# Strange way of solving a linear 2nd order DE

1. Nov 16, 2012

### Alexrey

1. The problem statement, all variables and given/known data
I was given a DE of the form: $$\Phi^{''}+(6/\eta)\Phi^{'}=0$$ where the next step was given as $$\Phi^{'} \propto \eta^{-6}$$ where the answer came out to be $$\Phi \propto \eta^{-5} + constant$$

3. The attempt at a solution
My attempt was to set $$\Phi^{'}=x$$ where I would then get $$x^{'}=-(6/\eta)x$$ and then solve for a seperable DE, but my answer was incorrect. Any help would be appreciated, thanks guys.

Any help would be appreciated, thanks guys.

2. Nov 16, 2012

### haruspex

To clarify, η is the independent variable, and differentiation is wrt that? If so $$dx/x=-6d\eta/\eta$$ yes? Isn't it straightforward from there?

3. Nov 16, 2012

### Alexrey

About 5 minutes after posting this I figured it out. :/ Thanks for replying though, I appreciate it, that's exactly what I got. Cheers!