Question

If $x=\cos \theta ,y=\sin 5\theta$, then
$(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}=$

• A. $-5y$
• B. $5y$
• C. $25y$
• D. $-25y$

Answer 1

The correct option is D $-25y$

We have
$x=\cos \theta ,y=\sin \theta$
$\therefore \frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=-\frac{5\cos 5\theta }{\sin \theta }$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=-5\frac{d}{d\theta }\left(\frac{5\cos 5\theta }{\sin \theta }\right).\frac{d\theta }{dx}$
$=\frac{-25\sin \theta \sin 5\theta -5\cos \theta \cos 5\theta }{{\sin }^3\theta }$
$\therefore (1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}$
$=(1-{\cos }^2\theta )\left[\frac{-25\sin \theta \sin 5\theta -5\cos \theta \cos 5\theta }{{\sin }^3\theta }\right]-\cos \theta \left[\frac{-5\cos 5\theta }{\sin \theta }\right]$
$=-25\sin 5\theta =-25y$