Find the isogonal trajectories, at constant angle α , of the…

Find the isogonal trajectories, at constant angle $\alpha$, of the family of lines through the origin $y=ax$.

Step 1: Re-derive the isogonal ODE

With $m_1=y/x$ (the family's slope) and $m_2=y'$ (the trajectory's slope), \texttt{sympy.solve}$(\tan\alpha=(m_2-m_1)/(1+m_1m_2),\,m_2)$ gives, with $k=\tan\alpha$, \[y'=\frac{kx+y}{x-ky},\] matching the source's equation (3) exactly.

Step 2: Solve the homogeneous equation independently

\texttt{sympy.dsolve} applied directly to $y'=(kx+y)/(x-ky)$ (with no reference to the source's hand steps) returns \[\ln x = C_1-\ln\sqrt{1+\left(\tfrac{y}{x}\right)^2}+\frac{\arctan(y/x)}{k},\] which rearranges to $\arctan(y/x)=k\ln\sqrt{x^2+y^2}-kC_1$, i.e. $2\arctan(y/x)=k\ln(x^2+y^2)+C$ — matching the source's boxed result. As an independent second check, \texttt{sympy.integrate} on the source's separated integrand $(1-kv)/(k(1+v^2))$ gives exactly $\arctan(v)/k-\tfrac12\ln(1+v^2)$, confirming the source's antiderivative was not simply asserted correctly by luck.

Step 3: Verify the special cases directly

$\alpha=0$: setting $k=0$ in $y'=(kx+y)/(x-ky)$ gives $y'=y/x$ identically — the original family's own ODE, confirming the trajectories degenerate back to the lines $y=ax$. $\alpha=90°$: \texttt{sympy.limit} of $(kx+y)/(x-ky)$ as $k\to\infty$ gives exactly $-x/y$, which is precisely the ODE of the orthogonal circle family $x^2+y^2=C$ (implicit differentiation of $x^2+y^2=C$ gives $y'=-x/y$) — confirming the source's claimed orthogonal case.

Step 4: Polar form

With $\theta=\arctan(y/x)$ and $r^2=x^2+y^2$, the boxed result becomes $2\theta=k\ln r^2+C=2k\ln r+C$, i.e. $\theta=k\ln r+C'$, equivalently $r=Ae^{\theta/k}$ — algebraically confirmed and matching the source's logarithmic-spiral form.