Solving Second-Order Nonlinear ODE Consider the ODE $\frac{d^2x}{dt^2} = a x + b x^2$, where $a$ and $b$ are some parameters. I wonder how I can solve this ODE?
Any comment/hint is greatly appreciated.
 A: Multiplying both sides by $\frac{dx}{dt}$ you get
$$\frac{1}{2} \frac{d}{dt}\left(\frac{dx}{dt}\right)^2 = \frac{d}{dt}\left(\frac{a}{2} x^2(t) +  \frac{b}{3} x^3(t)\right)$$ and by integration
$$\left(\frac{dx}{dt}(t)\right)^2 = a x^2(t) +  \frac{2b}{3} x^3(t)+ C$$ where $C$ is a real constant. By integration again you get
$$\int_{x_0}^{x(t)} \frac{du}{\sqrt{a u^2 +  \frac{2b}{3} u^3+ C}} = \pm(t-t_0)$$
A: There is another way to solve
$$x'' = a x + b x^2$$ Switch variables and write
$$-\frac{t''}{[t']^3}= a x + b x^2$$ Reduction of order $(p=t')$ gives
$$\frac {p'}{p^3}=-(ax+bx^2)\implies p=t'=\pm \frac{1}{\sqrt{ a x^2+\frac 23 b x^3+ c_1}}$$ Now, you will face elliptic integrals.
You can simplify the calculations writing
$$ a x^2+\frac 23 b x^3+ c_1=\frac{2 b}{3}\left(x^3+\frac{3 a x^2}{2 b}+\frac{3 c_1}{2 b} \right)=\frac{2 b}{3}(x-\alpha)(x-\beta)(x-\gamma)$$ where $(\alpha,\beta,\gamma)$ are the roots of the cubic equation. So
$$t'=\pm \sqrt{\frac{3}{2b}} \frac 1 {\sqrt{(x-\alpha)(x-\beta)(x-\gamma)}}$$
$$\int\frac {dx} {\sqrt{(x-\alpha)(x-\beta)(x-\gamma)}}=-\frac{2}{\sqrt{\beta -\alpha }}F\left(\sin ^{-1}\left(\frac{\sqrt{\beta -\alpha }}{\sqrt{x-\alpha
   }}\right)|\frac{\alpha -\gamma }{\alpha -\beta }\right)$$
So, you have the implicit equation
$$t+c_2=\pm \sqrt{\frac{6}{b(\beta-\alpha)}}F\left(\sin ^{-1}\left(\frac{\sqrt{\beta -\alpha }}{\sqrt{x-\alpha
   }}\right)|\frac{\alpha -\gamma }{\alpha -\beta }\right)$$
