Solve: (36D^2 - 24D + 13)y = sin^2 x - e^-x + 2?

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First solve homogeneous equation (36D² − 24D + 13)y = 0

36r² − 24r + 13 = 0
r = (24±√(24²−4*36*13))/(2*36)
r = (24±√−1296)/72
r = (24±36i)/72
r = 1/3 ± i/2

y_h = e^(x/3) (c₁ sin(x/2) + c₂ cos(x/2))

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Next we find particular solution of non-homogeneous equation using Method of Undetermined Coefficients.

(36D²−24D+13)y = sin²x − e^(−x) + 2
(36D²−24D+13)y = 1/2 (1−cos(2x)) − e^(−x) + 2
(36D²−24D+13)y = −1/2 cos(2x)) − e^(−x) + 5/2

y_p = A sin(2x) + B cos(2x) + C e^(−x) + D
y_p' = 2A cos(2x) − 2B sin(2x) − C e^(−x)
y_p'' = −4A sin(2x) − 4B cos(2x) + C e^(−x)

Plug into differential equation:

36 (−4A sin(2x) − 4B cos(2x) + C e^(−x)) − 24(2A cos(2x) − 2B sin(2x) − C e^(−x)) + 13(A sin(2x) + B cos(2x) + C e^(−x) + D)
= −1/2 cos(2x)) − e^(−x) + 5/2

(−131A + 48B) sin(2x) + (−48A−131B) cos(2x) + (73C) e^(−x) + (13D)
= −1/2 cos(2x)) − e^(−x) + 5/2

−131A + 48B = 0
−48A−131B = −1/2
73C = −1
13D = 5/2

A = 24/19465, B = 131/38930, C = −1/73, D = 5/26

y_p = 24/19465 sin(2x) + 131/38930 cos(2x) − 1/73 e^(−x) + 5/26

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General solution:

y = y_h + y_p

y = e^(x/3) (c₁ sin(x/2) + c₂ cos(x/2)) + 24/19465 sin(2x) + 131/38930 cos(2x) − 1/73 e^(−x) + 5/26

sin^2(x) = [1 - cos(2x)]/2

(5/2) - [cos(2x)/2] - [e^(-x)]

36*y'' - 24y' + 13y = (5/2) - [cos(2x)/2] - [e^(-x)]

y'' - (2y'/3) + (13y/36) = (5/72) - [cos(2x)/72] - [e^(-x)/36]

(u^2) - (2u/3) + 13/36 = 0; u = (1/3) +/- (i/2)

y_h = c_1*(e^(x/3))*cos(x/2) + c_2*(e^(x/3))*sin(x/2)

y_p = A*sin(2x) + B*cos(2x) + C*(e^(-x)) + D

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