# Differential equation help

1. Feb 12, 2016

### theone

I understand what is in the picture http://postimg.org/image/u5ib33kzb/
but the book goes on to say that the solution is thus of the form
$X_n = a_n sin \frac{n \pi x}{l}$
How does putting $β=σ^2=\frac{n^2π^2}{l^2}$ into (6.37) result in that?

2. Feb 12, 2016

### RUber

I will assume that you have a differential equation that looks like:
$x'' +\beta x = 0$
with boundary conditions:
$x(0)=x(l) = 0$
The general solution for the differential equation is
$x = A \sin( \sqrt{\beta} t ) + B \cos(\sqrt{\beta} t)$
And the boundary condition at $t=0$ forces B to go to zero and the boundary condition at $t = l$ forces $\beta$ to be the form you have above.

3. Feb 12, 2016

### theone

thats right, the differential equation is (X is X(x), a function of x) :

$X'' + \beta X = 0$

Assuming a general solution of $X(x) = A e^{ -\sqrt{-\beta}x} + B e^{+\sqrt{-\beta} x}$, that $\sqrt{-\beta}$ is complex (ie. $\beta =σ^2$) , and that the boundary conditions are $X(0)=0$ and $X(l)=0$, they found that $σ=\frac{n\pi}{l}$
What I want to know is how putting $σ=\frac{n\pi}{l}$ into the general solution results in $X_n=a_n\sin\frac{n\pi x}{l}$
Or how their general solution is equivalent to yours?

Last edited: Feb 12, 2016
4. Feb 12, 2016

### pasmith

$$\cos x = \frac{e^{ix} + e^{-ix}}2 \\ \sin x = \frac{e^{ix} - e^{-ix}}{2i}$$

5. Feb 12, 2016

### RUber

Applying your first boundary condition tells you that A = -B, giving $X(x) = A\left(e^{-i\sigma x}- e^{i\sigma x}\right)$
Noting what pasmith wrote above, this is equivalent to $C \sin (\sigma x )$.
Then, since any sigma of the form given can be a solution, your full solution might be an infinite sum:
$X(x) =\sum_{n=1}^\infty X_n(x) = \sum_{n=1}^\infty a_n \sin(\sigma_n x )$