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I need some help, I've tried to solve it since yesterday but i failed.As usually, I need to find $y(x)$ which is the solution of the differential equation. Here is the equation:
$ (1+x^2)y^3dx-(y^2-1)x^3dy = 0,\qquad y(1) = -1 $

It is supposed to be easy, but I didn't find the right theorem nor formula to use.

Edit: I already developed to get the following equation $ \frac {1+x^2}{x^3}dx = \frac {y^2-1}{y^3}dy $
and after integration got:
$ \frac {-1}{2x^2}+\ln(|x|) = \frac {1}{2y^2}+\ln(|y|) $
but how can I get $y(x)$ from there ?

2 Answers 2

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Please note that not every equations of the form $F(x,y)=0$ can be written as $y=f(x)$. So, here you should write the solution as $F(x,y)=0$.

It may be the case that you have found it difficult how to use the condition $y(1)=-1$. Assuming this is the case, I am giving a hint: where is your constant of integration?

So, let just add one "$C$" on the right hand side (of your last equation). The condition states that "$y$ is $-1$ when $x$ is $1$". So the value of $C$ is $-1$. Rearrange some terms to get the solution as $F(x,y)=0$.

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    Wow! As I said, it should be simple (cause it's just an exercise from a basic chapter). If the explicit solution contains complicated functions as Lambert W function, then I guess that the answer should be on $F(x,y)=0$ form. Not really sure, but gonna accept the answer.2012-06-24
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Hint: divide by $x^3y^3$ and the variables separate.

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    I forgot to mention that I already did that, I've just edited the question2012-06-24