Applying the method of eigenfunction expansion to the PDE $$u_t -c^2u_{xx}=F(t)$$ $$0<x<L, t>0$$ $$u(x,0)=f(x)$$ $$u_x(0,t)u_x(L,t)=0$$
for the homogenous part of this equation ($L[v(x,t)]=0$) we get the following eigenvalues: $$\lambda_n = \left( \frac{n\pi x}{L} \right)^2$$ and the following eigenfunctions: $$X(x) = \cos\left(\frac{n\pi x}{L}\right)$$
We can now assume the following general solution: $$u(x,t)=\sum_{n=0}^\infty a_n(t) \cos\left(\frac{n\pi x}{L}\right)$$
$$v(x,0) = \sum\limits_{n=0}^\infty a_n(0)\cos\left(\frac{n\pi x}{L}\right)=f(x)$$ Thus $$a_n(0)=\frac{2}{L}\int_0^L f(x)\cos\left(\frac{n\pi x}{L}\right)dx$$
Differentiating and substituting to the original PDE, we get: $$\sum\limits_{n=0}^\infty \left[ a_n'(t)+a(t)\left( \frac{n\pi}{L} \right)^2c^2 \right]\cos\left(\frac{n\pi x}{L}\right)=F(t)$$
Now comes the most surprising part:
$$a_n'(t)+a(t)\left( \frac{n\pi c}{L} \right)^2=\frac{2}{L}F(t)\int\limits_{0}^L \cos\left(\frac{n\pi x}{L}\right)dx = 0$$
Does this mean that $F(t)$ vanishes? I think it shouldn't be the case. Somehow I'm doing something correctly. Can someone please help me clarify this?
$$\sum\limits_{n=0}^\infty \left[ a_n'(t)+a(t)\left( \frac{n\pi}{L} \right)^2c^2 \right]\cos\left(\frac{n\pi x}{L}\right)=F(t)$$
$$\implies \begin{cases} a_n'(t)+a(t)\left( \frac{n\pi c}{L} \right)^2=0 \qquad n>0\\ a'_0(t)=F(t) \end{cases}$$
$$\begin{cases} a_n(t)=A_n\exp \left(-\left( \frac{n\pi c}{L} \right)^2 t\right) \qquad n>0\\ a_0(t)=A_0+\int_0^t F(\tau)d\tau \end{cases}$$
$$u(x,t)=A_0+\int_0^t F(\tau)d\tau + \sum_{n=1}^\infty A_n \exp \left(-\left( \frac{n\pi c}{L} \right)^2 t\right)\cos\left(\frac{n\pi x}{L}\right)$$
$$u(x,0)=A_0 +\sum_{n=1}^\infty A_n \cos\left(\frac{n\pi x}{L}\right) = f(x)$$
This is the Fourier series of $f(x)$ which determines the coefficients $A_0\:,\:A_1\:,\: ...\:A_n\:,\: ...$.