The coursework is available here.
I've been asked what shape a cone with a square base is. It is the same as a pyramid. I have now amended the coursework to make this (I hope) clear.
The period of the solutions to a linear differential equation is easy to find: just solve it (sines and cosines) and see what the period is.
If the equation is not linear (like the pendulum \(\ddot x = - \sin x\) ), one needs to approximate it by a linear equation.
Method: Do a Taylor series expansion of all the terms, and ignore all nonlinear terms.
Artificial example: Consider \[\ddot x + x\dot x +x\cos x - \dot x^2 =0.\] (Not solvable!) To "linearize" expand everything as Taylor series (the only thing that needs expanding here is \(\cos x\)), and then ignore all nonlinear terms. We get \[\ddot x + x\dot x + x(1-\frac12 x^2+\frac1{4!}x^4-\dots) - \dot x^2 =0.\] The only linear terms are the first, and the \(x.1\) from the cos series, to get \[\ddot x + x=0.\] The general solution to this equation is \[x(t) = A\cos t + B\sin t,\] and this has period \(2\pi\).
Another method
(Step 1) Let \(\varepsilon\) be a constant (small) and replace \(x\) by \(\varepsilon u\). So put \(x=\varepsilon u\) and \(\dot x = \varepsilon\dot u\) etc, to get
\[\varepsilon\ddot u +\varepsilon^2 u\dot u +\varepsilon u\cos(\varepsilon u) - \varepsilon^2\dot u^2=0.\]
(Step 2) Now divide by \(\varepsilon\) (notice it's a factor here- but you might have to take Taylor series to get it as a factor) to get
\[\ddot u +\varepsilon u\dot u + u\cos(\varepsilon u) - \varepsilon\dot u^2=0.\]
(Step 3) Put \(\varepsilon=0\). We then get (with \(\cos(0)=1\))
\[\ddot u + u=0.\]
(Step 4) Solve this as before.
Note: the two methods are essentially the same, the advantage of the second is that it makes systematic what one means by the linear terms