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This is the question, I have solved it but I need someone to double check my solution. Question: Find the temperature $u(x, t)$ in a rod of length $L$ if the initial temperature is $f(x)$ throughout and if the ends $x=0$ and $x=L$ are insulated. $F(x)= x, 0$


For an insulated rod the solution $$ X(x,t)= \frac{a_0}{2}+\sum B_0 \frac{\cos(n\pi)x}{Le}−\frac{n^2\pi^2\alpha^2}{L}t $$ I found $a_0= 1$ and $$ B_n= −\frac{2}{n\pi}\sin\left(\frac{n\pi}{2}\right)+\frac{2}{n\pi}2\cos\left(\frac{n\pi}{2}\right)−\left(\frac{2}{n\pi}\right)^2 $$ then just plug in the coefficients into the sum. I am just not sure if these are the correct values for the $a_0$ and $B_0$ coefficients.

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Your solution can't be right because it has no dependence on the initial condition, f(x). Also, you should edit your question. There are a couple of typos, e.g. in your solution the e shouldn't be in the denominator of the first fraction, it should be outside the fraction with the second fraction being its argument. –  in_wolframAlpha_we_trust Apr 13 '12 at 6:59
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(As another user pointed out), your formula for the solution should be written:

$$ X(x,t)= \frac{a_0}{2}+\sum B_{\mathbf{n}} \frac{\cos(n\pi x)}{L}\mathbf{e}^{−\frac{n^2\pi^2\alpha^2}{L}t} $$

Also, you don't have the correct formulas for the coefficients.

Recall $a_0 = (2/L)\int_0^L f(x) dx, \ $ $B_n = (2/L)\int_0^L \cos(n\pi x / L) f(x) dx$. So, with $f(x) = x$, we have

$$a_0 = (2/L)\int_0^L x dx = L$$

$$b_n = (2/L)\int_0^L \cos(n\pi x / L) x dx = (2/L)[x (L/n\pi) \sin(n \pi x / L) + (L/n\pi )^2 \cos(n \pi x / L)]_{x=0}^L$$ $$ = 2L/(n\pi)^2(\cos(n\pi) - 1) $$ ( = 0 if n even, $-4L/(n\pi)^2$ if n odd)

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