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ammmmgggg its zep!

+_+

:d

Oh I'm watching the video :O
He did this kind of backwards.. hmm interesting

Yeah I did it that way

My substitution was \[v = y^{-2} \implies y = \frac{ 1 }{ \sqrt{v} }\]

\[y' = - \frac{ 1 }{ 2 }v^{-3/2} \frac{ dv }{ dt }\]

Yeah haha

It didn't work out so well

I think I'll just stick to \[\large\rm y^{-n}y'+p(x)y^{1-n}=q(x)\] haha

ya, at least for this one :)

Anyways, thanks zepdrix!