Separation of variables , but for 2nd order

1. Jul 13, 2012

ericm1234

"separation of variables", but for 2nd order

Ok, I know how to separate variables in solving an ODE. I am unable to understand a solution I have for a problem which was the result of reduction of order- we end up with u''*sinx-2u'*cosx=0
so turn this into u''/u'=-2cosx/sinx
At this point I don't understand the process. I've tried looking at it as making the left side a "u substitution", or rewriting it d(du/dt)/dt / du/dt and integrating both sides,
also, tried making a substition w=u', but this seems hard to get u back in the end.
Can someone explain thoroughly what the process looks like in solving this?

2. Jul 13, 2012

tiny-tim

hi ericm1234!
well, then that's w'/w = -2cosx/sinx,

so ln(w) = … ?

3. Jul 13, 2012

ericm1234

Re: "separation of variables", but for 2nd order

Thanks, got it now, not as complicated as I was making it.