Find the general and singular solutions of x(y')^2=y

Find the general and singular solutions of $x(y')^2=y$.

Step 1: Separate and integrate

$y'=\pm\sqrt{y/x}\Rightarrow \dfrac{dy}{\sqrt y}=\pm\dfrac{dx}{\sqrt x}$, so $2\sqrt y=\pm2\sqrt x+C_0$, giving $\sqrt y = \pm\sqrt x+C$ and, squaring, $y=(\sqrt x+C)^2$.

Step 2: Verify the general solution by substitution

$y=(\sqrt x+C)^2\Rightarrow y'=\dfrac{\sqrt x+C}{\sqrt x}$, so $x(y')^2=(\sqrt x+C)^2=y$. Residual $\equiv0$ (confirmed symbolically). $\boxed{y=(\sqrt x+C)^2}$ verified.

Step 3: $p$-discriminant candidate

$F=xp^2-y$, $F_p=2xp=0\Rightarrow p=0$ (for $x\neq0$); substituting into $F=0$ gives $y=0$. Check: $y\equiv0\Rightarrow y'=0$, and $x\cdot0-0=0$, satisfied for all $x$.

Step 4: C-discriminant confirms it is a true envelope

$G=y-(\sqrt x+C)^2=0$, $\partial G/\partial C=-2(\sqrt x+C)=0\Rightarrow C=-\sqrt x$. Substituting back: $G=y-(\sqrt x-\sqrt x)^2=y=0$. The elimination reproduces $y=0$ exactly (confirmed symbolically), and every member of the family actually touches $y=0$ at the finite point $x=C^2$ with matching (vertical, in the $x=g(y)$ sense) contact. Both tests agree.