Here is the Problem: Solve $\frac{\partial T(x,t)}{\partial t} = \frac{\partial^{2} T(x,t)}{\partial x^{2}} +2xe^{-t} $ with the following boundary conditions $T(0,t)=10, and \frac{\partial T}{\partial x} (1,t)= 0 $ and initial condition $T(x,0)=10$
The attempt:
We need to make the PDE homogeneous and the Boundary Conditions homogeneous by using the eigenfunction expansion method. Assume $T(x,t) = v(x,t) - r(x,t)$ which $r(x,t)$ is found by the equilibrium temperature solution. What I have is $r(x,t) = T_e(x) = -\frac{1}{3}x^{3} e^{-t}+xe^{-t} +10$, which the PDE, BC's and IC becomes
$\frac{\partial v}{\partial t} = \frac{\partial^{2} v}{\partial x^{2}} - \frac{1}{3} x^{3} e^{-t} + xe^{-t}$
$v(0,t) = 0$
$\frac{\partial v}{\partial x} (1,t) = 0$
$v(x,0) = 1/3 x^{3} e^{-t} -xe^{-t} $
The Eigenvalues and eigenfunctions for the homogeneous PDE are
$\lambda_n= \big(\frac{(2n-1)\pi }{2}\big)^{2} $
$\phi_n(x) = \sin(\frac{(2n-1) \pi x}{2})$
From the Generalized Principle of Superposition, this would mean that
$v(x,t) = \sum_{n=1}^{\infty} h_n(t) \phi_n(x)$ which $h_n(t) = C_n e^{-\lambda_n t}$
Now I did solve for $h_n(0)$ but it is too hard to type it on this software.
So I do plug in the sub into the PDE and the way to find $q_n(t)$ is by getting an ODE from this. How do I do that in this scenario? Does this look okay so far?
Thank you very much. I really appreciate your thoughts and inputs.
I am obtaining the following solution