Strange way of solving a linear 2nd order DE

[Alexrey]

Homework Statement

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

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.

 

[haruspex]

x^{'}=-(6/\eta)x

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?

 

[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!