Solve $y=xy'+\frac{1}{2}y'^2$ (need approach) $y=xy'+\frac{1}{2}y'^2$
Can someone explain me how to solve these type of equations it's my first time "seeing" an equation that has $y'$ to some degree, no need for a complete solution just the general method.
EDIT:
I read some articles and came up with this:
$y=xp+\frac{1}{2}p^2$ now if we differentiate in respect to $x$ we get:
$p=xp'+p+pp'$
$pp'+xp'=0$
$p'(p+x)=0$
$p'=0$ then $p=C$ and if $p+x=0$ then $p=-x$
so we have $y_1 = Cx+K$ and $y_2=\frac{-x^2}{2}+K$, where C and K are const.

This is a type of ODE named "Clairaut's equation".
Suppose you have an equation in the form of:
$$y=xy'+\psi(y')$$
Defining $p=y'$ we get
$$y=xp+\psi(p)$$
If you take the derivative of the equation you get:
$$p=1\cdot p+x\cdot \frac{dp}{dx} + \psi'(p)\cdot \frac{dp}{dx}$$
or
$$\frac{dp}{dx}(x+\psi'(p))=0$$
So you get two solutions. Either $\frac{dp}{dx}=0$ and $p$ is constant (hence, you can see that $y=ax+\psi(a)$), or
$$x=-\psi'(p)$$
that's an algebraic equation that will give $p(x)$ and you can substitute it in the original equation:
$$y=xy'+\psi(y')=-\psi'(p)\cdot p + \psi(p) $$
where $p$ is a known function of $x$.
