Solving inhomogeneous PDEs with ODEs I am trying to understand a general method of solving an inhomogeneous PDE. I have begun with the Heat equation but am stuck with the last step.
For instance, solving 
$$\begin{cases}u_t=u_{xx}+g(x)\\
u(0,t)=0=u(L,t)\\
u(x,0)=f(x)
\end{cases}$$
So first of all I break the functions into two Fourier series:
$$f(x)=\sum a_n\sin\left(\frac{n \pi x}{L}\right ) \qquad g(x)=\sum b_n\sin\left(\frac{n \pi x}{L}\right )$$
Then, letting the particular solution be
$$u(x,t)=\sum U_n(t)\sin\left(\frac{n \pi x}{L}\right )$$
This can be reduced to an ODE
$$\dot{U_n(t)}=-\left(\frac{n\pi}{L}\right)^2U_n(t)+b_n$$
with $U_n(0)=a_n.$
When this is solved, it should be able to give the function of $t$ in the Fourier series.
This is where I am stuck. How do I solve an ODE like this? And is this the best method to solve the PDE?
 A: Starting from 
$$U_n'(t)+\left(\frac{n\pi}{L}\right)^2 U_n(t)=b_n$$
with $U_n(0)=a_n$, we can find a solution for $U_n(t)$ using a variety of methods.  Here, we will use LaPlace Transforms.
Let $\hat U_n(s)$ be the Laplace transform of $U_n(t)$.  Taking the Laplace transform of the ODE reveals
$$(s\hat U_n(s)-U_n(0))+\left(\frac{n\pi}{L}\right)^2 \hat U_n(s)=\frac{b_n}{s}$$ 
Now, solving for $\hat U_n(s)$, using $U_n(0)=a_n$, and expanding in partial fractions yields
$$\hat U_n(s)=\frac{a_n-\frac{b_n}{(n\pi/L)^2}}{s+(n\pi/L)^2}+\frac{b_n}{(n\pi/L)^2}\frac{1}{s}$$
whereby inverting the Laplace Transform $\hat U_n(s)$, we see that
$$U_n(t) = \frac{b_n}{(n\pi/L)^2}+\left(a_n-\frac{b_n}{(n\pi/L)^2}\right)e^{-(n\pi/L)^2t}$$
Finally, the solution for $u(x,t) is given by
$$\begin{align}
u(x,t) &=\sum_{n=1}^{\infty} \left(\frac{b_n}{(n\pi/L)^2}+\left(a_n-\frac{b_n}{(n\pi/L)^2}\right)e^{-(n\pi/L)^2t}\right)\sin\left(\frac{n\pi x}{L}\right)\\\\
=&\sum_{n=1}^{\infty} a_ne^{-(n\pi/L)^2t}\sin\left(\frac{n\pi x}{L}\right)\\\\
&+\sum_{n=1}^{\infty}   b_n\left(\frac{1-e^{-(n\pi/L)^2t}}{(n\pi/L)^2}\right) \sin\left(\frac{n\pi x}{L}\right)
\end{align}$$
A: Going on the question you asked MV in his comments, I'll do it for you by separation of variables (note that using the Laplace Transform method MV used is a much better choice for this problem). We have
$$U' = b_{n} - \bigg( \frac{n \pi}{L} \bigg)^{2} U$$
We will call 
$$m = \frac{n \pi}{L}$$
Separating and integrating
$$\begin{align}
\implies \int \frac{1}{ b_{n} - m^{2} U} dU &= \int dt \\
\implies - \bigg( \frac{1}{m} \bigg)^{2} \ln ( b_{n} - m^{2} U ) &= t + K \\
\implies \ln ( b_{n} - m^{2} U ) &= - m^{2} (t + K) \\
\implies b_{n} - m^{2} U &= \exp (- m^{2} (t + K) ) \\
&= \exp (- m^{2} t) \exp (- m^{2} K) \\
\implies m^{2} U &= b_{n} - \exp (- m^{2} t) \exp (- m^{2} K) \\
\implies U &= \frac{b_{n}}{m^{2}} - \frac{\exp (- m^{2} t) \exp (- m^{2} K)}{m^{2}} \ \ (*) \\
\implies U(0) &= \frac{b_{n}}{m^{2}} - \frac{\exp ( - m^{2} K)}{m^{2}} \\
&= a_{n} \\
\implies \frac{b_{n}}{m^{2}} - a_{n} &= \frac{\exp (- m^{2} K)}{m^{2}} \ \ (**) \\
\end{align}$$
Inserting $(**)$ into $(*)$, we get
$$\begin{align}
U (t) &= \frac{b_{n}}{m^{2}} - \exp (- m^{2} t) \bigg[ \frac{b_{n}}{m^{2}} - a_{n} \bigg] \\
&= \frac{b_{n}}{m^{2}} + \bigg[ a_{n} - \frac{b_{n}}{m^{2}} \bigg] \exp (- m^{2} t)
\end{align}$$
which is the same as MV when you replace $m$ with $\frac{n \pi}{L}$. As you can see though, some methods are easier to implement.
