# differential equation resolution

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 ?

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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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If you really want explicit solutions, this one can be solved using the Lambert W function. – Robert Israel Jun 24 '12 at 19:07
I have not noticed it. Please feel free to edit my post/& write your own answer/comment. – Tapu Jun 24 '12 at 19:16
Note, by the way, that the differential equation is singular on the line $y=-1$, corresponding to your curve $F(x,y)=0$ having a vertical tangent at $(1,-1)$. There will be two solutions for $x \ge 1$ and none for $x < 1$. – Robert Israel Jun 24 '12 at 19:20
Those solutions are, according to Maple, $$-{{\rm e}^{1/2\, \left( {\it LambertW} \left( -{{\rm e}^{-{\frac {-1+2 \,\ln \left( x \right) {x}^{2}+2\,{x}^{2}}{{x}^{2}}}}} \right) {x}^{2 }-1+2\,\ln \left( x \right) {x}^{2}+2\,{x}^{2} \right) {x}^{-2}}}$$ and $$-{{\rm e}^{1/2\, \left( {\it LambertW} \left( -1,-{{\rm e}^{-{\frac {- 1+2\,\ln \left( x \right) {x}^{2}+2\,{x}^{2}}{{x}^{2}}}}} \right) {x} ^{2}-1+2\,\ln \left( x \right) {x}^{2}+2\,{x}^{2} \right) {x}^{-2}}}$$ – Robert Israel Jun 24 '12 at 19:21
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. – mhfff32 Jun 24 '12 at 19:48

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 question – mhfff32 Jun 24 '12 at 15:49