The problem states to use the fourier series of the function f(t) defined as follows:
$f(t)= t+1 , -1<t<0 $
$f(t)=1-t , 0<t<1$
to solve the differential equation: x''+4x=f(t), x(0)=1, x'(0)=0
I have found the fourier series of f(t) to be:
$\frac1 2 + \sum_{n=odd}^\infty \frac{cos(\pi n t)}{n^2}$
the complementary solution for the homogeneous equation is: $x_c= C_1 cos(2t) + C_2 sin(2t)$
The particular solution corresponding to the constant 1/2 is 1/8
I then proceed to find the coefficient of each n-th particular solution corresponding to the n-th $cos(n \pi t)$
this turns out to be $\frac{1}{n^2(4-n^2 \pi^2)}$
The solution should be a sum of the complementary and particular solutions, thus:
$x=C_1 cos(2t) + C_2 sin(2t)+\frac{1}{8} + \sum_{n=odd}^\infty \frac{cos(\pi n t)}{n^2(4-n^2 \pi^2)}$
differentiating this termwise yields: $x'=-2C_1 sin(2t) + 2C_2 cos(2t) + \sum_{n=odd}^\infty \frac{-\pi n sin(\pi n t)}{n^2(4-n^2 \pi^2)}$
using the initial condition x'(0)=0 it is apperent $C_2$=0
thus:
$x=C_1 cos(2t) +\frac{1}{8} + \sum_{n=odd}^\infty \frac{cos(\pi n t)}{n^2(4-n^2 \pi^2)}$
This is where I am lost, how do I find the remaining coefficient?
My intuition tells me to use the initial condition x(0)=1 which yields: $C_1 + \frac{1}{8} + \sum_{n=odd}^\infty \frac{1}{n^2(4-n^2 \pi^2)}$=1
but this implies: $C_1 = \frac{-7}{8} - \sum_{n=odd}^\infty \frac{1}{n^2(4-n^2 \pi^2)}$
is such a solution allowed or is there a different way of doing this? I appreciate all the help in advance, I have a big test on this and really need to know!