So, I have this pair of differential equations:
$$\frac{\mathrm{d}^2x}{\mathrm{d}t^2} + \frac{2g(H/L)x}{4(H/L)^2x^2+1} = 0$$ $$\frac{\mathrm{d}^2y}{\mathrm{d}t^2} + \frac{4g(H/L)y}{4(H/L)y+1} = 0$$ I am looking for two functions, $x(t),\ y(t)$ and the initial conditions are:
$$x(0) = L,\ y(0) = H,\ x'(0) = 0,\ y'(0)=0$$
I got these by applying Newton's second law to a specific problem.
I don't have the necessary background to solve this equation but am really curious as to what the solution would look like.
If it couldn't be solved analytically, I would love to see a numerical solution.
$$\frac{\mathrm{d}^2x}{\mathrm{d}t^2} + \frac{2g(H/L)x}{4(H/L)^2x^2+1} = 0$$ $2\frac{\mathrm{d}^2x}{\mathrm{d}t^2}\frac{\mathrm{d}x}{\mathrm{d}t} + \frac{4g(H/L)x}{4(H/L)^2x^2+1}\frac{\mathrm{d}x}{\mathrm{d}t} = 0$
$\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \frac{gL}{2H}\ln\big|4(H/L)^2x^2+1\big| = c_1$
$\frac{\mathrm{d}x}{\mathrm{d}t} = \pm\sqrt{-\frac{gL}{2H}\ln\big|4(H/L)^2x^2+1\big| + c_1}$ $$t(x)=\pm\int \frac{\mathrm{d}x}{\sqrt{-\frac{gL}{2H}\ln\big|4(H/L)^2x^2+1\big| + c_1}} +c_2$$ As far as I know, there is no closed form for this integral in terms of a finite number of standard functions. A-fortiori, there is no closed form for the inverse function $x(t)$.
$$\frac{\mathrm{d}^2y}{\mathrm{d}t^2} + \frac{4g(H/L)y}{4(H/L)y+1} = 0$$ $2\frac{\mathrm{d}^2y}{\mathrm{d}t^2}\frac{\mathrm{d}y}{\mathrm{d}t} + \frac{8g(H/L)y}{4(H/L)y+1}\frac{\mathrm{d}y}{\mathrm{d}t} = 0$
$\left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2 + \frac{g}{4H}\left(4Hy -L\ln\big|4Hy+L\big| \right)= c_1$
$\frac{\mathrm{d}y}{\mathrm{d}t}= \pm\sqrt{-\frac{g}{4H}\left(4Hy -L\ln\big|4Hy+L\big| \right)+ c_1}$ $$t(y)=\pm\int \frac{\mathrm{d}y}{\sqrt{-\frac{g}{4H}\left(4Hy -L\ln\big|4Hy+L\big| \right)+ c_1}}+c_2$$ Same comment than for the above integral.