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I need some help on the following PDE using separation of variables:

$$\frac {\delta U}{\delta t}+U =\frac {\delta^2U}{\delta x^2 }, 0<x<\pi, t>0 $$ Given that $$ U(0,t)=U(\pi,t)=0, t>0 $$ $$U(x,0)=x(\pi-x), 0<x<\pi$$ My attempt, let; $$ U = X T $$ Then $$ XT'+XT=X''T $$ $$\frac {T'+T}{T}=\frac{X''}{X}=-k^2 $$ Solving ODE with $X$ $$X(x)=A\cos kx+B\sin kx$$ $$X(0)=A=0$$ $$X(\pi)=B\sin k\pi=0$$ $B=0$ or $\sin k\pi=0$ take the non trivial case $\sin k\pi=0$ $$k=n, n=1,2,3....$$ $$X(x)=B\sin nx$$ Solving ODE with $T$ $$\frac {T'+T}{T}=\frac{T'}{T}+1=-k^2 $$ $$T(t)=Ce^{-t(n^2+1)}$$ $$U(x,t)=B\sin nx \cdot Ce^{-t(n^2+1)}, BC=C$$ By the principle of superposition: $$U(x,t) = \sum C_n \sin nx \cdot e^{-t(n^2+1)}, $$ $$U(x,0) = \sum C_n \sin nx = x(\pi-x)$$ And now solve using fourier sine series. If this is complete garbage could somebody point me in the right direction or perhaps where I have gone wrong only just started PDE's so would be nice to have some constructive input, thanks as always and feel free to edit my late night mistakes.

pi-e
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Using $$\int_{0}^{\pi} x(\pi - x) \, \sin(n x) \, dx = \frac{2(1-(-1)^{n})}{n^{3}}$$ then \begin{align} U(x,t) &= \frac{2}{\pi} \, \sum_{n=1}^{\infty} \frac{1 - (-1)^{n}}{n^{3}} \, \sin(n x) \, e^{- (n^{2}-1) \, t} \\ &= \frac{4}{\pi} \, \sum_{n=0}^{\infty} \frac{1}{(2n+1)^{3}} \, \sin((2n+1) \, x) \, e^{- ((2n+1)^{2}-1) \, t} \\ &= \frac{4}{\pi} \, \sum_{n=0}^{\infty} \frac{1}{(2n+1)^{3}} \, \sin((2n+1) \, x) \, e^{- 4 \, n (n+1) \, t} \end{align}

Leucippus
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  • Would there not be a $\frac {1}{\pi}$ as a constant multiple of the integral for the fourier sine series? or is that what you've included outside the sum in your second term? – pi-e Nov 18 '15 at 23:32
  • @pi-e There is, but here I only stated the integral and value. I added the $\frac{1}{\pi}$ value when used in the series. – Leucippus Nov 18 '15 at 23:35
  • Yes that's fair enough, why did you choose to re-evaluate the term n=1 in the series to n=0 is that common? could we not just say $U(x,t) = 0$ when n is even and otherwise when not? – pi-e Nov 18 '15 at 23:39
  • @pi-e It is common to reduce $1 \pm (-1)^{n}$ to the most most compact form. Only the odd terms are non-zero in this series. This is why the index changed. – Leucippus Nov 19 '15 at 00:18
  • Thanks again for your insight Leucippus much appreciated I wasn't aware that reduction was commonplace. – pi-e Nov 19 '15 at 14:30