A question about finding the integrating factors

Question

$$\frac {dy}{dx}=\frac {(xy-y+y^2)}{(x+2y)}$$ how to solve this equation by finding the integrating factors? I tried to find it but it seems didn't work.

Answer 1

we start with ste Standard trick here $$y(x)=xv(x)$$ then we have $$x\frac{dv(x)}{dx}+v(x)+\frac{(x+xv(x)-1)v(x)}{2v(x)+1}$$ and this simplifies to $$\frac{dv(x)}{dx}=\frac{(x-2)(v(x)+1)v(x)}{x(2v(x)+1)}$$ $$\frac{dv(x)}{dx}\frac{(2v(x)+1)}{(v(x)+1)v(x)}=\frac{x-2}{x}$$ and we can integrate $$\int\frac{\frac{dv(x)}{dx}(2v(x)+1)}{(v(x)+1)v(x)}dx=\int\frac{x-2}{x}dx$$

Answer 2

Hint

$$dy/dx=\frac{(xy-y+y^2)}{(x+2y)}$$ $$-(xy-y+y^2)dx+(x+2y)dy=0$$ $$\partial_yP-\partial_xQ=-(x+2y)$$ Integrating factor depends only on x $$\frac {d\mu}{\mu}=-\int dx \implies \mu(x)=e^{-x}$$