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2y

You mean separable, right? :D

and lol, @openstudy11 It is unfortunately not that simple :D

@Study23 Shall we? :)

Yup :D

dy = y^2 dx; I was doing this but I got confused because you couldn't integrate this could you?

dy/y^2=dx (?)

So, ln|y^2|=x ?

Sorry, my bad
\[\huge \int ax^n = \frac{ax^{n+1}}{n+1}\]

So, I get \(\ \Huge -y^-1 + C_1 = x \) ?

That should be \(\ y^{-1} \)

Mother of sizes... o.O
LOL
\[\huge -y^{-1}+C = x\]

But on the whole, good job :)

So, my textbook has this interns of Y= ...

Waaaay ahead of you ^

In terms (autocorrect)

Already done... ^

It just needs some algebraic manipulation is all.

I don't know what exam that is... better consult someone more familiar with it :)

Oh haha okay. Thanks!

dy/dx=y^2
dy/y^2=dx
then integrate both the side
-1/y=x+c