Solution of the differential equation $x^2(y-x\frac{dy}{dx})=y(\frac{dy}{dx})^2$ which does not contain singular solution Solution of the differential equation $x^2\big(y-x\frac{dy}{dx}\big)=y\big(\frac{dy}{dx}\big)^2$ which does not contain singular solution is
A) $x^2(y-xc)=yc^2$
B) $y=cx+c^2$
C) $y^2=cx^2+c^2$
D) $xy=cx^2+c$
My Attempt:
I put $\frac{dy}{dx}=p$ and tried differentiating w.r.t $x$, but couldn't conclude.
I also tried to rearrange the equation but couldn't form the Clairaut's equation.
 A: D'Alembert's differential equation
$$x^2\Big(y-x\frac{dy}{dx}\Big)=y\Big(\frac{dy}{dx}\Big)^2$$
Substitute $u=x^2$
$$y-2u\frac{dy}{du}=4y\Big(\frac{dy}{du}\Big)^2$$
$$y(1-4y'^2)=2uy'$$
$$y=u\dfrac {2y'}{(1-4y'^2)}$$
This is D'Alembert's differential equation
$$y=uf(y')+g(y')$$
With $g=0$

Clairaut's differential equation
I also tried to rearrange the equation but couldn't form the Clairaut's equation.
$$x^2\Big(y-x\frac{dy}{dx}\Big)=y\Big(\frac{dy}{dx}\Big)^2$$
Multiply by $y$
$$x^2\Big(y^2-xy\frac{dy}{dx}\Big)=y^2\Big(\frac{dy}{dx}\Big)^2$$
Substitute $w=y^2$
$$w-\dfrac x2w'=\dfrac 1 {4x^2}(w')^2$$
Substitute $u=x^2$
$$w-uw'=(w')^2$$
This is Clairaut's differential equation:
$$w=uw'+w'^2$$
The general solution is:
$$w=uC+C^2$$
$$\boxed {y^2=x^2C+C^2}$$
A: Substitute : \begin{cases} w=y^{2} \\\\
z=x^{2}\\\\
\end{cases}
With this variable change, the equation becomes Clairaut's differential equation
$w=zw'+w'^{2}$
let: $w'=p$
$w=zp+p^{2}$
$p'(z+2p)=0$
\begin{cases} p'=0 \\\\
z+2p=0\\\\
\end{cases}
so
\begin{cases} w=cz+c^{2} \\\\
w=\frac{-z^{2}}{4}\\\\
\end{cases}
so
\begin{cases} y^{2}=cx^{2}+c^{2} \\\\
y^{2}=\frac{-x^{4}}{4}\\\\
\end{cases}
