= Solution
Take axes fixed in a <rigid body>, chosen along its <principal axes>, with moments of inertia $I_1,I_2,I_3$. The <angular momentum> components are $H_i=I_i\omega_i$. Newton's rotational equation is $\boldsymbol\tau=(d\mathbf H/dt)_{\rm inertial}$, while the rotating-frame transport rule gives
$$
\left(\frac{d\mathbf H}{dt}\right)_{\rm inertial}
=\left(\frac{d\mathbf H}{dt}\right)_{\rm body}+\boldsymbol\omega\times\mathbf H.
$$
Thus <Euler's equations (rigid body dynamics)> are
$$
\boxed{\begin{aligned}
I_1\dot\omega_1+(I_3-I_2)\omega_2\omega_3&=\tau_1,\\
I_2\dot\omega_2+(I_1-I_3)\omega_3\omega_1&=\tau_2,\\
I_3\dot\omega_3+(I_2-I_1)\omega_1\omega_2&=\tau_3.
\end{aligned}}
$$
The nonlinear cross terms arise because the <principal axes> rotate. Even with no external <torque>, the body-frame components of <angular momentum> can vary while the <angular momentum> <vector> stays fixed in inertial space.
With zero <torque>, the equations conserve
$$
E=\tfrac12\sum_i I_i\omega_i^2,\qquad
H^2=\sum_iI_i^2\omega_i^2.
$$
Uniform <rotation> about any <principal axis> is a torque-free solution. If $I_1<I_2<I_3$, <linearization> about the intermediate-axis solution $\omega_2=\Omega$ gives
$$
\delta\ddot\omega_1=\Omega^2\frac{(I_2-I_1)(I_3-I_2)}{I_1I_3}\delta\omega_1.
$$
The positive coefficient gives exponential growth: intermediate-axis <rotation> is unstable, whereas <rotation> about the largest or smallest moment is linearly stable. For a symmetric body, <Euler's equations (rigid body dynamics)> also describe the <precession> of the transverse <angular velocity> around its <symmetry> axis. <Torques> from gravity or an applied couple enter through the right-hand sides rather than being absorbed into the inertial cross terms.
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