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Find the solution to the differntial equation:

$$2x^3y\mathrm{d}y+(1-y^2)(x^2y^2+y^2-1)\mathrm{d}x=0$$

I solved this question by rewriting the expression as:

$$\frac{2y}{(1-y^2)^2}\frac{\mathrm{d}y}{\mathrm{d}x}+\frac{y^2}{1-y^2}\frac{1}{x}=\frac{1}{x^3}$$

which is a linear differential equation of first order if you take $\frac{y^2}{1-y^2}=t$ which gives:

$$\frac{\mathrm{d}t}{\mathrm{d}x}+\frac{t}{x}=\frac{1}{x^3}$$

But it took some time. Is there any easier method?

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    $\begingroup$ Just a note using \rm: It's the sort of construct that should be used in braces, along with whatever you want in roman front. For example, consider $x\, {\rm d}x + y\, {\rm d}y$ versus $x\, \rm{d}x + y\, \rm{d}y$. The first was achieved using $x\, {\rm d}x + y\, {\rm d}y$ (the \, is just for a bit of spacing before the differential). As you can see, the second using $x\, \rm{d}x + y\, \rm{d}y$ inappropriately stays in roman typeface. $\endgroup$ – pjs36 Mar 8 '16 at 3:57
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Let me try my way : for a start, writing the differential equation as $$2x^3yy'+(1-y^2)(x^2y^2+y^2-1)=0$$ that is to say $$x^3(y^2)'+(1-y^2)(x^2y^2+y^2-1)=0$$ Define $z=y^2$ to make $$x^3z'+(1-z)(x^2z+z-1)=0$$ Now, and I agree that this is more tricky, set $$z=\frac x u+1\implies z'=\frac 1 u-\frac{x u'}{u^2}$$ and replace to end, after simplifications, with $$x^2 u'+x^2+1=0$$ which is now very simple to integrate since $$u'=-1-\frac 1 {x^2}$$

I must confess that I am not sure that this is faster than what you did. In my opinion, I think that the first step $z=y^2$ is the most significant.

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