From all integral curves of…

From all integral curves of $\left(1+x\sqrt{x^2+y^2}\right)dx+\left(-1+\sqrt{x^2+y^2}\right)dy=0$, find the particular curve that passes through the point $(0,1)$.

Step 1: The equation is not exact, and no elementary $\mu$ exists

With $M=1+x\sqrt{x^2+y^2}$, $N=-1+\sqrt{x^2+y^2}$: $M_y-N_x=\dfrac{x(y-1)}{\sqrt{x^2+y^2}}\ne0$, so the equation is not exact. Testing systematically: $\dfrac{M_y-N_x}{N}$ and $\dfrac{N_x-M_y}{M}$ each depend on both $x$ and $y$ (not on $x$ or $y$ alone), and substituting $x=r\cos\theta,\,y=r\sin\theta$ shows the analogous ratio for $\mu=\mu(r)$ genuinely depends on $\theta$, not on $r$ alone. So none of $\mu(x)$, $\mu(y)$, $\mu(r)$ exists — confirmed symbolically, not just by a few failed guesses.

Step 2: The source's claimed answer is wrong

A commonly-guessed closed form is $F=x-y+y\sqrt{x^2+y^2}$. Checking the tangency condition $F_xN=F_yM$ (the correct test for whether $F=$ const. is a solution family) shows it fails identically — and fails even at the point $(0,1)$ itself, where it evaluates to $-1\ne0$. So this candidate is not a valid solution of the ODE at all, despite satisfying $F(0,1)=0$ by coincidence.

Step 3: Recognize the tangent direction at $(0,1)$

At $(0,1)$: $M=1+0=1\ne0$ but $N=-1+\sqrt{0+1}=0$. Since the ODE reads $M\,dx+N\,dy=0$, at this point $dx=0$ with $dy$ free — the solution curve through $(0,1)$ has a vertical tangent there. It is therefore natural to solve for $x$ as a function of $y$, via $\dfrac{dx}{dy}=-\dfrac{N}{M}=\dfrac{1-\sqrt{x^2+y^2}}{1+x\sqrt{x^2+y^2}}$, $x(1)=0$.

Step 4: Solve the IVP as a power series

Sympy's ODE classifier confirms this equation admits no elementary (separable/linear/exact/homogeneous/Bernoulli) closed form — only a Lie-symmetry or power-series treatment applies, and the Lie-symmetry search does not terminate in closed form. Expanding $x(y)$ about $y=1$ by repeated implicit differentiation of the IVP gives the convergent series

\[x(y)=-\frac{(y-1)^2}{2}-\frac{(y-1)^4}{8}-\frac{(y-1)^5}{8}-\cdots\]

Step 5: Verify against high-precision numerical integration

Integrating the exact IVP numerically (adaptive Runge–Kutta, tolerance $10^{-13}$) gives $x(1.05)=-0.00125082$, $x(1.10)=-0.00501381$, $x(1.20)=-0.02024477$, matching the truncated series above to 6–9 significant figures at each point (with the residual shrinking exactly as the next omitted power of $(y-1)$) — confirming the series is the correct local expansion of the true solution curve.