Solve the differential equation (xy' - y)^2 = (y')^2…

Solve the differential equation \[ (xy' - y)^2 = (y')^2 - \frac{2yy'}{x} + 1. \]

1. Expand and reduce to a quadratic in \(p=y'\)

Expanding \((xy'-y)^2\) and moving everything to one side gives \[ (x^2-1)p^2 - \frac{2y(x^2-1)}{x}\,p + (y^2-1) = 0, \] which after dividing by \(x^2-1\) (for \(x^2\neq 1\)) is \[ p^2 - \frac{2y}{x}p + \frac{y^2-1}{x^2-1} = 0.\qquad(\star) \] A direct symbolic check (Sympy \(\texttt{Poly}\) on the fully expanded original equation) confirms these are exactly the coefficients \((x^2-1,\,-2y(x^2-1)/x,\,y^2-1)\), so \((\star)\) is correct.

2. Solve the quadratic for \(p\)

The discriminant of \((\star)\) simplifies to \(\dfrac{4(x^2-1)(x^2-y^2)}{x^2}\), so \[ p = \frac{y}{x} \pm \frac{\sqrt{x^2-y^2}}{x\sqrt{x^2-1}}. \qquad (4) \] This was checked with Sympy by solving \((\star)\) directly for \(p\) and confirming the two roots agree with \((4)\) (the difference vanishes identically on the domain \(x^2\ge 1,\ x^2\ge y^2\) where both square roots are real). This part of the original derivation is correct — the error in the source was giving up immediately after this point.

3. Trigonometric substitution \(y = x\sin\theta\)

Both radicals \(\sqrt{x^2-y^2}\) and \(\sqrt{x^2-1}\) suggest a mixed trig/inverse-secant substitution. Put \(y = x\sin\theta(x)\), so \(\sqrt{x^2-y^2}=x\cos\theta\) (for \(\cos\theta\ge0\)). Then \[ y' = \sin\theta + x\cos\theta\,\theta'. \] Substituting into \((4)\) — verified symbolically with Sympy — the \(\sin\theta\) terms cancel and, after dividing by \(x\cos\theta\) (\(\cos\theta\neq0\)), each branch collapses to the strikingly simple separable equation \[ \theta' = \frac{\pm 1}{x\sqrt{x^2-1}}. \]

4. Integrate to get the two general-solution families

Sympy evaluates \(\displaystyle\int \frac{dx}{x\sqrt{x^2-1}} = -\arcsin(1/x)+\text{const}\) for \(x>1\). Hence \(\theta = \mp\arcsin(1/x) + C\), i.e. \(\arcsin(y/x) \pm \arcsin(1/x) = C\). This was verified the rigorous way — not just by integrating, but by taking the implicit relation \(\arcsin(y/x) + s\arcsin(1/x) = C\) (\(s=\pm1\)) and differentiating it implicitly in Sympy: the resulting \(y'\) reproduces \(\dfrac{y}{x} + s\dfrac{\sqrt{x^2-y^2}}{x\sqrt{x^2-1}}\) exactly, term for term, confirming both families solve equation \((4)\) — and hence the original ODE — with no residual error.

5. The envelope: why \(y=\pm x\) appeared from a lucky guess

The source only found \(y=\pm x\) by blindly trying \(y=kx\). Applying the standard \(p\)-discriminant (envelope) test to the quadratic \(F(x,y,p)=(x^2-1)p^2-\frac{2y(x^2-1)}{x}p+(y^2-1)=0\) is the rigorous way to explain this: eliminate \(p\) between \(F=0\) and \(\partial F/\partial p=0\). Sympy gives \(\partial F/\partial p=0 \Rightarrow p=y/x\); substituting back into \(F\) yields \(F = x^2-1\)-free expression \(\dfrac{y^2}{x^2}-1=0\), i.e. \(y=\pm x\) exactly. So \(y=x\) and \(y=-x\) are not accidental particular solutions — they are the genuine \(p\)-discriminant locus (the envelope tangent to every curve of both one-parameter families above), which is exactly why the naive trial \(y=kx\) stumbled onto them.