Show that $y^2 \frac{d^2y}{dx^2}+1=0$ for a set of parametric equations.

Question

A function of $x$ is defined parametrically by $x=t-\sin(t)$ and $y=1-\cos(t).$

How do I answer this question, then?

Show that $$y^2 \dfrac{d^2y}{dx^2}+1=0.$$

Answer 1

$dx = dt - \cos t dt, dy = \sin t dt$ so $\dfrac{dy}{dx} = \dfrac{\sin t}{ 1 - \cos t}$.

now use the fact
$$\dfrac{d^2y}{dx^2} = {\dfrac{d}{dt}\left(\frac{dy}{dx}\right) \over \frac{dx}{dt}}$$