This (combined with the last question) is somewhat of a classic problem on examining the stability of free rotations. Our goal is to solve Euler's equations in the special case where $A < B$, $C = A + B$, and the body begins at
\begin{gather}
\omega_2(0) = 0 \label{bc1} \\
(A + B) \omega_3^2(0) = (B-A) \omega_1^2(0). \label{bc2}
\end{gather}

What do we expect? Well $A < B < C$. Imagine flipping a book about one of its principal axes:

*Figure from Applied Dynamics with Applications by Moon, p.202*

The depth of the book (the thinnest side) corresponds to the smallest moment of inertia, $A$, so $\mb{e}_1$ corresponds to what you see as the $x$-axis (out of the page). The largest moment of inertia, $C$, corresponds to $\mb{e}_3$, which is the height of the book. The middle moment of inertia corresponds to the width of the book.

What you demonstrated form the previous question was that the stable points are $\omega = (\Omega, 0, 0)$, and $\omega = (0, 0, \Omega)$, so you can flip the book around the $x$-axis or the $z$-axis. But spinning the book around the $y$-axis is unstable.

The goal of this question is to see what happens if you begin spinning the book with an angular velocity vector that points in the $(x,z)$ plane. Euler's equations will be supplemented by conservation of kinetic energy and angular momentum:
\begin{gather}
2T = A\omega_1^2 + B \omega_2^2 + C\omega_3^2 = \text{constant} \label{const1} \\
L^2 = A^2\omega_1^2 + B^2 \omega_2^2 + C^2 \omega_3^2 = \text{constant}. \label{const2}
\end{gather}

Basically, \eqref{const1} and \eqref{const2} should allow us to solve for $\omega_2$ and $\omega_3$ in terms of $\omega_1$, and then we can use the first Euler equation to solve for $\omega_1$. Before we do this, we need to first figure out what $T$ and $L$ are. We now use the initial conditions \eqref{bc1} and \eqref{bc2} to obtain
\begin{gather}
2T = A\omega_1^2(0) + 0 + (B-A)\omega_1^2(0) = B\omega_1^2(0) \\
L^2 = A^2\omega_1^2(0) + 0 + (A+B)(B-A)\omega_1^2(0).
\end{gather}

Eliminating $\omega_1^2(0)$ within these two equations gives
\begin{gather}
L^2 = 2BT \\
2CT - L^2 = AB(\omega_1^2 + \omega_2^2).
\end{gather}

The first line relates $T$ with $L$. We are left with one constant still being unknown (say $T$) because $\omega_1(0)$ is not specified by the problem. The second line allows us to solve for $\omega_2$:
\begin{equation} \label{om2}
\omega_2^2 = \frac{2T}{B} - \omega_1^2.
\end{equation}

In a similar way, we can obtain $\omega_3$ from \eqref{const1} and \eqref{const2} as
\begin{equation}
\omega_3^2 = \frac{B-A}{A+B} \omega_1^2.
\end{equation}

Substitution of both these results into the first Euler Equation gives
\[
A^2 \dot{\omega_1}^2 = A^2 \omega_2^2 \omega_3^2 = A^2 D \omega_1^2 \left(\frac{2T}{B}-\omega_1^2\right)\
\]

where
\[
D = \left(\frac{B-A}{A+B}\right).
\]

*The question claims that this quantity is squared. Is this an error?*. To solve this, we set $\omega_1 = c_1 \text{sech}^2(c_2 t)$ into the equation. This gives (after simplifying)
\[
c_2^2 - \frac{2DT}{B} + \left[ c_1^2 D - c_2^2 \right] \text{sech}(c_2 t)^2 = 0,
\]

so
\[
c_1^2 = \frac{2T}{B}, \qquad c_2^2 = \frac{2T}{B} D = \frac{2T}{B} \left(\frac{B-A}{A+B}\right).
\]

Now the question is what happens to $\omega_2$ as $t\to \infty$. According to \eqref{om2}, we should have
\[
\omega_2^2 = \frac{2T}{B} \left[ 1 - \text{sech}(c_2^2 t)\right] \sim \frac{2T}{B}.
\]

where we have chosen the constants for $\omega_1$ so that $\omega_2(0) = 0$. In other words, if the book is flipped with angular velocity in the $(x,y)$ plane which is pointing *just right*(?), it should tend to the configuration with $\omega_2$ fixed as $t \to \infty$.

There is a bit of geometric analysis you can do known as the Poinsot Construction. Basically, the idea is that because kinetic energy is constant, then \eqref{const1} tells us that the angular velocity vector $\omega$ must be confined to the surface of an ellipsoid. The paths of the possible trajectories, $\omega(t)$ can then be plotted on the ellipsoid using the equation
\[
\frac{\omega_1^2}{2T/A} + \frac{\omega_2^2}{2T/B} + \frac{\omega_2^2}{2T/C} = 1.
\]

These are shown below.

*Figure from Applied Dynamics with Applications by Moon, p.204*

Compare the numerical simulations at the top of this page with what you see in this video: Rotating solid bodies in microgravity.