= Solution
The quadratic <first integrals> of the torque-free <Euler equations for a torque-free rigid body> are
$$
\boxed{2E=I_1\omega_1^2+I_2\omega_2^2+I_3\omega_3^2,\qquad
L^2=I_1^2\omega_1^2+I_2^2\omega_2^2+I_3^2\omega_3^2.}
$$
<Differentiation> and substitution of all three Euler equations make each <derivative> vanish: the coefficients respectively sum to $(I_2-I_3)+(I_3-I_1)+(I_1-I_2)=0$ and $I_1(I_2-I_3)+I_2(I_3-I_1)+I_3(I_1-I_2)=0$. At either endpoint these constants are $2E=I_2\Omega^2$, $L^2=I_2^2\Omega^2$, proving $L^2=2EI_2$.
Eliminating the other squared components gives, with $s=\omega_2/\Omega$,
$$
\omega_1^2=\frac{I_2(I_3-I_2)}{I_1(I_3-I_1)}\Omega^2(1-s^2),\qquad
\omega_3^2=\frac{I_2(I_2-I_1)}{I_3(I_3-I_1)}\Omega^2(1-s^2).
$$
Hence $|s|\le1$ and the second Euler equation gives $ds/d\tau=\pm(1-s^2)$, where $\tau=\Omega t\sqrt{(I_2-I_1)(I_3-I_2)/(I_1I_3)}$ increases with $t$. The specified orbit goes from $s=1$ to $s=-1$, so its sign must be negative. The sign in the original PDF's reduced equation is therefore inconsistent with its endpoint conditions. The corrected heteroclinic satisfies
$$
\boxed{\frac{ds}{d\tau}=s^2-1,\qquad s(\tau)=-\tanh(\tau-\tau_0).}
$$
Integrating by separation gives this family for $|s|<1$; the integration constant is a time translation. The printed positive-sign equation instead has the increasing solutions $s=\tanh(\tau-\tau_0)$ and cannot meet the stated endpoints. For completeness its other real solutions are $s=\pm1$ and the branches $s=\coth(\tau-\tau_0)$ on intervals avoiding their pole. These latter branches are excluded by the conserved-energy constraint $|s|\le1$. The scalar heteroclinic is unique up to translation. For the full angular-velocity vector there are two sign-related branches: simultaneously reversing $\omega_1$ and $\omega_3$ preserves all three Euler equations and both endpoints. Full-vector uniqueness therefore also requires choosing one of those branches.
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