I have the following second order equation, where the first derivative is missing, and I am asked to find its solution:
$$x(t)''=-\omega^2x(t)+a(x(t))^2$$ I don't know how to solve it. I have tried with to substitute $f(x) = -\omega^2x(t)+a(x(t))^2 = v'$ and then set $x(t)' = v$ but it doesn't seem to help me...
Is my approach for this ODE wrong? Are there any other solution methods I could try out?
Would be thankful for any advice!
\begin{align} & x''=-\omega^2x+ax^2 \\ \cdot x'\implies & x'x''=-\omega^2xx' + ax^2x' \\ \int dt\implies & \frac{1}{2}x'^2=-\frac{\omega^2}{2}x^2 + \frac{a}{3}x^3 + C\\ \cdot2 \implies & x'^2=-\omega^2x^2 + \frac{2a}{3}x^3 + C \\ \text{solve for } x' \implies & \frac{x'}{\sqrt{\frac{2a}{3}x^3 -\omega^2x^2 + C}} = \pm 1 \end{align} If you can integrate it, of course, you get a solution