Lagrange Differential equation I have problems proceeding in solving the following differential equation $$xy' + y + (y')^2 = 0.$$
After solving for $\frac{dy}{dx}$ in the quadratic and using the substitution $u^2 = x^2 - 4y$ in the discriminant, I obtain $\frac{du}{dx} = 2 \frac{x}{u} -1$. Please how can I proceed? 
 A: $$xy+y+y’^2=0$$
Let :  $y’=p$
$$y=-xy’-y’^2=-xp-p^2$$
$$ \frac{dy}{dp} =-p\frac{dx}{dp}-x-2p$$
$$p=y’=\frac{dy}{dx}=\frac{dy}{dp} \frac{dp}{dx} =-p-(x+2p) \frac{dp}{dx}$$
$$2p=-(x+2p) \frac{dp}{dx}$$
$$2p\frac{dx}{dp}+x=-2p$$
The solution of this first order linear ODE is :
$$x=-\frac{2}{3}p+\frac{C}{2\sqrt{p}}$$
$$y=-xp -p^2= \frac{2}{3}p^2-\frac{C}{2}\sqrt{p} –p^2 =-\frac{1}{3}p^2-\frac{C}{2}\sqrt{p}$$
Finally, the solution of $xy+y+y’^2=0$ expressed on parametric form with parameter $p$ is :
$$\begin{cases}
    x=-\frac{2}{3}p+ \frac{C}{2}\frac{1}{\sqrt{p} }     \\
    y =-\frac{1}{3}p^2-\frac{C}{2}\sqrt{p} \\
  \end{cases}$$
If one want to find the explicit function $y(x)$ the parameter $p$ has to be eliminated from the system $\left(x(p),y(p)\right)$. This is possible, but will lead to complicated equations.
A: First make the substitution $p=y'$ to get  $$px+y+p^2=0$$
Now differentiate wrt $x$ $$p'x+2p+2pp'=0$$
$$p'=\frac{-2p}{x+2p}$$
Put $p=vx\implies p'=v+xv'$
$$v+xv'=\frac{-2v}{1+2v}$$
$$xv'=\frac{-3v-2v^2}{1+2v}$$
$$\frac{1+2v}{3v+2v^2}dv=-\frac{dx}x$$
Now integrating,
$$\ln v+2\ln(2v+3)+3\ln x=c$$
Now put back $v=\frac px$ to get
$$\ln p+2\ln(2p+3x)=c\implies p(2p+3x)^2=c$$
I have no idea how to solve that last equation but I think someone else might be able to do it.
