Determine the orthogonal trajectories of the curves…
Determine the orthogonal trajectories of the curves \[ (x^2+y^2)^2 + 2b^2(y^2-x^2) = a, \] where \(b\) is a constant and \(a\) is a variable parameter.
Explanation of Method
The source correctly reaches the differential equation of the orthogonal trajectories, \(dy/dx = y(x^2+y^2+b^2)/\big[x(x^2+y^2-b^2)\big]\) — re-derived here from scratch with Sympy and confirmed to match — then abandons a Cartesian solution attempt ("this is complex, let me try a different approach") and starts a polar-coordinates computation which it also abandons after writing down \(r^4-2b^2r^2\cos2\theta=a\), guessing a final answer with no polar integration actually carried out. The guess, \((x^2+y^2)^2-2b^2(x^2-y^2)=C\), is shown below to be algebraically identical to the original family (not a new family at all, let alone the orthogonal one) — its \(y'\) is verified with Sympy to equal the original family's \(y'\), not the orthogonal one. The correct trajectories are instead derived by actually carrying out the polar computation the source started: the substitution \(s=r^2\) turns the orthogonal ODE into a Bernoulli equation in \(s(\theta)\), which linearizes via \(w=1/s\) and integrates exactly with Sympy, landing on \(x^2-y^2-2kxy=b^2\). This is then verified two independent ways: directly, by implicit differentiation in Cartesian coordinates against the target ODE, and separately, by implicit differentiation of its polar form against the orthogonal ODE's own polar version — both residuals reduce to exactly \(0\).
1. Re-derive the differential equations from scratch
Differentiating \((x^2+y^2)^2+2b^2(y^2-x^2)-a\) implicitly with Sympy and solving for \(y'\) gives the original family's ODE \[ y' = \frac{x(b^2-x^2-y^2)}{y(b^2+x^2+y^2)}, \] matching the source. Replacing \(y'\) by \(-1/y'\) gives the orthogonal-trajectory equation \[ y' = \frac{y(x^2+y^2+b^2)}{x(x^2+y^2-b^2)}. \qquad(3) \]
2. The source's guess is not even a new family
Since \(2b^2(y^2-x^2)=-2b^2(x^2-y^2)\), the source's guessed \((x^2+y^2)^2-2b^2(x^2-y^2)=C\) is literally the same equation as the original family, just with the constant renamed. Sympy confirms this concretely: implicitly differentiating the guess gives exactly \(y'=x(b^2-x^2-y^2)/\big[y(b^2+x^2+y^2)\big]\) — the original family's slope, not the orthogonal slope \((3)\). So the guess fails outright; it isn't a near-miss, it's the starting family again.
3. Carry out the polar computation the source abandoned
With \(x=r\cos\theta,\,y=r\sin\theta\), \(x^2+y^2=r^2\) and \(x^2-y^2=r^2\cos2\theta\), so the original family is \(r^4-2b^2r^2\cos2\theta=a\) (matching the source's last written line). Differentiating implicitly and solving for \(dr/d\theta\), then applying the polar orthogonality substitution \(dr/d\theta\to -r^2\big/(dr/d\theta)\), Sympy gives the orthogonal-trajectory equation in polar form: \[ \frac{dr}{d\theta} = \frac{r\big(r^2-b^2\cos2\theta\big)}{b^2\sin2\theta}. \] Substituting \(s=r^2\) (\(ds/d\theta=2r\,dr/d\theta\)) turns this into a Bernoulli equation, \[ \frac{ds}{d\theta} + 2\cot(2\theta)\,s = \frac{2}{b^2\sin2\theta}\,s^2, \] and \(w=1/s\) linearizes it: \[ \frac{dw}{d\theta} - 2\cot(2\theta)\,w = -\frac{2}{b^2\sin2\theta}. \] Sympy confirms the needed antiderivatives \(\int 2\cot2\theta\,d\theta=\ln(\sin2\theta)\) and \(\int d\theta/\sin^2 2\theta = -\tfrac12\cot2\theta\), giving integrating factor \(1/\sin2\theta\) and \[ w = \frac{\cos2\theta}{b^2} + K\sin2\theta. \]
4. Back-substitute and convert to Cartesian
Since \(w=1/s=1/r^2\), \(1=r^2\cos2\theta/b^2+Kb^2r^2\sin2\theta\); multiplying by \(b^2\) and writing \(r^2\cos2\theta=x^2-y^2\), \(r^2\sin2\theta=2xy\), and relabelling the constant, this becomes \[ \boxed{\ x^2-y^2-2kxy=b^2\ } \] (equivalently \(2xy=k'(x^2-y^2-b^2)\) with \(k'=-1/k\)).
5. Two independent verifications against ODE (3)
Cartesian check: implicitly differentiating \(F=x^2-y^2-2kxy-b^2=0\) with Sympy and solving for \(y'(x)\) gives \(y'=(k x-y)/(ky+x)\); eliminating \(k\) using \(F=0\) itself (\(k=(x^2-y^2-b^2)/(2xy)\)) and simplifying reduces this to exactly \(y(x^2+y^2+b^2)\big/\big[x(x^2+y^2-b^2)\big]\) — the target \((3)\), with zero residual.
Polar check: writing the same family in polar form, \(r^2\sin2\theta=k(r^2\cos2\theta-b^2)\), implicitly differentiating, eliminating \(k\), and comparing against the polar orthogonal ODE \(r'=r(r^2-b^2\cos2\theta)/(b^2\sin2\theta)\) derived in Step 3 gives a residual that Sympy simplifies to \(0\) as well. Both routes — Cartesian and polar — confirm the same family, independently.
Result
Final answer:
The orthogonal trajectories, verified two independent ways (Cartesian and polar implicit differentiation against the target ODE), form the family of rectangular hyperbola-type conics \[ \boxed{\ x^2 - y^2 - 2kxy = b^2\ } \] for arbitrary constant \(k\) — not the source's guess \((x^2+y^2)^2-2b^2(x^2-y^2)=C\), which is shown in Step 2 to coincide with the original family rather than its orthogonal trajectories.