Compute the inertial tensor and then solve the equation? If the $J_{\Omega}$ is the following matrix, which is solved by ja72 in How to compute the inertia tensor ${\bf{J}} _{\Omega}$ of a body of revolution:
$${\bf J} = \rho\, \begin{bmatrix}
\frac{\pi}{2}\int_a^b{f(x)}^4\,{\rm d}x&0&0\\
0&\frac{\pi}{4}\int_a^b\left({f(x)}^4+4 x^2 {f(x)}^2\right)\,{\rm d}x
&0\\0&0&\frac{\pi}{4}\int_a^b\left({f(x)}^4+4 x^2 {f(x)}^2\right)\,{\rm d}x
\end{bmatrix}$$
Now denote the element on the diagonal as $J_1,J_2,J_3$. It is easy to find that $J_2=J_3$.
My problem is to solve $J_{\Omega}\dot w=(J_{\Omega}w \times w)$.
My attempt is as follows:
$J_{\Omega}\dot w=(J_{\Omega}w \times w)$ can be written as 
$\left\{ \begin{array}{l}
J_1\dot w_1=(J_2-J_3)w_2w_3\\
J_2\dot w_2=(J_3-J_1)w_3w_1\\
J_3\dot w_3=(J_1-J_2)w_1w_2
\end{array} \right.$
Since $J_2=J_3$, then I got 
$$\dot w_1=0 \tag{1},$$ 
and 
$$w_3\dot w_3+w_2\dot w_2=0 \tag{2}$$.
By my Lagrange doesn't depend on t and Noether's theorem, I was also given 
$\left\{ \begin{array}{l}
J_{\Omega}w \cdot w=2E_0\\
|J_{\Omega}w|^2=A_0^2
\end{array} \right.$
Then what to do next to solve the equation?
 A: The equations of motion are 
$${\bf J} \dot{\vec{\omega}} = \left( {\bf J} \vec{\omega} \right) \times \vec{\omega}$$
 so just invert the diagonal ${\bf J}$ to get the components of $\dot{\vec{\omega}}$
$$ \dot{\vec{\omega}} ={\bf J}^{-1} \left( {\bf J} \vec{\omega} \right) \times \vec{\omega}$$
$$ \dot{\vec{\omega}} = \begin{pmatrix} 0 \\ J_2^{-1} (J_2-J_1) \omega_x \omega_z \\ J_2^{-1} (J_1-J_2) \omega_x \omega_y \end{pmatrix} $$
Which leads to $\dot{\vec{\omega}} = (0, K \omega_z, -K \omega_y)$ where $K=\omega_x - \frac{J_1}{J_2} \omega_x$. This has a solution
$$\vec{\omega} = \begin{pmatrix} \omega_x \\ \Omega \sin(K t+\varphi) \\ \Omega \cos(K t+\varphi) \end{pmatrix} $$
with $\Omega$ and $\varphi$ and $\omega_x$ to satisfy the initial conditions.
A: Integrating your equations (1) and (2) you get that 
$(\text i) \ \omega_1 = C_1$
a constant in time, while  another constant in time. This equality you can re-write as
$(\text {ii}) \ \omega_2^2 + \omega_3^2 = C_2^2$,
As to the vector $J_{\Omega}\omega$ you know that its components are $\{J_1\omega_1, J_2\omega_2, J_2\omega_3\}$, therefore 
$(\text {iii}) \ J_{\Omega}\omega = J_1\omega_1^2 + J_2(\omega_2^2 + \omega_3^2) = 2E_0$
$(\text {iv}) \ J_1^2 \omega_1^2 + J_2^2(\omega_2^2 + \omega_3^2) = A_0^2$.
Introducing in (iii) and (iv) the results (i) and (ii), and then manipulating with your equations for $\dot {\omega}_2$ and $\dot {\omega}_3$, I hope that from now on you'll manage.
