I am not sure how to continue with the following question:
Find the periodic solutions of the differential equations (a)$\frac{dy}{dx}+ky=f(x)$, (b) $\frac{d^3y}{d^3x}+ky=f(x)$
where k is a constant and f(x) is a $2\pi$-periodic function.
Consider a Fourier series expansion for $f(x)$ using the complex form, $f(x)=\sum\limits_{n=-\infty}^{n=+\infty} f_ne^{inx}$
and try a solution of the form: $y(x)=\sum\limits_{n=-\infty}^{n=+\infty} y_ne^{inx}$
So far, I've taken the first, second and third derivative of $y(x)$, and tried to substitute these into the given differential equations. Since the summation is from $-\infty$ to $\infty$, I figured no rewriting of the summation terms is needed. Therefore, $y'(x)$ and $y''(x)$ and so on would be easier to write down. Now for the solution, what I get for a) is that
$$iny_n+ky_n=f_n$$
From this point on, I'm not sure how to continue, but I think you have to solve the equation for $y(x)$ given the fact that $f(x)$ is periodic. I know how to find the solutions of differential equations, but for this periodic one, I am not sure what I should find and how I should do it.
Perhaps you can help. I only need a general way of solving these kinds of problems, and maybe a tiny example to illustrate how it's done. Anyway, I'd appreciate it if at least someone could help me with question a). I'd be able to do b) that way.
Thanks in advance!
~Adam
To me, this looks like just equating terms after differentiating.
If $y(x) =\sum\limits_{n=-\infty}^{n=+\infty} y_ne^{inx} $, then $y' =\sum\limits_{n=-\infty}^{n=+\infty} iny_ne^{inx} $, $y'' =\sum\limits_{n=-\infty}^{n=+\infty} -n^2y_ne^{inx} $, and $y''' =\sum\limits_{n=-\infty}^{n=+\infty} -in^3y_ne^{inx} $.
For the second equation, $f_n = -in^3y_n+y_n =y_n(1-in^3) $ so $y_n =\dfrac{f_n}{1-in^3} =\dfrac{f_n(1+in^3)}{1+n^6} $.
As to how to simplify this, that is up to you.