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$y=xy'+\frac{1}{2}y'^2$

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.

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    i would solve this quadratic equation for $y'$ – 2017-01-28
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    something like $y'=t$ and we get $t^2+2xt-y=0$? – 2017-01-28
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    yes that's what i meant – 2017-01-28
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    can you verify if my approach was correct I edit the answer to include it. Thanks! – 2017-01-28

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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 $x$.

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    And if you have an equation of the form $y=x\cdot \varphi(y')+\psi(y')$ it's called "Lagrange's equation" (but $\varphi(y')\neq y'$) – 2017-01-28
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    I edited my question beforehand and it seems that I applied that approach, is it correct? – 2017-01-28
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    Yes, you applied this approach correctly. Well done! Don't forget that if you have a Lagrange equation the solution is different. – 2017-01-28