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The Ordinary Differential Equation $y(x^4-y^2)dx+x(x^4+y^2)dy=0$ can be solve using Integrating Factor by Inspection. Here is the solution using the inspection method:

By expanding the equation, we have,$$x^4ydx-y^3dx+x^5dy+xy^2dy=0$$ $$x^4(ydx+xdy)+y^2(xdy-ydx)=0$$ Multiplying $\frac{1}{x^4}$,$$(ydx+xdy)+\frac{y^2}{x^2}\frac{xdy-ydx}{x^2}=0$$ $$d(xy)+\left(\frac{y}{x}\right)^2d\left(\frac{y}{x}\right)=0$$ $$xy+\frac{y^3}{3x^3}=C ,\ where\ C\ is\ a\ constant\ no.$$ $$3x^4y+y^3=Cx^3$$

I wanna know if it is also possible to use other method to solve this equation, because I tried it using the "General Method for finding the Integrating Factor" but my answer is not complete, and if it is, what are those possible method? If possible with your solution attached.

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Looking at the equation, I first change $y=x^2 \,z$ which makes $y'=2x \,z+x^2\,z'$. Replacing and simplifying leads to $$x \left(z^2+1\right) z'+z \left(z^2+3\right)=0$$ which is separable $$\frac{x'} x=-\frac{z^2+1}{z \left(z^2+3\right)}=-\frac{2 z}{3 \left(z^2+3\right)}-\frac{1}{3 z}$$ which integrates very simply and leads to $$x=\frac{C}{\sqrt[3]{z(z^2+3)}}$$

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  • $\begingroup$ Does the answer remains the same using different methods? $\endgroup$ – H. Vigilia Feb 25 '16 at 16:46
  • $\begingroup$ I think it must ! This is exactly your answer. Cheers. $\endgroup$ – Claude Leibovici Feb 25 '16 at 17:09

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