Prove that it is a solution to the differential equation

Question

Being, $$x=a\big(\theta-\sin(\theta)\big),y=a\big(1-\cos(\theta)\big)$$ prove that it is a solution of the differential equation $$1+\big(y'\big)^{2}+2yy''=0$$

My solution: I already got that, $$\frac{dy}{dx}=\frac{\sin(\theta)}{1-\cos(\theta)},\hspace{2mm}\frac{dy^{2}}{dx^{2}}=-\frac{1}{1-\cos(\theta)},\hspace{5mm}y=a\big(1-\cos(\theta)\big)$$using the derivation of the parametric equations, but at the time of replacement it is supposed that it should be annulled ... but I have left $$\big(1-\cos(\theta)\big)\big[1-a+a\cos(\theta)\big]=0$$

Answer 1

$$\frac{dy}{dx}=\frac{\sin(\theta)}{1-\cos(\theta)}$$is correct but $$\hspace{2mm}\frac{d^2y}{dx^{2}}=-\frac{1}{1-\cos(\theta)}$$is not correct.

Notice that $$ \frac{d^2y}{dx^{2}} = \frac {d}{dx} (\frac {dy}{dx}) =\frac {d}{d\theta} (\frac {dy}{dx})/ (\frac {dx}{d\theta})$$

You can take over from here.

Answer 2

Observe that $$y=\frac {dx}{d\theta}$$ And also we have that $$1+\big(y'\big)^{2}+2yy''=0$$ $$ \implies 1+2(y'y)'-y'^2=0$$ With derivative taken wrt $\theta$ $$ \implies 1+\frac 2y\frac {d^2y}{d \theta^2}-\left (\frac 1y \frac {dy}{d \theta} \right )^2=0$$ $$ \implies \left( \frac {dy}{d \theta}\right)^2-2y\frac {d^2y}{d \theta^2}=y^2$$ With $$\frac {dy}{d \theta}=a\sin \theta \quad;\frac {d^2y}{d \theta^2}=a \cos \theta$$

Answer 3

You can separate the ODE as $$ \frac{2y'y''}{1+y'^2}=-\frac{y'}{y}\implies 1+y'^2=\frac{C}{y} $$ which should make verifying the solution easier.