I want to solve the following:
$$ u_{tt} - c^2u_{xx} = F(x)\cos(\omega t) \\ u(0, t) = u_x(\pi, t) = u(x, 0) = u_t(x,0) = 0$$
First of all I looked for a solution of the form $ u(x, t) = X(x)\cdot T(t)$
and got the following equations:
1. $$ X''(x) + \lambda X=0 \\ X(0) = X'(\pi) = 0 $$ 2. $$ \frac{T''(t)}{C^2 T(t)} + \lambda C^2T(t) = 0 \\ T(0) = T(0) = 0 $$
Solving the first system I get :
$$ \lambda _n = (n+\frac{1}{2})^2$$ $$ X_n(X) = \sin((n+\frac{1}{2}) x) $$
so: $$ u(x, t) = \sum _{n=0}^{\infty} T_n(t)\cdot \sin((n+\frac{1}{2}) x)$$
Now I don't know how to procceed, Usually I take the non-homogeneous part of the equation and substitute the fourier series of it in the equation and find some sort of ODE on $T_n(t)$.
Help would be appreciated!
**Edit: **
I thought I'd go along with my thoughts:
$$ F(x) = \sum_{n=0}^{\infty} a_n\cdot \sin((n+\frac{1}{2}) x)$$ $$ u(x, t) = \sum _{n=0}^{\infty} T_n(t)\cdot \sin((n+\frac{1}{2}) x) = F(x)\cdot \cos(\omega t) = \sum_{n=0}^{\infty} a_n\cdot \sin((n+\frac{1}{2}) x)\cos(\omega t)$$ where $ a_n = \frac{2}{\pi}\int_0^\pi F(x)\sin((n+\frac{1}{2}) x) $
comparing coefficients I get :
$$ T_n(t)'' + c^2(n+\frac{1}{2})^2\ T_n(t) = a_n\ \cos(\omega t) $$ solving this I got: $$ \xi = c(n+\frac{1}{2}) $$ $$ T_n(t) = T_{n_H}(t) + T_{n_P}(t) $$ $$ T_{n_H}(t) = C_1 \cos(\xi \ t) + C_2\sin(\xi \ t) $$ $$ T_{n_P}(t) = A \cos(\omega \ t) + B\sin(\omega \ t) $$
with the initial conditions $ T(0) $ I got:
$$ B = 0, A = \frac{a_n}{\xi^2 - \omega^2}, C_2 = 0, C_1 = -\frac{a_n}{\xi^2 - \omega^2} $$
** Edit ** solved, I will post the solution as soon as I can