# Thread: non-homogeneous equation, method of undetermined coefficients

1. ## non-homogeneous equation, method of undetermined coefficients

Find the solution of:

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

2. using the initial conditions (maybe Im not supposed to use them here?)
I find a = 9, b = -3
don't use the initial conditions here, use it at the end

$\displaystyle y(x)=-2\sin(3t)-6\cos(3t)+a+be^{-3t}$

so

$\displaystyle y(0)=0=-6+a+b$ so $\displaystyle a=6-b$ and
$\displaystyle y^{\prime}(t)=-6\cos(3t)+18\sin(3t)-3be^{-3t}$ and
$\displaystyle y^{\prime}(0)=9=-6-3b$ so $\displaystyle b=-5$ and $\displaystyle a=11$

then finally you end up with

$\displaystyle y(x)=-2\sin(3t)-6\cos(3t)+11-5e^{-3t}$