Approximating the solution to \begin{align*} \frac{d^2x}{dt^2} = \frac{-981x}{50\sqrt{1+4x^2}}, \; x(0)=-1, \; x'(0)=0 \end{align*}

Question

The Problem:

In an exploration of a ball rolling inside a parabola, I discovered and needed to solve or approximate the following differential equation: \begin{align*} \end{align*} \begin{align*} \frac{d^2x}{dt^2} = \frac{-981x}{50\sqrt{1+4x^2}}, \; x(0)=-1, \; x'(0)=0 \end{align*} \begin{align*} \end{align*} where $x$ is a function of $t$. \begin{align*} \end{align*}

Attempts:

Trying to solve this analytically lead me to dead ends, so I figured the best way to solve this DE is numerically. Alternatively, I tried using a Taylor series centred about $x = 0$. I managed to find the first 15 non-zero terms. \begin{align*} \end{align*} \begin{align*} \end{align*} However, the series appears to converge quite slowly and this took a lot of tedious, repetitive calculation. So, trying to obtain an accurate approximation of the solution this way feels futile. Are there any other ways of approximating this DE more efficiently and accurately, or maybe is it solvable analytically? Thanks.

Answer 1

You can turn it into two first-order DEs, the first is doable, and the second you might feed into your graphics calculator.
Let $v=\frac{dx}{dt}$, then $$\frac{d^2x}{dt^2}=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}$$ You integrate both sides to get $v=f(x)$. Then, recall $v=dx/dt$, so $$t=\int\frac{dx}{f(x)}$$