Solve the I.V.P for the heat equation with time-dependent source $$ \frac{du}{dt} = k \frac{d^2x}{dx^2} + Q(x,t)$$
Subject to the following B.C:
$u(0,t) =0, \frac{du}{dx}(L,t) = 0$
I know that the eigen values for this equation would be $\lambda= (x\pi)^2$ with the corresponding eigenfunction $sin= nx\pi$ 
I sort of know how to solve for the dirichlet b.c but this isnt dirichlet nor neuman b.c. Please, can someone help me solve this?
 A: Use the method of image.
Reflect Q with respect to $x=0$ and $x=L$ such that the resulting superposition, say, $\tilde Q$is an even function with respect to $x=L$ (so that the spatial partial derivative there vanished) and an odd function with respect to $x=0$ (so that the value there vanishes). 
Convolve the heat kernel (or Green's function, or Gaussian distribution) with $\tilde Q$ then integrate over time $t$. That is a special solution for the inhomogeneous problem.
Solve the homogeneous initial value problem ($Q=0$) by again reflecting the heat kernel about $x=0$ and $x=L$ in the same fashion described above and convolve the resulting superposition function against the initial value.
The sum of the inhomogeneous special solution and the homogeneous solution is the solution you seek.
A: I believe there is a typo since the eigenvalues should be numbers, something like $\lambda_n=(n\pi)^2$, let's find out what they should be by setting $u_h=X(x)T(t)$ and considering the problem with homogeneous right hand side, $\partial_tu-k\partial^2_{xx}u=0$
so $X''(x)+\lambda X(x)=0$, and $T'(t)=-\lambda kT(t)$.
Thus $X(x)=A\cos(\sqrt{\lambda}x)+B\sin(\sqrt{\lambda}x)$, and $X'(x)=-A\sqrt{\lambda}\sin(\sqrt{\lambda}x)+B\sqrt{\lambda}\cos(\sqrt{\lambda}x)$, so 
$u(0,t)=X(0)T(t)=0\Rightarrow X(0)=0$, and $\partial_xu(L,t)=X'(L)T(t)=0\Rightarrow X'(L)=0$ 
Thus $A=0$, $B\sqrt{\lambda}\cos(\sqrt{\lambda}L)=0$, so we need $\sqrt{\lambda}L=\pi/2+n\pi,n\in\Bbb N$, so 
$\lambda_n = (\pi/2+n\pi)^2/L$, $n\in\Bbb N$
with eigenfunctions $\varphi_n(x)=\sin(\sqrt{\lambda_n}x)$.
so $X_n(x)=B_n\varphi_n(x)$, and $T_n(t)=C_ne^{-\lambda_nkt}$
so $u_h=\sum_{n=0}^\infty D_n\varphi_n(x)e^{-\lambda_nkt}$
Now we have the homogeneous solution, to find $D_n$ for $n\in\Bbb N_0$, use the initial data, i.e. $u(x,0)=u_0$
Now let $u_c$ be the solution to the to the problem with non homogeneous right hand side, and Homogeneous BC's, test the equation with $u_h$, use integration and use integration by parts.
Then the full solution $u=u_c+u_h$.
