Originally Posted by

**sataq** Hi guys. I am going through an example in my book, and I am stuck on the last step concerning the initial condition.

The coefficient derived from the initial condition is:

$\displaystyle B_n = 4 \frac{1 - (-1)^n}{\pi n^3}$

and the linear sum of the particular solutions is:

$\displaystyle

u(x, t) = \displaystyle\sum_{n=1}^\infty B_n e^{(-a^2 k_n^2 t)} sin (\frac{n \pi x}{L}) $

I think then $\displaystyle B_n$ is substituted into the equation$\displaystyle u(x, t)$ above to get:

$\displaystyle u(x, t) = \frac{8}{\pi} \displaystyle\sum_{n=1}^\infty \frac{e^{- (2n-1)^2 a^2 t} }{(2n-1)^3} sin((2n-1)x) $

I am having difficulty working out the algebra on this one. Can anyone show it to me step by step. I can see that there is a factor (2n-1) but I don't know where they got that from. Can anyone explain? I am finding this really difficult to work out.