# Separation of Variables

1. Nov 4, 2013

### kgal

1. The problem statement, all variables and given/known data

Hey guys,

I have this problem which I am having a hard time solving.

$$u_{tt} -x^2u_{xx} = 0$$
$$1<x<2 \hspace{4mm} t>0$$
$$u(x,0)=0$$
$$u_t(x,0)=g(x)$$
$$u(1,t)=0=u(2,t)$$

2. Relevant equations

$$u_{tt} -x^2u_{xx} = 0$$
$$1<x<2 \hspace{4mm} t>0$$
$$u(x,0)=0$$
$$u_t(x,0)=g(x)$$
$$u(1,t)=0=u(2,t)$$

3. The attempt at a solution

I used separation of variables to reach:

$$x^2X'' - \lambda X=0$$
$$T''- \lambda T = 0$$

Solving for $$X$$ since the determinant $$a^2-4b > 0$$ (since $$x$$ is between 1 and 2)
and the roots are both real, we have to use the form of a solution:
$$X(x) = Ae^{rx} + Be^{rx}$$

solving for the roots yields $$r = \pm \frac{\lambda}{x}$$

Plugging the roots into the solution gives:
$$X(x) = Ae^{\lambda} + Be^{\lambda}$$

but this solution does not $$x$$ at all!

Where did I go wrong?!

2. Nov 4, 2013

### pasmith

I assume you're thinking of $aX'' + bX = 0$, but assuming a solution of the form $e^{rx}$ doesn't work here because $a = x^2$ is not a constant.

Try a solution of the form $x^{\alpha}$ instead.

Or set $X(x) = Z(\log x) = Z(z)$, which does give you a second-order linear ODE with constant coefficients for $Z(z)$.

Last edited: Nov 4, 2013