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Solve the heat equation

$\displaystyle k\frac{\partial^2 u}{\partial x^2} = \frac{\partial u}{\partial t}, 0< x < L, t > 0$

$\displaystyle u(0,t) = 0, u(L,t) = 0, t > 0$

$\displaystyle u(x,0) = x(L-x), 0 < x < L$

using

$u(x,t)$		$= \frac{2}{L}\displaystyle\sum\limits_{n=1}^\infty \left( \int_0^L u(x,0)\sin\left( \frac{n\pi}{L}x\right) dx\right) e^{\left( -k\frac{n^2\pi^2}{L^2}t\right)} \sin\left( \frac{n\pi}{L}x\right)$
		$=\frac{2}{L}\displaystyle\sum\limits_{n=1}^\infty b_n e^{\left( -k\frac{n^2\pi^2}{L^2}t\right)} \sin\left( \frac{n\pi}{L}x\right)$

$\displaystyle\int_0^L u(x,0)\sin\left( \frac{n\pi}{L}x\right) dx$

$=$ $\displaystyle \int_0^L$

$\hskip 3pt dx$

$u =$

$dv =$

$du =$

$v =$

$=$

$\Bigg\vert_0^L+\frac{L}{n\pi}\displaystyle\int_0^L$

$dx$

$u =$

$dv =$

$du =$

$v =$

$=$

$+$

$\Bigg\vert_0^L+2\left(\frac{L}{n\pi}\right)^2\displaystyle\int_0^L$

$\hskip 3pt dx$

$=$

$+$

$\Bigg\vert_0^L$

$=$

You can earn partial credit on this problem.