Can someone explain to me, how to find the periodic solution to this differential equation? $d^3y/dx^3+ky=f(x)$
I am given that $k$ is a constant and $f(x)$ is a $2\,\pi$-periodic function.
So I assume that the solution can be represented as a Fourier series expansion, and thus I could fill the $y$ and $y'$ I found in in the original function.. However, would this be the right approach?
if I let $f(x) = \sum_{n=-\infty}^{\infty} y_n\cdot e^{inx} $ , then obviously
$f'(x) = \sum_{n=-\infty}^{\infty} iny_n\cdot e^{inx}$
What do I do next? How do I solve this? Thanks!
You have $$\frac{d^{3}}{dx^{3}}y(x)+ky(x)=f(x)$$ Where $f(x)=\sum_{n\in\mathbb{Z}}f_{n}e^{inx}$. We let $y(x)=\sum_{n\in\mathbb{Z}}y_{n}e^{inx}$, which leads to $$\sum_{n\in\mathbb{Z}}\{[-in^{3}+k]y_{n}-f_{n}\}e^{inx}=0$$ Multiplying by $e^{-imx}$ and integrating over $[0, 2\pi]$ we get by completeness $$[-im^{3}+k]y_{m}=f_{m}$$ So that $$y(x)=\sum_{n\in\mathbb{Z}}\frac{f_{n}e^{inx}}{k-in^{3}}$$