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

Question

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:

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

Answer 1

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 : $$I.F.=x^2y^4$$ $$x^2y^4\left((x^2y-2xy^2)\,dx+(x^3-3x^2y)\,dy\right)=0$$ $$(x^4y^5-2x^3y^6)\,dx+(x^5y^4-3x^4y^5)\,dy=0$$ $$d(\frac15 x^5y^5-\frac12 x^4y^6)=0$$ $$\frac15 x^5y^5-\frac12 x^4y^6=C$$

Answer 2

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?

Answer 3

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.