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

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

Step 1: Expand and find $p$-discriminant candidates

$F=y^2p^2-2xyp-x^2+2y^2=0$. $F_p=2y^2p-2xy=0\Rightarrow p=x/y$; substituting back, $F=x^2-2x^2-x^2+2y^2=2(y^2-x^2)=0\Rightarrow y=\pm x$. Direct substitution of $y=x$ ($y'=1$) and $y=-x$ ($y'=-1$) into the original equation gives residual $0$ in both cases (sympy-verified) — these are genuine solution curves.

Step 2: Homogeneous substitution $y=vx$

Since the equation is homogeneous of degree 2, set $y=vx$, $p=v+xv'$. Substituting into $F=0$ and dividing by $x^2$ (sympy expansion) gives, after solving the resulting quadratic in $v'$ via sympy: \[v'=\frac{1-v^2\pm\sqrt{2-2v^2}}{xv}.\]

Step 3: Separate and integrate

This separates as $\dfrac{v\,dv}{1-v^2\pm\sqrt{2-2v^2}}=\dfrac{dx}{x}$. With $u=\sqrt{1-v^2}$ (so $v\,dv=-u\,du$), sympy confirms \[\int\frac{v\,dv}{1-v^2+\sqrt{2-2v^2}}=-\ln\!\left(u+\sqrt2\right)+\text{const},\qquad \int\frac{v\,dv}{1-v^2-\sqrt{2-2v^2}}=-\ln\!\left(\sqrt2-u\right)+\text{const}\] (each verified by differentiating back to the integrand — residual $0$). Integrating the right side ($\ln x$) and exponentiating gives $x\left(u\pm\sqrt2\right)=C$ i.e. $x\, u = C\mp\sqrt2\,x$.

Step 4: Back-substitute $u=\sqrt{1-v^2}=\sqrt{x^2-y^2}/x$

Since $v=y/x$, $u=\sqrt{1-y^2/x^2}=\sqrt{x^2-y^2}/x$ (for $x>0$), so $xu=\sqrt{x^2-y^2}$, giving the implicit general solution \[\sqrt{x^2-y^2}\;\pm\;\sqrt2\,x=C.\]

Step 5: Verify the general solution directly

Implicit differentiation of $\Phi=\sqrt{x^2-y^2}\pm\sqrt2 x-C=0$ gives $y'=\dfrac{x\pm\sqrt{2x^2-2y^2}}{y}$ (matching the source's quadratic-formula root for $p$ exactly). Substituting this $y'$ into $yy'(yy'-2x)-(x^2-2y^2)$, sympy simplifies the residual to $0$ identically — for any value of $C$, confirming this is a genuine two-branch general solution.

Step 6: Confirm $y=\pm x$ are genuinely singular

Setting $y=x$ in $\sqrt{x^2-y^2}\pm\sqrt2 x=C$ forces $\pm\sqrt2\,x=C$, true only at one point per fixed $C$, not for all $x$ — so $y=\pm x$ are not members of the general family for any fixed $C$. Together with the $p$-discriminant derivation in Step 1, this confirms $y=\pm x$ are true singular (envelope) solutions.