Solve the differential equation…

Solve the differential equation $y'=\dfrac{y-x^2\sqrt{x^2-y^2}}{xy\sqrt{x^2-y^2}+x}$.

Step 1: Exactness / integrating-factor test (independent of the source)

Writing the equation as $M\,dx+N\,dy=0$ with $M=y-x^2\sqrt{x^2-y^2}$, $N=-x(y\sqrt{x^2-y^2}+1)$, \texttt{sympy} gives \[\partial M/\partial y=1+\frac{x^2y}{\sqrt{x^2-y^2}},\qquad \partial N/\partial x=\frac{-2x^2y+y^3-\sqrt{x^2-y^2}}{\sqrt{x^2-y^2}},\] which are unequal, so the equation is not exact. Testing $(M_y-N_x)/N$ and $(N_x-M_y)/M$ with \texttt{sympy.simplify}, neither reduces to a function of $x$ alone or $y$ alone — no simple integrating factor $\mu(x)$ or $\mu(y)$ exists. This dead end is confirmed, not merely asserted.

Step 2: Push $y=x\sin\theta$ through completely with \texttt{sympy}

Substituting $y=x\sin\theta$, $\sqrt{x^2-y^2}=x\cos\theta$ into the original equation and solving symbolically for $\theta'$ (rather than stopping at the source's unsimplified equation (4)), \texttt{sympy} returns \[\theta'=\frac{2x(\cos^2\theta-2)}{x^2\sin2\theta+2}=\frac{-x(1+\sin^2\theta)}{x^2\sin\theta\cos\theta+1},\] which is algebraically identical to the source's equation (4) — so the source's algebra up to that point was actually correct; it simply stopped short of finishing the analysis, wrongly declaring it "too complex."

Step 3: Recognize a linear equation in $x^2$ (inverting the roles, as in Problem 216)

This $\theta$-equation is not separable as written, but inverting it to $dx/d\theta$ and substituting $u=x^2$ (using $2x\,dx/d\theta=du/d\theta$), \texttt{sympy} reduces it to \[\frac{du}{d\theta}+\underbrace{\frac{2\sin\theta\cos\theta}{1+\sin^2\theta}}_{P(\theta)}\,u=\underbrace{\frac{-2}{1+\sin^2\theta}}_{Q(\theta)},\] a genuine first-order linear equation in $u(\theta)$ — exactly the same trick that solves Problem 216, applied here after inverting the independent variable.

Step 4: Solve the linear equation and back-substitute

Since $\dfrac{d}{d\theta}(1+\sin^2\theta)=2\sin\theta\cos\theta$, the integrating factor is $\mu(\theta)=1+\sin^2\theta$, and \texttt{sympy.dsolve} confirms \[u(1+\sin^2\theta)=-2\theta+C_1.\] Since $u=x^2$ and $y=x\sin\theta$, note $u\sin^2\theta=x^2\sin^2\theta=y^2$, so $u(1+\sin^2\theta)=x^2+y^2$ and $\theta=\arcsin(y/x)$. This gives the remarkably clean implicit solution \[x^2+y^2+2\arcsin\!\left(\frac{y}{x}\right)=C.\]

Step 5: Full independent verification — symbolic and numeric

Symbolic: implicitly differentiating $x^2+y(x)^2+2\arcsin(y(x)/x)=C$ with \texttt{sympy}, solving for $y'$, substituting into the original right-hand side $\dfrac{y-x^2\sqrt{x^2-y^2}}{xy\sqrt{x^2-y^2}+x}$, and calling \texttt{sympy.simplify(radsimp(...))} on the difference gives exactly $\mathbf{0}$ — an exact symbolic confirmation, not an approximation.
Numeric (independent cross-check): parametrizing $x(\theta)=\sqrt{(C-2\theta)/(1+\sin^2\theta)}$, $y(\theta)=x(\theta)\sin\theta$ and computing $dy/dx$ by finite differences at several $(C,\theta_0)$ points — $(C,\theta_0)=(5,\,0.3)$, $(10,\,-0.7)$, $(3,\,1.0)$, $(20,\,0.05)$ — the numeric $dy/dx$ matches the original ODE's right-hand side to within $10^{-9}$–$10^{-10}$ (finite-difference error) in every case. This is a genuine closed-form solution the source missed entirely.