (1 - cos 2x)2 /4

= (1 - 2 cos 2x + cos22x)/4 = (3 - 4 cos 2x + cos 4x)/8

on the right-hand side of the differential equation, and then substitute the trial solution p cos 2 sin 2 cos 4 sin 4 .y A x B x C x D x E= + + + + We find that A = –1/10, B = 0,

C = –1/56, D = 0, E = 1/24. The resulting general solution is

y = c1cos 3x + c2sin 3x + 1/24 - (1/10)cos 2x - (1/56)cos 4x.

46. By the formula for 3cos x in Problem 43, the differential equation can be written as

3 14 4cos cos3 .y y x x x x′′ + = +

The complementary solution is yc = c1cos x + c2sin x, so we substitute the trial solution

( ) ( ) ( ) ( )p cos sin cos3 sin 3 .y x A Bx x C Dx x E Fx x G Hx x = ⋅ + + + + + + +

We find that 3 /16, 0, 3 /16, 0, 1/ 32, 3 /128, 0.A B C D E F G H= = = = = = − = = Hencethe general solution is given by y = yc + y1 + y2 where

y1 = (3x cos x + 3x2sin x)/16 and y2 = (3 sin 3x - 4x cos 3x)/128.

In Problems 47-49 we list the independent solutions 1 2andy y of the associated homogeneous

equation, their Wronskian 1 2( , ),W W y y= the coefficient functions

2 11 2

( ) ( ) ( ) ( )( ) and ( )

( ) ( )

y x f x y x f xu x dx u x dx

W x W x= − =⌠ ⌠

⌡ ⌡

in the particular solution p 1 1 2 2y u y u y= + of Eq. (32) in the text, and finally yp itself.

47. y1 = e-2x, y2 = e-x, W = e-3x

u1 = -(4/3)e3x, u2 = 2e2x,

yp = (2/3)ex

48. y1 = e-2x, y2 = e4x, W = 6e2x

u1 = -x /2, u2 = -e-6x/12,

yp = -(6x + 1)e-2x /12

49. y1 = e2x, y2 = xe2x, W = e4x

u1 = -x2, u2 = 2x,

yp = x2e2x

50. The complementary function is y1 = c1cosh 2x + c2sinh 2x, so the Wronskian is

W = 2 cosh22x - 2 sinh22x = 2,

so when we solve Equations (31) simultaneously for 1 2and ,u u′ ′ integrate each andsubstitute in yp = y1u1 + y2u2, the result is

y x x x dx x x x dxp = − +� �(cosh ) (sinh )(sinh ) (sinh ) (cosh )(sinh )2 2 2 2 2 212

12 .

Using the identities 2 sinh2x = cosh 2x - 1 and 2 sinh x cosh x = sinh 2x, we evaluatethe integrals and find that

yp = (4x cosh 2x - sinh 4x cosh 2x + cosh 4x sinh 2x)/16,

yp = (4x cosh 2x - sinh 2x)/16.

51. 1 2cos 2 , sin 2 , 2y x y x W= = =

Liberal use of trigonometric sum and product identities yields

1 1(cos5 5cos ) / 20, (sin 5 5sin ) / 20u x x u x x= − = −

yp = -(1/4)(cos 2x cos x - sin 2x sin x) + (1/20)(cos 5x cos 2x + sin 5x sin 2x)

= -(1/5)cos 3x (!)

52. y1 = cos 3x, y2 = sin 3x, W = 3

1 1(6 sin 6 ) / 36, (1 cos 6 ) / 36u x x u x= − − = − +

yp = -(x cos 3x)/6

53. y1 = cos 3x, y2 = sin 3x, W = 3

1u′ = -(2/3)tan 3x, 2u′ = 2/3

yp = (2/9)[3x sin 3x + (cos 3x) ln cos3x ]

54. y1 = cos x, y2 = sin x, W = 1

1u′ = -csc x, 2u′ = cos x csc2x

yp = -1 - (cos x) ln csc cotx x−

55. y1 = cos 2x, y2 = sin 2x, W = 2

1u′ = -(1/2)sin2x sin 2x = -(1/4)(1 - cos 2x)sin 2x

2u′ = (1/2)sin2x cos 2x = (1/4)(1 - cos 2x)cos 2x

yp = (1 - x sin 2x)/8

56. y1 = e-2x, y2 = e2x, W = 4

u1 = -(3x - 1)e3x /36, u2 = -(x + 1)e-x /4

yp = -ex(3x + 2)/9

57. With y1 = x, y2 = x-1, and f(x) = 72x3, Equations (31) in the text take the form

x 1u′ + x-12u′ = 0,

1u′ - x-22u′ = 72x3.

Upon multiplying the second equation by x and then adding, we readily solve first for

1u′ = 36x3, so u1 = 9x4

and then

2u′ = -x21u′ = -36x5, so u2 = -6x6.

Then it follows that

yp = y1u1 + y2u2 = (x)(9x4) + (x-1)(-6x6) = 3x5.

58. Here it is important to remember that — for variation of parameters — the differentialequation must be written in standard form with leading coefficient 1. We thereforerewrite the given equation with complementary function yc = c1x

2 + c2x3 as

y″ - (4/x)y′ + (6/x2)y = x.

Thus f(x) = x, and W = x4, so simultaneous solution of Equations (31) as in Problem50 (followed by integration of 1u′ and 2u′ ) yields

y x x x x dx x x x x dx

x dx x x dx x x

p = − ⋅ ⋅ + ⋅ ⋅

= − + = −

− −� �

� �

2 3 4 3 2 4

2 3 31 1( / ) (ln ).

59. y1 = x2, y2 = x2 ln x,

W = x3, f(x) = x2

1u′ = -x ln x, 2u′ = x

yp = x4 /4

60. 1/ 21 ,y x= 3/ 2

2y x=

f(x) = 2x-2/3; W = x

5/ 61 12 / 5,u x= − 1/ 6

2 12u x−= −4 /3

p 72 / 5y x= −

61. y1 = cos(ln x), y2 = sin(ln x), W = 1/x,

f(x) = (ln x)/x2

u1 = (ln x)cos(ln x) - sin(ln x)

u2 = (ln x)sin(ln x) + cos(ln x)

yp = ln x (!)

62. y1 = x, y2 = 1 + x2,

W = x2 - 1, f(x) = 1

1u′ = (1 + x2)/(1 - x2), 2u′ = x /(x2 - 1)

yp = -x2 + x ln|(1 + x)/(1 - x)| + (1/2)(1 + x2)ln|1 - x2|

63. This is simply a matter of solving the equations in (31) for the derivatives

′ = − ′ =uy x f x

W xu

y x f x

W x12

21( ) ( )

( )

( ) ( )

( )and ,

integrating each, and then substituting the results in (32).

64. Here we have y x x y x x W x f x x1 2 1 2( ) cos , ( ) sin , ( ) , ( ) sin= = = = , so (33) gives

y x x x x dx x x x dx

x x dx x x x dx

x x x x x x

x x x x x

y x x x x

p

p

( ) (cos ) sin sin (sin ) cos sin

(cos ) ( cos ) (sin ) (sin ) cos

(cos )( sin cos ) (sin )(sin )

cos (sin )(cos sin )

( ) cos sin

= − ⋅ + ⋅

= − − + ⋅

= − − +

= − + += − +

� �

� �

2 2

1 2 2

2

2 2

But we can drop the term sin x because it satisfies the associated homogeneousequation 0.y y′′ + =