The differential equation for the pendulum is $$\ddot{\theta}=-\frac{g}{L}\sin\theta$$ But physics professors (on youtube at least) turn this equation into $\ddot{\theta}=-\frac{g}{L}\theta$ and $\theta$ is assumed to be small. So what if the angle could be as big as you want? What is the real solution to $\ddot{\theta}=-\frac{g}{L}\sin\theta$? Is it unsolvable (and if it is, please show why. I don't understand Differential Algebra yet but I just want to familiarize myself with it)? I will accept any solution, non-elementary or not, but I don't want an analytic solution.
Edit: @mr_e_man said that a change of variables could transform the differential equation into $$\frac{d^2\theta}{d\tau^2}-\sin\theta=0$$
With $\theta=\theta(x)$ $$\theta''=-\frac{g}{L}\sin(\theta)$$ Switch variables $$\frac {x''}{[x']^3}=\frac{g}{L}\sin(\theta)$$ Reduction of order $p=x'$ gives $$\frac {p'}{p^3}=\frac{g}{L}\sin(\theta)\quad \implies \quad \frac 1{p^2}+c_1=\frac{g}{L}\cos(\theta)\quad \implies \quad p=x'=\pm \sqrt { \frac{L}{c_1+g \cos (\theta )}}$$ One more integration gives $$x+c_2=\pm \sqrt L\int \frac{d \theta}{\sqrt{c_1+g \cos (\theta )}}$$ Using the tangent half-angle substitution $$\int \frac{d \theta}{\sqrt{c_1+g \cos (\theta )}}=\frac{2}{\sqrt{c_1+g}}\,F\left(\frac{\theta }{2}|\frac{2 g}{c_1+g}\right)$$ where appears the elliptic integral of the first kind.
The problem is now to inverse the result to obtain $\theta(x)$.
But, if $\theta$ is small $$F\left(\frac{\theta }{2}|\frac{2 g}{c_1+g}\right)=\frac{\theta }{2}\left(1+\frac{g }{12 (c_1+g)}\theta ^2+\frac{g(7 g-2 c_1)}{480 (c_1+g)^2} \theta ^4 +O\left(\theta ^6\right) \right)$$ and we could use power series reversion.
Let $$y=\pm\frac{(x+c_2) \sqrt{c_1+g}}{2 \sqrt{L}}$$ and an approximation will be $$\theta=2 y-\frac{2 g }{3 (c_1+g)}y^3+\frac{g (2 c_1+3 g)}{15 (c_1+g)^2}y^5+O\left(y^7\right)$$
All of the above would seriously simply after inreoducing the boudary conditions.