Find the orthogonal trajectories of the family (2a-x)y^2=x^3…

Find the orthogonal trajectories of the family $(2a-x)y^2=x^3$, with $a$ the parameter.

Step 1: Re-derive the family ODE independently

With $F=(2a-x)y^2-x^3$, \texttt{sympy} gives $\dfrac{dF}{dx}=4ayy'-3x^2-2xyy'-y^2$. Solving the original equation $F=0$ for $a$: $a=\dfrac{x^3}{2y^2}+\dfrac{x}{2}$. Substituting this into $dF/dx=0$ and simplifying with \texttt{sympy} gives, after clearing denominators, \[\frac{2x^3y'}{y}-3x^2-y^2=0\ \Longrightarrow\ y'=\frac{y(3x^2+y^2)}{2x^3},\] which matches the source's family ODE exactly — this step, the one flagged as highest-risk, checks out.

Step 2: Orthogonal ODE

Negative reciprocal (trivial): $y'=-\dfrac{2x^3}{y(3x^2+y^2)}$, matching the source.

Step 3: Test the source's guessed integral — it fails

Write the orthogonal ODE as $2x^3\,dx+(3x^2y+y^3)\,dy=0$, i.e. $M=2x^3,\ N=3x^2y+y^3$. Implicit differentiation of the source's claimed $2x^4+6x^2y^2+y^4=C$ gives, via \texttt{sympy}, $y'=\dfrac{x(-2x^2-3y^2)}{y(3x^2+y^2)}$. Comparing to the orthogonal ODE, the difference is \[\frac{x(-2x^2-3y^2)}{y(3x^2+y^2)}-\left(-\frac{2x^3}{y(3x^2+y^2)}\right)=-\frac{3xy}{3x^2+y^2}\ne0.\] The source's "verification by re-differentiating" was checking its guess against its own guess's implied slope, not against the true orthogonal ODE — it is wrong.

Step 4: Solve rigorously via an integrating factor (not a guess)

$M_y=0,\ N_x=6xy$, confirming non-exactness (source was right about that much). Testing systematically with \texttt{sympy}: $(M_y-N_x)/N=-6x/(3x^2+y^2)$ still depends on $y$ (no $\mu(x)$), and $(N_x-M_y)/M=3y/x^2$ still depends on $x$ (no $\mu(y)$). Since $M,N$ are both homogeneous of degree $3$, the standard homogeneous integrating factor $\mu=1/(xM+yN)=1/(2x^4+3x^2y^2+y^4)$ applies; \texttt{sympy} confirms $\partial_y(\mu M)=\partial_x(\mu N)$ exactly. Integrating $\mu M$ in $x$ and matching against $\mu N$ gives the potential \[\Phi=\ln(x^2+y^2)-\tfrac12\ln(2x^2+y^2)+\text{const},\] so $\Phi=$ const $\iff \dfrac{(x^2+y^2)^2}{2x^2+y^2}=$ const, i.e. \[(x^2+y^2)^2=C(2x^2+y^2).\]

Step 5: Confirm the corrected solution symbolically

Implicit differentiation of $(x^2+y^2)^2=C(2x^2+y^2)$ and elimination of $C$ (via the original relation) reproduces the orthogonal ODE $y'=-\dfrac{2x^3}{y(3x^2+y^2)}$ exactly — residual $0$ confirmed by \texttt{sympy}.

Step 6: End-to-end numerical cross-check

Take $a=1$: the point $(x,y)=(1,1)$ lies on $(2-x)y^2=x^3$. Implicit differentiation of the original equation (with $a=1$ fixed, not the derived ODE) gives family slope $y'=2$ at $(1,1)$. The corrected orthogonal candidate $(x^2+y^2)^2=C(2x^2+y^2)$ through $(1,1)$ requires $C=4/3$; implicit differentiation gives orthogonal slope $y'=-1/2$ there. Product: $2\times(-\tfrac12)=-1$ — exact confirmation. Repeating the same check with the source's wrong candidate $2x^4+6x^2y^2+y^4=C$ gives slope $-5/4$ at the same point, product $2\times(-\tfrac54)=-\tfrac52\ne-1$, definitively refuting it.