Question

If $x\cos \left(\frac{y}{x}\right)(ydx+xdy)=y\sin \left(\frac{y}{x}\right)(xdy-ydx),y\left(1\right)=2\pi$, then the value of $4\frac{y\left(4\right)}{\pi }\cos \left(\frac{y\left(4\right)}{4}\right)$ is

• A. 4
• B. 2
• C. 1
• D. 0

Answer 1

Answer: (B)

2

The given equation can be written as
$\frac{dy}{dx}=\frac{y\left(y\sin \right(y/x)+x\cos (y/x\left)\right)}{x\left(y\sin \right(y/x)-x\cos (y/x\left)\right)}$
Substitute $y=ux$ in the above equation, we have
$u+x\frac{du}{dx}=\frac{u(u\sin u+\cos u)}{u\sin v-\cos u}$
$x\frac{du}{dx}=\frac{u(u\sin u+\cos u)}{u\sin u-\cos u}-u=\frac{2u\cos u}{u\sin u-\cos u}$
$\Rightarrow \frac{u\sin v-\cos u}{v\cos u}dv=2\frac{dx}{x}\Rightarrow (\tan u-\frac{1}{u})dv=2\frac{dx}{x}$
$\Rightarrow \log \sec u-\log u=\log x^2+constant$
$\Rightarrow \frac{\sec u}{u{x}^2}=constant\Rightarrow \frac{1}{xy}\sec \frac{y}{x}=constant$
$\Rightarrow xy\cos \left(y/x\right)=constantsoy\left(1\right)=2\pi \Rightarrow constant$
$=2\pi$. Hence $xy\cos \left(y/x\right)=2\pi$. Substituting $x=4$, we have $4\frac{y\left(4\right)}{\pi }\cos \left(y\right(4\left)/4\right)=2$.