Question

If $y=\sin \left(\sin x\right)$, prove that $\frac{d^{2}y}{d{x}^2}+\tan x\frac{dy}{dx}+y{\cos }^{2}x=0$

Answer 1

$y=\sin \left(\sin x\right)$ ... $\left(i\right)$

$\frac{dy}{dx}=\cos \left(\sin x\right).\cos x$ ... $\left(ii\right)$

$\frac{d^{2}y}{d{x}^2}=-\sin \left(\sin x\right)\cos x.\cos x-\cos \left(\sin x\right)\sin x$

= $-\left[\sin \right(\sin x).{\cos }^{2}x+\cos (\sin x\left)\sin x\right]$

= $-[y.{\cos }^{2}x+\frac{dy}{dx}.\frac{1}{\cos x}.\sin .x]$ [using $\left(i\right)$ and $\left(ii\right)$]

$\frac{d^{2}y}{d{x}^2}+y{\cos }^{2}x+\tan x\frac{dy}{dx}=0$

or, $\frac{d^{2}y}{d{x}^2}+\tan x\frac{dy}{dx}+y{\cos }^{2}x=0$.