# Differential equation Fourier series

I am having trouble figuring out how to solve this problem using the Fourier series method. I'm not allowed to use methods of undetermined coefficient or variation of parameter. $$y'' + 2y' + y = 25\cos(2t)$$ The general solution is obvious but I'm unable to find the particular solution. I tried to use the Fourier series representation of $25\cos(2t)$ but I get 0 for my $a_0$, $a_n$, and $b_n$. The only value that's not 0 is when my $n=2$, then my value becomes $25$.

In the book I use, this is called the steady state solution and Theorem 1 is called Forced Oscillation. I'm unable to apply the theorem because I am unable to obtain the Fourier coefficients.

Solution: $$ce^{-t} + dte^{-t} -3\cos(2t)+4\sin(2t)$$

• If $a_n$ and $b_n$ are the coefficients of the $\sin$ and $\cos$ terms in the Fourier series expansion, it makes sense that $a_n = b_n = 0$ for all $n\ne 2$, does it not? Commented Sep 24, 2017 at 21:54

The idea is to assume that $$y(x) = \sum_{n=0}^\infty \left[ a_n \sin(nx) + b_n \cos(nx)\right]$$ Now compute $y', y''$ and equate it to the right-hand side to get a recurrence relationship for $a_n$ and $b_n$ which you can solve.