Find the general solution of each of the following different
Solution
First we find solution to associated homogeneous equation y\'\'  2y\' = 0
 
 We find solutions of characteristic equation:
 r²  2r = 0
 r (r  2) = 0
 r = 0, r = 2
 
 y_h = c e^(2x) + c e^(0x)
 y_h = c e^(2x) + c
 
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 Find particular solution to non-homogeneous equation: y\'\'  2y\' = 3  25 cos(2x)
 using method of undetermined coefficients.
 
 Now usually, we would set y_p = A + B sin(2x) + C cos(2x). But since constant term is part of solution to homogeneous solution, we multiply A by x
 
 y_p = Ax + B sin(2x) + C cos(2x)
 y_p\' = A + 2B cos(2x)  2C sin(2x)
 y_p\'\' = 4B sin(2x)  4C cos(2x)
 
 y_p\'\'  2 y_p\' = 3  25 cos(2x)
 4B sin(2x)  4C cos(2x)  2A  4B cos(2x) + 4C sin(2x) = 3  25 cos(2x)
 2A + (4B + 4C) sin(2x)  (4B + 4C) cos(2x) = 3  25 cos(2x)
 
 Matching coefficients we get
 2A = 3
 4B + 4C = 0
 4B + 4C = 25
 
 Solving, we get: A = 3/2, B = 25/8, C = 25/8
 
 y_p = 3/2 x + 25/8 sin(2x) + 25/8 cos(2x)
 
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 General solution:
 
 y = y_h + y_p
 
 y = c e^(2x) + c  3/2 x + 25/8 sin(2x) + 25/8 cos(2x)

