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y" + 3y' = 72sin(3t) + 36cos(3t)

where y(0) = 6 and y'(0) = 9

I first found the solution to the homogeneous eq:

the roots (R^2 + 3R = 0) are R = 0, -3

which gives the family of solutions:

y = a(1) + be^(-3t)

and y' = -3be^(-3t)

using the initial conditions (maybe Im not supposed to use them here?)

I find a = 9, b = -3

For the inhomogeneous eq:

I try (guess)

y = Asin(3t) + Bsin(3t)

y' = 3Acos(3t) - 3Bsin(3t)

y" = -9Asin(3t) - 9Bcos(3t)

substitute those values into the original equation (left hand side) I find

sin3t(-9A-9B) + cos3t(-9B-9A) = 72sin3t + 36cos3t

therefore

-9A - 9B = 72

-9B - 9A = 36

giving B = -6, A = -2

Therefore I get the solution:

y = 9 - 3e^(-3t) - 2sin3t - 6cos3t

What did I do wrong (this answer is incorrect)

Thanks