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:
$$I.F=\frac{1}{x^2y^2}$$
Best Answer
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 : $$I.F.=x^2y^4$$ $$x^2y^4\left((x^2y-2xy^2)\,dx+(x^3-3x^2y)\,dy\right)=0$$ $$(x^4y^5-2x^3y^6)\,dx+(x^5y^4-3x^4y^5)\,dy=0$$ $$d(\frac15 x^5y^5-\frac12 x^4y^6)=0$$ $$\frac15 x^5y^5-\frac12 x^4y^6=C$$