Question

If $x\sin\left( \frac{y}{x} \right)dy = \left\lbrack y\sin\left( \frac{y}{x} \right) - x \right\rbrack dx$ and $y(1) = \frac{\pi}{2}$, then the value of $\cos\left( \frac{y}{x} \right)$ is equal to

• A. x
• B. \frac{1}{x}
• C. log x
• D. e^{x}

Answer 1

Answer: (C)

log x

Answer 2

So$x\sin\left( \frac{y}{x} \right)dy = \left\lbrack y\sin\left( \frac{y}{x} \right) - x \right\rbrack dx$

$\Rightarrow$ $\frac{dy}{dx} = \frac{y\sin\left( \frac{y}{x} \right) - x}{x\sin\left( \frac{y}{x} \right)} = \frac{\frac{y}{x}\sin\left( \frac{y}{x} \right) - 1}{\sin\left( \frac{y}{x} \right)}$

Let $\frac{y}{x} = u$ and $\frac{dy}{dx} = x\frac{du}{dx} + u$

$\therefore x\frac{du}{dx} + u = \frac{u\sin u - 1}{\sin u}$

$\Rightarrow x\frac{du}{dx} = \frac{u\sin u - 1u\sin u}{\sin u} \Rightarrow \sin udu = \frac{1}{x}dx$

On integrating both sides, we get

$\cos u = \log x + C \Rightarrow \cos\left( \frac{y}{x} \right) = \log x + C$

$\therefore y(1) = \frac{\pi}{2}$ $\therefore$ $\cos^{p}2 = \log 1 + C \Rightarrow C = 0$

Thus, $\cos\left( \frac{y}{x} \right) = \log x$