# 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,

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

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

• Isn't $$I.F=\frac{1}{Mx-Ny}$$ what theorem you have used? – Nosrati Aug 8 '18 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. – Soumee Aug 8 '18 at 13:15
• What $M$ , $N$, maybe in $f=\dfrac{M}{N}$ – Nosrati Aug 8 '18 at 13:17
• @user108128 Here $M=(x^2y-2xy^2)$ and $N=(x^3-3x^2y)$ – Soumee Aug 8 '18 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. – JJacquelin Aug 8 '18 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 – Soumee Aug 13 '18 at 6:54
• Hi Soumee! Meanwhile you got an answer from Nosrati. I think that is sufficient to help you. – JJacquelin Aug 14 '18 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?

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.
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}