I am doing some work with differential equations. I have solved the following problem but am uncertain if I'm doing it correctly. Could someone look over it for me and check if I'm doing something wrong or not? Thanks in advance!
The problem is as follows:
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My Solution:
$\frac{\partial u(x,t)}{\partial t} = \sum_{n=1}^{\infty} \sin(n \pi x)\big(a_n 100\pi \cos(100\pi t)-b_n 100\pi \sin(100 \pi t)\big)$
From initial condition $\displaystyle\frac{\partial u(x,0)}{\partial t} = 0$:
$\frac{\partial u(x,0)}{\partial t} = \sum_{n=1}^{\infty} \sin(n \pi x)[a_n 100\pi] = 0 \\ \implies a_n = 0$
Therefore:
$u(x,t) = \sum_{n=1}^{\infty} \sin(n \pi x)\big(b_n\cos(100\pi t)\big).$
Since
$b_m = \frac{2}{L} \int_{0}^{L} u(x,0) \sin\left( \frac{m\pi x}{L}\right)dx, \quad\text{and}\quad u(x,0) = 4\sin(3\pi x)),$
it follows that
$b_m = \frac{2}{1} \int_{0}^{1} 4\sin(3\pi x) \sin\left( \frac{m\pi x}{1}\right)dx.$
Due to property of sine function where
$\int_{0}^{L} \ \sin\left( \frac{n\pi x}{L}\right) \sin\left( \frac{m\pi x}{L}\right)dx= \begin{cases}L/2 & \text{if } n=m \\ 0 & \text{if } n\ne m, \end{cases}$
when $m = 3$, $b_m = 2\cdot 4\cdot\frac{1}{2} = 4$ and $b_m=0$ otherwise.
Finally, then:
$ u(x,t)= 4 \sin(3\pi x)\cos(100 \pi t).$
If anyone could have a look over this it would be much appreciated. Thanks!