Non homogeneous Heat Equation PDE $$A_t-A_{xx} = \sin(\pi x)$$
$$A(0,t)=A(1,t)=0$$
$$A(x,t=0)=0$$
Find $A$.
I know I need to find the homogeneous and particular solutions. Im just not sure on this PDE.
 A: The solution may be accomplished using a Laplace transform.  Defining
$$\hat{A}(x,s) = \int_0^{\infty} dt \, A(x,t) \, e^{-s t}$$
and applying the initial condition, we get an ordinary differential equation in $x$:
$$\frac{d^2}{dx^2} \hat{A} - s \hat{A} = -\frac{1}{s} \sin{\pi x}$$
The zero boundary conditions in $x$ mean that the homogeneous solution is zero.  The solution then takes the form $\hat{a}(x,s) = P \sin{\pi x}$.  Plugging this into the equation, we get the solution
$$\hat{A}(x,s) = \frac{\sin{\pi x}}{s (\pi^2 + s)}$$
You can use partial fractions, or simply look up in a table of inverse LT's; the solution is
$$A(x,t) = \frac{1}{\pi^2} \sin{\pi x} \, (1-e^{-\pi^2 t})$$
A: Since the non-homogeneity depends only on $x$, we can assume a solution of the form $A(x,t)=u(x,t)+\phi(x)$.
Substituting this into the PDE gives
$$u_t-u_{xx}-\phi_{xx}=\sin(\pi x).$$
Choosing $\phi(x)$ such that $-\phi_{xx}=\sin(\pi x)$, means that $u$ only needs to satisfy a homogeneous PDE. 
Note that the boundary conditions on $u$ will change with this assumed solution.
A: You have to guess particular solution first.
$$
A^p = B\sin \pi x \\
-A^p_{xx} = B\pi^2\sin \pi x = \sin \pi x \\
B = \frac 1{\pi^2}
$$
so
$$
A^p = \frac 1{\pi^2} \sin \pi x
$$
General solution of inhomogeneous problem is a sum of general solution of homogeneous problem and particular solution. So
$$
A = A^h + A^p
$$
It'll be much easier if one solves homogeneous problem instead. So all you need to do is alter BCs as follows
$$
A(0,t) = A^h(0,t)+A^p(0,t) = A^h(0,t)+0 = \fbox{$A^h(0,t)=0$} \\
A(1,t) = A^h(1,t)+A^p(1,t) = A^h(1,t)+0 = \fbox{$A^h(1,t)=0$} \\
A(x,0) = A^h(x,0)+A^p(x,0) = A^h(x,0)+\frac 1{\pi^2}\sin \pi x = 0 \Leftrightarrow \fbox{$A^h(x,0) = -\frac 1{\pi^2} \sin \pi x$}
$$
So, now solve homogeneous heat equation with BCs provided above.
