[tex] \frac{\partial u}{\partial t} - k \frac{\partial^2 u}{\partial x^2} = 0 [/tex](adsbygoogle = window.adsbygoogle || []).push({});

for 0 <x < pi, t> 0

[itex] u(0,t) = u(\pi,t) = 0 [/itex]

[itex] u(x,0) = x (\pi - x) [/itex]

OK i know the boring part of getting u(x,t) = X(x) T(t)

the infinite series part is hard part

the coefficient [tex] c_{n} = \frac{2}{\pi} \int_{0}^{\pi} \sin(n(\pi -x)) \sin(nx) = \frac{-2}{\pi} \left(\frac{-1 + (-1)^n}{n}\right) \sin(nx) [/tex]

the -1^n is from the Cos n pi term taht would coem from the integration

thus n must be odd

[tex] c_{n} = \frac{4}{\pi} \frac{1}{2n-1} \sin(nx) [/tex]

is this good so far?

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# Homework Help: More PDE fun

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