What is the general solution to, $x \frac{dy}{dx} = 2xy + y^2$ using the substitution $y=vx$

Question

I get as far as $\frac{dv}{dx} = 2v + v^2 - \frac{v}{x}$ but then I have no idea what to do from there.

The result you get may be left as implicit

Answer 1

A typo? Did you mean $$ \dfrac{dy}{dx} = \dfrac{2xy+y^2}{x^2}?$$

Standard substitution $ y=v x$ then works.

Answer 2

You don't need to use that substitution cause its reserved for Homogeneous cases, this isn't one of them.

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

It has a Bernoulli Form, rearrange in this way $\ y^{-2}y'-2y^{-1}=x^{-1}\ $ and the substitution $\ z=y^{-1}\ $ derive with respect to x, cause you will substitute in $y'$.

Then $\ z'+2z=-x^{-1}\ $, after an Integration by Parts; solve for $z$

$$z=\frac14(1-2x)+Ce^{-2x}\rightarrow \ y^{-1}=\frac14(1-2x)+Ce^{-2x}$$

$$y(x)=\frac{4}{Ce^{-2x}-2x+1}$$