# Check my solution - Modelling of a spring with Differential Equation

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:

$\hskip 1in$

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!

-
+1 for showing your work. Your solution looks correct to me. (Some $\LaTeX$ tips: you can use double dollar signs $$ like this $$ to create display-style equations, as I've done for you, and you can put inline equations within text instead of on separate lines. You can use \left( blah \right) to create flexibly large parentheses around expressions, and use \sin and \cos to make upright-looking functions, as in $\sin,\cos$. I've also used \text{} and {cases} in my edit that you can look at.) –  anon Jun 21 '12 at 1:06
@anon thank you for your help, and i'll definitely keep those tips in mind next time i ask a question. –  Synia Jun 21 '12 at 4:49