Nonlinear Differential Equation - Part 2

the equation can be simplified to $\large{\dfrac{d^2x}{dt^2}} + kx^2 = 0 \to v\dfrac{dv}{dx} =-kx^2\to \int_0^v v\text{dv} =-\int_A^x kx \text{dx}\to v^2 = \dfrac{2}{3} k(A^3-x^3)$ since $x$ must decrease from some positive $A$ to 0, $v$ has to be the negative square root, i.e $\large v = - \sqrt{\dfrac{2}{3} kA^3} \sqrt{1-\dfrac{x^3}{A^3}}\to \int_A^0 \left(1-\dfrac{x^3}{A^3}\right)^{-1/2} \text{dx} = - \int_0^T\sqrt{\dfrac{2}{3} kA^3}\text{dt}\\ \dfrac{A}{3}\int_1^0 u^{-2/3} (1-u)^{-1/2} \text{du}=-\sqrt{\dfrac{2}{3} kA^3} T\to T = \boxed{\dfrac{\Beta\left(\dfrac{1}{3},\dfrac{1}{2}\right)}{\sqrt{6Ak}}}$

note that for the diff eq $\large \dfrac{d^2x}{dt^2} + kx^n = 0$ with the same initial conditions, the value would be $\large T = \boxed{\dfrac{\Beta\left(\dfrac{1}{n+1},\dfrac{1}{2}\right)}{\sqrt{2(n+1)A^{n-1}k}}}$ which can be shown by the same method used here.