Find the orthogonal trajectories of the family of curves…

Find the orthogonal trajectories of the family of curves $r^2(\pi-\varphi)=a^2\varphi$.

Step 1: Re-derive the family ODE independently

\texttt{sympy} differentiates $r^2(\pi-\varphi)$ in $\varphi$ to get $2r r'(\pi-\varphi)-r^2$, and this must equal the derivative of the right side, $a^2$ (a constant along a single curve). Substituting $a^2=r^2(\pi-\varphi)/\varphi$ from the original equation and solving for $r'$, \texttt{sympy} returns \[\frac{dr}{d\varphi}=\frac{r\pi}{2\varphi(\pi-\varphi)},\] 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 \[\frac{dr}{d\varphi}=-\frac{2r\varphi(\pi-\varphi)}{\pi},\] matching the source.

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

Calling \texttt{sympy.dsolve} directly on this separable ODE (not re-checking the source's by-hand integration) returns \[r(\varphi)=C_1\exp\!\left(\varphi^2\left(\frac{2\varphi}{3\pi}-1\right)\right)=C_1\exp\!\left(\frac{2\varphi^3}{3\pi}-\varphi^2\right),\] reproducing the source's boxed answer exactly, via an independent solution path.

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

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

Step 5: Numeric cross-check

At $\varphi=1,\ r=1$: the original relation fixes $a^2=\pi-1$, giving family slope $dr/d\varphi=\dfrac{\pi}{2(\pi-1)}\approx0.7335$. The orthogonal-family member through the same point needs $C\approx2.1985$, giving orthogonal slope $\approx-1.3634$ there. Product $\approx-1.0000$, matching $-r^2=-1$ exactly.