Differential equation, substitution By means of the substitute $y = v(x)Y (x)$, where $Y (x)$ is to be specified, solve the differential equation:

$$\dfrac{dy}{dx}+\dfrac{y}{x}=\dfrac{y^2}{x}$$ 

with $y=2$ at $x=1$

Anyone can solve it for me with explaining the steps please, I have no idea how to do it. Thank you very much
$$Y(x)=x$$


$\therefore y=vx\implies \dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$

$\therefore\dfrac{dy}{dx}+\dfrac{y}{x}=\dfrac{y^2}{x}\\\implies 2v+x\dfrac{dv}{dx}=v^2x\\\implies \dfrac{dv}{dx}+2\left(\dfrac{1}{x}\right)v=v^2$

Which is in Bernoulli's Form.
$$
y' = v'Y + vY' = -\frac{vY}{x} + \frac{v^2Y^2}{x}
$$
if we re-write
$$
v'Y = -vY' -\frac{vY}{x} + \frac{v^2Y^2}{x} = -\left(Y' + \frac{Y}{x}\right)v + \frac{v^2Y^2}{x}
$$
set the brackets term to zero i.e.
$$
Y' = -\frac{Y}{x}\implies Y(x) = \frac{C}{x} 
$$
then we obtain
$$
v' = \frac{1}{Y}\frac{v^2Y^2}{x} = \frac{Y}{x}v^2 = \frac{C}{x^2} v^2
$$
which is separable.
Thus
$$
v = \frac{x}{C+C_1x}\implies y(x) = \frac{C}{C+C_1x}
$$
put it back into the equation we find
$$
y' +\frac{y}{x} = \frac{-C C_1}{\left(C+C_1x\right)^2} + \frac{C}{\left(C+C_1x\right)x} = \\
\frac{-C}{\left(C+C_1x\right)^2x}\left[C_1 x -\left(C+C_1x\right)\right] = \frac{C^2}{\left(C+C_1x\right)^2x} = \left(\frac{C}{\left(C+C_1x\right)}\right)^2\frac{1}{x} = \frac{y^2}{x}
$$