I've been asked to use Fourier's method to obtain the following solution;
$$u(x,t) = \sum_{n=1}^{\infty} B_n e^{-(n \pi C / L)^2 t} \sin(\frac{n \pi x}{L})$$ $$B_n = \frac{2}{L} \int_0^L \sin(\frac{n \pi x}{L}) f(x) dx$$
for the 1-D heat equation for $t>0$, on the region $0 \leqslant x \leqslant L$ with the boundary conditions $u(0,t) = 0 = u(L,t)$, $t > 0$ and initial condition $u(x, 0) = f(x)$, $0 \leqslant x \leqslant L$.
So, to achieve this, I started off with the 1-D heat equation; $u_t = c^2 u_{xx}$, and set $u(x,t) = X(x) T(t)$. Then, with a bit of working, I've found; $$T(t) = Ae^{-c^2 t}$$ $$X(x) = B \sin(c x) + C \sin(c x)$$ Now, using my initial conditions on $x$, I was able to get $X(x) = B \sin(\frac{n \pi x}{L})$, with $c = \frac{n \pi}{L}$. Thus, this made $T(t) = Ae^{-(\frac{n \pi}{L})t}$. Putting this all together, I've got... $$u(x,t) = D e^{-(\frac{n \pi}{L})^2t} \sin(\frac{n \pi x}{L})$$ With $D = AB$.
Now, from here, I'm a bit confused as to what results I should be getting. If I use the initial condition $u(x,0) = f(x)$, I get something that resembles the $B_n$ I'm meant to acquire. But, I'm not really sure where to go from here. Any help would be fantastic!! :)