Solve the differential equation : $(x^2y-2xy^2)dx-(x^3-3x^2y)dy=0.$
My solution goes like this:
This is a first order and first degree differential equation of the form $Mdx+Ndy=0,$ where $M=x^2y-2xy^2,N=-(x^3-3x^2y)=3x^2y-x^3.$ Now, since, $Mx+Ny=x^2y^2\neq 0,$ and homogeneous as well, so the integrating factor will be $\frac{1}{Mx+Ny}=\frac{1}{x^2y^2}.$ Multiplying $\frac{1}{x^2y^2}$ on both sides of the given equation we get, an exact differential equation i.e $$\frac{1}{x^2y^2}[(x^2y-2xy^2)dx-(x^3-3x^2y)dy]=0.$$ Now, the solution of this exact differential equation is $$\int [(\frac{x^2y-2xy^2}{x^2y^2})]dx+\int \frac 3y dy=c\implies \frac xy - 2\log x+3\log y=c,$$ where $c$ is an arbitary constant.
Is the above solution correct? If not, where is it going wrong ?
Standard form : $M(x,y)dx +N(x,y) dy =0\tag{1}$
Since $(1) $ is a homogenous non exact ode and $xM+yN=x^2y^2\neq 0$ , integrating factor $\frac{1}{x^2y^2}$
$\begin{align}&\int [(\frac{x^2y-2xy^2}{x^2y^2})]dx+\int \frac{ 3}{y}dy=c\\&\implies \frac{x}{y}-2\log x+{3}\log{y}=c\end{align}$
Alternative approach :
$(x^2y-2xy^2)dx-(x^3-3x^2y)dy=0$
$\frac{dy}{dx}=\frac{x^2y-2xy^2}{x^3-3x^2y}\tag{1}$
Let $y=vx$
$\frac{dy}{dx}=v+x\frac{dv}{dx}$
$\begin{align}v+x\frac{dv}{dx}&=\frac{vx^3-2v^2x^3}{x^3-3vx^3}\\&=\frac{v-2v^2}{1-3v}\end{align}$
$\begin{align}x\frac{dv}{dx}&=\frac{v-2v^2}{1-3v}-v\\&=\frac{v^2}{1-3v}\end{align}$
$\begin{align}&\implies \frac{1-3v}{v^2}dv=\frac{dx}{x}\\&\implies -\frac{1}{v}-3\log v=\log x+c\end{align}$
Now substituting $v=\frac{y}{x}$ , we have
$\frac{x}{y}+3\log\frac{y}{x}+\log x=c$
$\implies \frac{x}{y}+3\log{y}-3\log {x}+\log x=c$ $\implies \frac{x}{y}+3\log{y}-2\log {x}=c$