Find the orthogonal trajectories of the family of curves…

Find the orthogonal trajectories of the family of curves $r^2=\ln\tan\varphi+c$.

Step 1: Re-derive the family ODE independently

\texttt{sympy} differentiates $r^2=\ln\tan\varphi+c$ in $\varphi$: $2r\,r'=\dfrac{\sec^2\varphi}{\tan\varphi}$, which \texttt{sympy.simplify} reduces to $\dfrac{2}{\sin2\varphi}$. Hence \[\frac{dr}{d\varphi}=\frac{1}{r\sin2\varphi},\] matching the source's boxed family ODE exactly (residual $0$).

Step 2: Form the orthogonal ODE

Applying $(dr/d\varphi)_{\text{orth}}=-r^2/(dr/d\varphi)_{\text{orig}}$: \texttt{sympy.simplify} gives \[\frac{dr}{d\varphi}=-r^3\sin2\varphi,\] matching the source.

Step 3: Solve independently with \texttt{sympy.dsolve}

Calling \texttt{sympy.dsolve} on $r'=-r^3\sin2\varphi$ directly returns $r(\varphi)=\pm\sqrt{-1/(C_1+\cos2\varphi)}$, i.e. \[r^2=\frac{-1}{C_1+\cos2\varphi}=\frac{1}{C-\cos2\varphi}\quad(C=-C_1),\] so $\dfrac{1}{r^2}=C-\cos2\varphi$ — reproducing the source's boxed answer exactly, via an independent solution path.

Step 4: Symbolic orthogonality identity (exact, not numeric)

Multiplying $\left(\frac{dr}{d\varphi}\right)_1=\dfrac{1}{r\sin2\varphi}$ by $\left(\frac{dr}{d\varphi}\right)_2=-r^3\sin2\varphi$, \texttt{sympy.simplify} collapses the product to exactly $-r^2$, residual $0$.

Step 5: Numeric cross-check

At $\varphi=1$, take $r_0=1.2$ on the original family: this fixes $c\approx0.9970$, giving family slope $dr/d\varphi=\dfrac{1}{1.2\sin2}\approx0.9165$. The orthogonal-family member through the same point needs $C\approx0.2783$, giving orthogonal slope $\approx-1.5713$ there. Product $\approx-1.4400$, matching $-r_0^2=-1.44$ exactly.