Solving $y'=f(x,y)$ when $f$ is a rational function $f(x,y)=\frac{y^2+2xy}{x^2y-x^2}$

Question

Solve the initial value problem $$\frac{dy}{dx}=\frac{y^2+2xy^2}{x^2y-x^2},~~y(1)=1$$

My attempt: I checked whether $(x^2y-x^2)dy-(y^2+2xy)dx=0$ is exact, but it was not exact. Also the substitution $y=ux$ makes $$x\frac{du}{dx}=\frac{u^2+ux-u}{ux-1}$$ and stucked.

Answer 1

$\frac{dy}{dx}=\frac{y^2+2xy^2}{x^2y-x^2}=\frac{(1+2x)y^2}{(y-1)x^2}\\ $

Your diff eq is separable.

$\int \frac {y-1}{y^2}\ dy = \int \frac {1+2x}{x^2} \ dx$

Answer 2

$$(x^2y-x^2)dy-(y^2+{2xy})dx=0$$ You forgot the square for $y$: $$(x^2y-x^2)dy-(y^2+\color {red}{2xy^2})dx=0$$ $$\dfrac {(y-1)}{y^2}dy-\dfrac {(1+2x)}{x^2}dx=0$$ The differential is exact. $$Mdx+Ndy=0$$ $$\partial_y M =\partial_x N=0$$ Note that the DE in the title and the DE in the body are different since you wrote: $$f(x,y)=\frac{y^2+2xy}{x^2y-x^2}$$