Consider the ODE $\frac {d^2u}{dx^2} + 2\pi\frac {du}{dx} + \frac 54\pi^2u = g(x)$
where g is a periodic fuction with period 1 given by $g(x) = e^{\pi x}$ , $ 0 \le x \lt 1$.
It is desired to find the steady state solution (ie the periodic solution with period 1) of this equation.
Briefly explain how you would use a discrete Fourier transform to solve this problem, and derive an explicit forumla for the discrete Fourier coefficients of u in this case.
I've found the relation $F[u](n) = \frac{F[u''](n)}{-4\pi^2n^2}$
Hint: with the notation $F[u](n)$ for the $n^t{}^h$ Fourier coefficient, you get $$ F[u'](n) = 2i\pi n F[u](n) $$ via an integration by parts.\
proof: $$ F[u'](n) = \int_0^1 u'(t) e^{-2i\pi n t }dt = [u(t) e^{-2i\pi n t }]_0^1 - \int_0^1 u(t) (-2i\pi n e^{-2i\pi n t })dt $$Here, the $[.]_0^1$ part is $0 $ because of periodicity. It remains:
$$ F[u'](n) = 2i\pi n \int_0^1 u(t)e^{-2i\pi n t }dt = 2i\pi n F[u](n) $$