System differential equation

Question

Well I stumbled upon the next differential equation system: $$ \begin{cases} \dot{\omega}_1 + a \omega_2 \omega_3 = t_1 \\ \dot{\omega}_2 + b \omega_1 \omega_3 = t_2 \\ \dot{\omega}_3 + c \omega_1 \omega_2 = t_3\end{cases}$$ with $a,b,c \in \mathbb{R}$. If necessary you can assume $t_2 = t_3 = 0$. I don't know if I ever solved this or not, but I do not remember how to do it.

Answer 1

It's a no linear system, needing its own trick to solve, if any. I'll try first the case with $t_2=t_3=0$. It's possible to separate variables.

$$\begin{cases} \dot{\omega}_1 + a \omega_2 \omega_3 = t_1 \\ \dot{\omega}_2 + b \omega_1 \omega_3 =0 \\ \dot{\omega}_3 + c \omega_1 \omega_2 =0 \end{cases}$$

$$\frac{\mathrm d\omega_3/\mathrm dt}{\mathrm d\omega_2/\mathrm dt}= \frac{c\omega_2}{b\omega_3}\implies\frac{\mathrm d\omega_3}{\mathrm d\omega_2}=\frac{c\omega_2}{b\omega_3}\implies\omega_3\mathrm d\omega_3=(c/b)\omega_2\mathrm d\omega_2$$

$$\omega_3^2=(c/b)\omega_2^2+c_1$$

$$\frac{\mathrm d\omega_2/\mathrm dt}{\mathrm d\omega_1/\mathrm dt}= \frac{-b\omega_1\omega_3}{t_1-a\omega_2\omega_3}\implies\frac{\mathrm d\omega_2}{\mathrm d\omega_1}=\frac{-b\omega_1\sqrt{(c/b)\omega_2^2+c_1}}{t_1-a\omega_2\sqrt{(c/b)\omega_2^2+c_1}}\implies$$

$$\implies\frac{t_1-a\omega_2\sqrt{(c/b)\omega_2^2+c_1}}{\sqrt{(c/b)\omega_2^2+c_1}}\mathrm d\omega_2=-b\omega_1\mathrm d\omega_1$$

$$(a/b)\omega_2^2-\frac{2t_1\ln\left|\sqrt{\frac{c}{bc_1}}\omega_2+\sqrt{1+\frac{c}{bc_1}\omega_2^2}\right|}{\sqrt{c/b}}+c_2=\omega_1^2$$

And now, things get increasingly "messy" when you try to recover the t dependence:

$$\dot\omega_2=-b\left((a/b)\omega_2^2-\frac{2t_1\ln\left|\sqrt{\frac{c}{bc_1}}\omega_2+\sqrt{1+\frac{c}{bc_1}\omega_2^2}\right|}{\sqrt{c/b}}+c_2\right)^{1/2}\left((c/b)\omega_2^2+c_1\right)^{1/2}$$

At least, it's separable too. I am not able to discuss the assimptotic behavior neither the orbits, if any, the critical points...

Anyway, I hope this helps.