Particular solution to a non-homogenous differential equation Consider the following ODE:
$$\frac{d^2 x}{dt^2} + kx = f(t)$$
where $$f(t) = \frac{1}{2} -2 \pi \sum_{n = 1}^{\infty}{n\sin\left(\frac{n \pi t}{2}\right)}$$ was derived using fourier series.
What is the particular solution to this? The sum symbol on the right hand side confuses me and I don't know how to handle a sum in this case.
I thought to try this for a guess:
$$x(t) = A_2t + A_0 - Bcos(\frac{n \pi t}{2}) - Csin(\frac{n \pi t}{2})$$
 A: My personal favorite method for finding particular solutions, because it's more or less algorithmic, is variation of parameters.  You can look up the details, but basically if $x_1(t)$ and $x_2(t)$ are the two elementary solutions to the associated homogeneous equation, then the particular solution may be written
$$x_p(t) = \left(-\int\frac{x_2(t)f(t)}{W(t)}dt\right)x_1 + \left(\int\frac{x_1(t)f(t)}{W(t)}dt\right)x_2 $$
where $W(t) = x_1x'_2 - x'_1x_2$ is the Wronskian of the two solutions.  For this problem, the elementary solutions are by inspection
$ x_1(t) = \sin(\sqrt{k}t) $ and $ x_2(t) = \cos(\sqrt{k}t) $.  So the first term in the formula I gave would be
$$ -\frac{1}{\sqrt{k}}\int\cos(\sqrt{k}t)dt\left( \frac{1}{2} - 2\pi\sum_{n=0}^\infty n\sin\left(\frac{n\pi t}{2}\right) \right)dt $$
$$ =-\frac{1}{2\sqrt{k}}\int\cos(\sqrt{k}t)dt+\frac{2\pi}{\sqrt{k}}\sum_{n=0}^\infty n\int\cos(\sqrt{k}t)\sin\left( \frac{n\pi t}{2} \right)dt $$
Which you can integrate explicitly.  An exactly similar calculation follows from the second term in $x_p(t)$.
