Find the general and singular solutions (if any) of…

Find the general and singular solutions (if any) of $xy'=y+\sqrt{y^2-x^2}$.

Step 1: Homogeneous substitution

With $v=y/x$ (so $y'=v+xv'$), for $x>0$ the equation reduces to $x\,v'=\sqrt{v^2-1}$.

Step 2: Solve for $v^2>1$

$\dfrac{dv}{\sqrt{v^2-1}}=\dfrac{dx}{x}\Rightarrow\ln\big|v+\sqrt{v^2-1}\big|=\ln|x|+\text{const}\Rightarrow v+\sqrt{v^2-1}=Kx$. Solving this for $v$ (isolate the radical and square) gives $v=\dfrac{K^2x^2+1}{2Kx}$, hence $y=vx=\dfrac{K^2x^2+1}{2K}$; writing $C=1/(2K)$ gives $y=\dfrac{x^2}{4C}+C$.

Step 3: Domain of validity

Direct substitution shows $y=\dfrac{x^2}{4C}+C$ satisfies the original equation $xy'=y+\sqrt{y^2-x^2}$ (with the $+$ branch of the square root) only where $x^2\ge4C^2$, i.e. $|x|\ge2|C|$ — the region consistent with $v^2\ge1$ used in Step 2 (this domain restriction was glossed over, and never actually confirmed, in the AI-generated source's own "verification" step, which broke down into an unresolved computation). On $|x|<2|C|$ the same algebraic formula solves the companion equation $xy'=y-\sqrt{y^2-x^2}$ instead.

Step 4: Envelope / singular solutions

The boundary case $v^2=1$ gives $y=x$ and $y=-x$; both satisfy the original equation directly ($\sqrt{y^2-x^2}=0$). Eliminating $C$ from $G=y-\frac{x^2}{4C}-C=0,\ G_C=\frac{x^2}{4C^2}-1=0$ gives $C=\pm x/2$, and substituting back reproduces exactly $y=x$ and $y=-x$ — confirming both are envelopes of the parabola family, and neither is a member of it for any single fixed $C$.