Question

$x=\cos \theta ,y=\sin 5\theta \Rightarrow \left(1-x^2\right)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}$ is equal to

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

Answer 1

Answer: (D)

Given, $x=\cos \theta ,y=\sin 5\theta$
$\frac{dx}{d\theta }=-\sin \theta ,\frac{dy}{d\theta }=5\cos 5\theta$
$\therefore \frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }=-\frac{5\cos 5\theta }{\sin \theta }$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{d\theta }\left(\frac{dy}{dx}\right)\cdot \frac{d\theta }{dx}$
$=\frac{d}{d\theta }\left(\frac{-5\cos 5\theta }{\sin \theta }\right)\cdot \frac{1}{-\sin \theta }$
$=\left(\frac{\sin \theta \sin 5\theta \cdot 25+5\cos 5\theta \cos \theta }{{\sin }^2\theta }\right)\cdot \frac{1}{-\sin \theta }$
$=-\frac{25\sin 5\theta }{{\sin }^2\theta }-\frac{5\cos 5\theta \cos \theta }{{\sin }^3\theta }$
Now,
$\left(1-x^2\right)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}$
$=\left(1-{\cos }^2\theta \right)\left(\frac{-25\sin 5\theta }{{\sin }^2\theta }-\frac{5\cos 5\theta \cos \theta }{{\sin }^3\theta }\right)$
$-\cos \theta \left(\frac{-5\cos 5\theta }{\sin \theta }\right)$
$={\sin }^2\theta \left(\frac{-25\sin 5\theta }{{\sin }^2\theta }-\frac{5\cos \theta \cos 5\theta }{{\sin }^3\theta }\right)$
$+\frac{5\cos \theta \cos 5\theta }{\sin \theta }$
$=-25\sin 5\theta -\frac{5\cos \theta \cos 5\theta }{\sin \theta }+\frac{5\cos \theta \cos 5\theta }{\sin \theta }$
$=-25\sin 5\theta$
$=-25y$