Question

${\cos }^{-1}\left(\frac{y}{b}\right)=2\log \left(\frac{x}{b}\right),x>0\Rightarrow x^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}=$

• A. 4y
• B. -4y
• C. 0
• D. -8y

Answer 1

Answer: (B)

-4y

$y=bcos\left(\log \frac{x^2}{b^2}\right)$ .... (1)

$\Rightarrow \frac{dy}{dx}=-bsin\left(\log \frac{x^2}{b^2}\right)\cdot (\frac{b^2}{x^2}\cdot \frac{2x}{b^2})$

$\Rightarrow \frac{dy}{dx}=\frac{-2b}{x}sin\left(\log \frac{x^2}{b^2}\right)$ ...(2)

$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{2b}{x^2}sin\left(\log \frac{x^2}{b^2}\right)-\frac{2b}{x}cos\left(\log \frac{x^2}{b^2}\right)\cdot (\frac{b^2}{x^2}\cdot \frac{2x}{b^2})$

$\Rightarrow \frac{x^2{d}^{2}y}{d{x}^2}=2bsin\left[\log \right(\frac{x^2}{b^2}\left)\right]-4bcos\left[\log \frac{x^2}{b^2}\right]$ ...(3)

from (1), (2) and (3)

$\therefore \frac{x^2{d}^{2}y}{d{x}^2}+\frac{xdy}{dx}=-4y$