# Differential Equation (2nd Order Linear)

1. Apr 19, 2014

### turpy

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

(x2)d2y/dx2 + 2*x*(dy/dx) + w2*x2*y=0

Where w is a constant

2. Relevant equations

3. The attempt at a solution

I am having a really hard time figuring out how to solve this. Usually for second order linear ODEs I start with assuming a solution of form y=eλx, substitute into the equation, find the two roots and get the solution as y= c1*y1+ c2*y2. It doesn't work in this case though.

From another solver I found that the solution has the form of C1*sin(wx)/x + C2*cos(wx)/x.
With sin and cos present, usually the method I mentioned above has to yield complex conjugate roots. But I'm really puzzled by the x's in the denominators. What kind of solution form would I have to assume to get that?

Any help would be appreciated, thanks!

2. Apr 19, 2014

### Dick

The solution you got should be a good clue. Try assuming y=exp(λx)/x.

3. Apr 19, 2014

### turpy

Works out great, thanks!! I just wonder how I would ever think to assume that solution form without being given the solution first...

4. Apr 19, 2014

### Dick

Don't know. I'm not that great with ODE's. I'd just say trial and error. But maybe somebody else knows a rule that applies here.

5. Apr 19, 2014

### LCKurtz

Without seeing anything but the DE I would think a series solution would be the first thing to try. Because x=0 is a regular singular point, I would try$$y=\sum_{n=0}^\infty a_nx^{n+r}$$With luck one might recognize the resulting series.

6. Apr 21, 2014

### epenguin

You have a product of a function of x and y as last term on LHS. When you have that sort of thing you can try to recognise the sort of terms that differentiation of products gives you. Leibniz' theorem is often useful. In this case divide all by x and I think you can recognise you have a differential equation in xy.

7. Apr 21, 2014

Nice.