Find the solution to the differntial equation:


I solved this question by rewriting the expression as:


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


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

  • 1
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.