Fourier series solution of differential equation (Dirac Delta)

Question

I have been given the following:

$$y''(x)+\omega^2y(x)=s(x),$$

$$s(x)= \delta(x)-\delta(x-\frac{1}{2})$$ for $-\frac{1}{4}<x<\frac{3}{4}$. (Periodically repeating for x outside this interval.)

Question: Find the Fourier series solution y(x) of the above differential equation. I am completely lost, as I am unsure of how to expand s(x) into a Fourier series. Thanks :)

Answer 1

HINT: The Fourier series for a function $f(x)$ on the interval $[-L, L]$ is defined as $$f(x) \sim\frac {a_0}{2} + \sum_{n=1}^\infty{a_n \cos \frac{n \pi x}{L} + b_n \sin \frac{n \pi x}{L}}$$ where $$a_n = \int_{-L}^{L} f(x) \cos \frac{n \pi x}{L}$$ and $$b_n = \int_{-L}^{L} f(x) \sin \frac{n \pi x}{L}$$ And if $f(x)$ is continuous on $(-L, L)$ then this is an equality.

Answer 2

So, write $$y(x)=\sum_{n\in\mathbb{Z}}y_{n}e^{2\pi{inx}}$$ Plug into the equation $$\sum_{n\in\mathbb{Z}}\big(\omega^{2}-(2\pi{n})^{2}\big)y_{n}e^{2\pi{inx}}=\delta(x)-\delta(x-1/2)$$ Multiply by $e^{-2\pi{imx}}$ and integrate over $[-1/4, 3/4]$ $$\sum_{n\in\mathbb{Z}}\big(\omega^{2}-(2\pi{n})^{2}\big)y_{n}\int_{-1/4}^{3/4}e^{2\pi{i(n-m)x}}dx=1-e^{-\pi{i}m}$$ $$\int_{-1/4}^{3/4}e^{2\pi{i(n-m)x}}dx=e^{-\frac{\pi}{2}{i(n-m)}}\int_{0}^{1}e^{2\pi{i(n-m)z}}dz=\delta_{mn}$$ Thus, $$\big(\omega^{2}-(2\pi{m})^{2}\big)y_{m}=1-e^{-\pi{i}m}$$ hence $$y(x)=\sum_{n\in\mathbb{Z}}\frac{e^{2\pi{inx}}-e^{2\pi{in(x-1/2)}}}{\omega^{2}-(2\pi{n})^{2}}$$