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Find the integrating factor of the differential equation: $$(x^2y-2xy^2)\,dx+(x^3-3x^2y)\,dy=0$$

What I tried:

This is a homogeneous equation.

Therefore,

$$I.F=\frac{1}{Mx+Ny}=\frac{1}{(x^2y-2xy^2)x+(x^3-3x^2y)y}=\frac{1}{x^3y-2x^2y^2+x^3y-3x^2y^2}$$

However, the given answer is:

$$I.F=\frac{1}{x^2y^2}$$

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  • $\begingroup$ Isn't $$I.F=\frac{1}{Mx-Ny}$$ what theorem you have used? $\endgroup$ – Nosrati Aug 8 '18 at 13:11
  • $\begingroup$ @user108128 If $ M$ and $N$ are both homogeneous functions in $x,y$ of same degree and $Mx+Ny \ne 0 $, then $\frac{1}{Mx+Ny}$ is an integrating factor. $\endgroup$ – Soumee Aug 8 '18 at 13:15
  • $\begingroup$ What $M$ , $N$, maybe in $f=\dfrac{M}{N}$ $\endgroup$ – Nosrati Aug 8 '18 at 13:17
  • $\begingroup$ @user108128 Here $M=(x^2y-2xy^2)$ and $N=(x^3-3x^2y)$ $\endgroup$ – Soumee Aug 8 '18 at 13:20
  • $\begingroup$ @Soumee : I suggest to you to write in your question the exact wording of the theorem that you use, without forgetting to fully specify the conditions of validity for its application. $\endgroup$ – JJacquelin Aug 8 '18 at 15:13
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I don't know if you apply the so-called "theorem" on convenient condition, but the equation that you found is false. Just check it with the correct integrating factor : $$I.F.=x^2y^4$$ $$x^2y^4\left((x^2y-2xy^2)\,dx+(x^3-3x^2y)\,dy\right)=0$$ $$(x^4y^5-2x^3y^6)\,dx+(x^5y^4-3x^4y^5)\,dy=0$$ $$d(\frac15 x^5y^5-\frac12 x^4y^6)=0$$ $$\frac15 x^5y^5-\frac12 x^4y^6=C$$

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  • $\begingroup$ Sir, can you please help me with this question: math.stackexchange.com/q/2881062/394202 $\endgroup$ – Soumee Aug 13 '18 at 6:54
  • $\begingroup$ Hi Soumee! Meanwhile you got an answer from Nosrati. I think that is sufficient to help you. $\endgroup$ – JJacquelin Aug 14 '18 at 9:39
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Hint: Making the Substitution $$y(x)=xv(x)$$ then we get

$$\frac{dv(x)}{dx}=\frac{-5v(x)^2+2v(x)}{x(3v(x)-1)}$$ and this is $$\frac{\frac{dv(x)}{dx}(3v(x)-1)}{-5v(x)^2+2v(x)}=\frac{1}{x}$$ and we can integrate

$$\int \frac{\frac{dv(x)}{dx}(3v(x)-1)}{-5v(x)^2+2v(x)}dx=\int \frac{1}{x}dx$$ Can you finish?

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However I'm not familiar with your method, but perhaps this This method for homogeneous equation may help. We have $$f(x,y)=-\dfrac{x^2y-2xy^2}{x^3-3x^2y}=y'$$ is homogeneous in sense of $$\color{red}{f(tx,ty)=t^\alpha f(x,y)}$$ then $$y'=-\dfrac{\frac{y}{x}-2(\frac{y}{x})^2}{1-3\frac{y}{x}}$$ with $u=\dfrac{y}{x}$ $$u'x+u=-\dfrac{u-2u^2}{1-3u}$$ which is separable.

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