In this problem we explore using Fourier series to solve nonhomogeneous boundary value problems. For $u_n$ type un, for derivatives use the prime notation $u_n^\prime,u_n^{\prime\prime},\ldots$.


Solve the heat equation



To find a series solution u we first must write the function $xe^{-3t}$ as a Fourier series
$xe^{-3t} = \displaystyle\sum\limits_{n=1}^\infty F_n\sin\left(\frac{n\pi}{5}x\right)$
Therefore

$\displaystyle F_n = \frac{2}{5}\displaystyle\int_0^{5}$$\displaystyle\hskip 2pt dx$

$\displaystyle\hskip 15pt =$

Now we try to find a solution $u$ of the form $u(x,t) = \displaystyle\sum\limits_{n=1}^\infty u_n(t)\sin\left(\frac{n\pi}{5}x\right)$
Using this series in the PDE we get


$\displaystyle \frac{\partial u}{\partial t}-\frac{\partial^2 u}{\partial x^2}$

$\displaystyle\hskip 15pt =\displaystyle\sum\limits_{n=1}^\infty$ $\displaystyle\sin\left(\frac{n\pi}{5}x\right)$

Since we want $\displaystyle\frac{\partial u}{\partial t}-\frac{\partial^2 u}{\partial x^2} = xe^{-3t}$ their Fourier coefficients must be equal:
which gives us an ODE in $u_n$ which we solve using constant $c_n$, to get
$u_n(t) =$

Now that we have a general form for u, we can find the constants $c_n$ by using the initial condition $u(x,0) = x$. Plugging the formula we just derived for $u_n(t)$ into the series for u we get
$u(x,0) =\displaystyle\sum\limits_{n=1}^\infty$ $\displaystyle\sin\left(\frac{n\pi}{5}x\right) = x$
Recognizing that this is a Fourier series for $x$, we can solve for $c_n$:


$\displaystyle c_n =\frac{2}{5}\displaystyle\int_0^{5}$ $\displaystyle\hskip 2pt dx -$

$\displaystyle\hskip 12pt =$