Solve y'^2+4xy'-y^2-2x^2y=x^4-4x^2 for y' and integrate

Solve $y'^2+4xy'-y^2-2x^2y=x^4-4x^2$ for $y'$ and integrate.

Step 1: View as quadratic in $p$

With $p=y'$: $p^2+4xp-(y^2+2x^2y+x^4-4x^2)=0$, a quadratic in $p$ with $a=1,\ b=4x,\ c=-(y^2+2x^2y+x^4-4x^2)$.

Step 2: Discriminant is a perfect square (verified in sympy)

\[b^2-4ac=4x^4+8x^2y+4y^2=4(x^2+y)^2.\] This is confirmed symbolically — the discriminant factors exactly as claimed, so $\sqrt{b^2-4ac}=2(x^2+y)$ (up to sign).

Step 3: Solve the quadratic

\[p=\frac{-4x\pm 2(x^2+y)}{2}=-2x\pm(x^2+y),\] giving two branches (sympy's solve returns these directly): \[\text{(A)}\quad p=y+x^2-2x,\qquad \text{(B)}\quad p=-y-x^2-2x.\]

Step 4: Solve branch (A) by integrating factor

$y'-y=x^2-2x$ is linear; $\mu=e^{-x}$ gives, after integration by parts (verified with sympy's dsolve), \[y=-x^2+C_1e^{x}.\]

Step 5: Solve branch (B) by integrating factor

$y'+y=-x^2-2x$ is linear; $\mu=e^{x}$ gives, again verified by sympy, \[y=-x^2+C_2e^{-x}.\]

Step 6: Check the $p$-discriminant curve $y=-x^2$

Substituting $y=-x^2$ into the original equation gives $0=0$ identically (sympy confirms), so $y=-x^2$ solves the ODE. But it is already the $C_1=0$ member of family (A) and the $C_2=0$ member of family (B) — so it is not an independent singular solution, merely a shared special case. Source's conclusion is correct.