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

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:

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

• Isn't $$I.F=\frac{1}{Mx-Ny}$$ what theorem you have used? Aug 8, 2018 at 13:11
• @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. Aug 8, 2018 at 13:15
• What $M$ , $N$, maybe in $f=\dfrac{M}{N}$ Aug 8, 2018 at 13:17
• @user108128 Here $M=(x^2y-2xy^2)$ and $N=(x^3-3x^2y)$ Aug 8, 2018 at 13:20
• @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. Aug 8, 2018 at 15:13

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

• Sir, can you please help me with this question: math.stackexchange.com/q/2881062/394202 Aug 13, 2018 at 6:54
• Hi Soumee! Meanwhile you got an answer from Nosrati. I think that is sufficient to help you. Aug 14, 2018 at 9:39

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?

• Sir, what is wrong with my approach? Aug 8, 2018 at 13:08
• Sir, can you please help me with this question: math.stackexchange.com/q/2881062/394202 Aug 13, 2018 at 6:55

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.