How to decouple and solve these coupled differential equations While solving Euler's equation of motion for an asymmetric top, I cam across the three following equations :
$$\dot{\omega_1}+\omega_2 \omega_3=0$$
$$\dot{\omega_2}-\omega_1 \omega_3=0$$
$$\dot{\omega_3}+\frac{\omega_2 \omega_1}{\mu^2}=0$$
The only information that I have been provided is that at time $t=0$, $\omega=N\mu \hat{i}+N\hat{k}$. So I know $\omega_1, \omega_2 $ and $\omega_3$ at time $0$.
How can I decouple and solve the above set of equations ?
I've already been given the solution as follows :
$$\omega_1 =\frac{N\mu}{\cosh(Nt)}$$
$$\omega_2 ={N\mu}\tanh(Nt)$$
$$\omega_3 =\frac{N}{\cosh(Nt)}$$
How do I solve the equations myself ? I've solved linear coupled differential equations using matrices, however this one seems to be nonlinear because of the $\omega_i\omega_j$ terms.
 A: $$\dot{\omega_1}+\omega_2 \omega_3=0$$
$${\omega_1}\dot{\omega_1}+{\omega_1}\omega_2 \omega_3=0 \tag{1} $$
$$\dot{\omega_2}-\omega_1 \omega_3=0$$
$${\omega_2}\dot{\omega_2}-{\omega_2}\omega_1 \omega_3=0 \tag {2}$$
Add both DE and integrate.
$${\omega_1}^2+{\omega_2}^2=C_1$$
Third DE gives:
$$\dot{\omega_3}+\frac{\omega_2 \omega_1}{\mu^2}=0$$
$${\omega_3}\dot{\omega_3}+\frac{{\omega_3}\omega_2 \omega_1}{\mu^2}=0$$
$${\omega_3}\dot{\omega_3}-\frac{{\omega_1}\dot{\omega_1}}{\mu^2}=0$$
$${\omega_3}^2-\frac{{\omega_1}^2}{\mu^2}=C_2$$
A: You can combine the first two equations into one for a complex variable $z=ω_1+iω_2$, $z(0)=μN$
$$
\dot z=iω_3z\implies z(t)=z(0)\exp(i\int_0^tω_3(s)\,ds)=z(0)\bigl(\cos(u(t))+i\sin(u(t))\bigr)
$$
Then use, as per comment, $ ω_1\dot ω_1 = μ^2ω_3\dotω_3$, so that with initial conditions $ω_1^2=μ^2ω_3^2$, even more $ω_1=μω_3$. Then use $\dot u=ω_3$ and solve the remaining first-order equation
$$
\dot u(t)=N\cos(u(t)),~~u(0)=1.
$$

Of course knowing the initial conditions allows to introduce alternative parametrizations of the circle that are valid around the initial point, such as using
$$
(1-v^2)^2+(2v)^2=(1+v^2)^2
$$
to set
$$
ω_1=μN\frac{1-v^2}{1+v^2},~~~ω_2=μN\frac{2v}{1+v^2},~~~ω_3=N\frac{1-v^2}{1+v^2}
$$
so that
$$
-\frac{4v\dot v}{(1+v^2)^2}=-N^2\frac{(1-v^2)(2v)}{(1+v^2)^2}
$$
