Question

The solution of the differential equation $y\sin \left(\frac{x}{y}\right)dx=(x\sin \left(\frac{x}{y}\right)-y)dy$ satisfying $y\left(\frac{\pi }{4}\right)=1$ is

• A. $\cos \frac{x}{y}=-{\log }_{e}y+\frac{1}{\sqrt{2}}$
• B. $\sin \frac{x}{y}={\log }_{e}y+\frac{1}{\sqrt{2}}$
• C. $\sin \frac{x}{y}={\log }_{e}x-\frac{1}{\sqrt{2}}$
• D. $\cos \frac{x}{y}=-{\log }_{e}x-\frac{1}{\sqrt{2}}$

Answer 1

The correct option is A $\cos \frac{x}{y}=-{\log }_{e}y+\frac{1}{\sqrt{2}}$

Given differential equation is $\frac{y\sin \frac{x}{y}}{x\sin \frac{x}{y}-y}=\frac{dy}{dx}$

Now replace $x=vy$ such that,$\frac{dx}{dy}=v+y\frac{dv}{dy}$

$⟹v+y\frac{dv}{dy}=\frac{v\sin v-1}{\sin v}$

$⟹y\frac{dv}{dy}=\frac{-1}{\sin v}$

Separating the variables and integrating gives,

$\int \sin vdv=\int \frac{1}{y}dy$,

$⟹-\cos v+c=\ln y$

Given $y\left(\frac{\pi }{4}\right)=1$

$⟹c=\frac{1}{\sqrt{2}}$

$\therefore$ the particular solution is ,

$\cos \frac{x}{y}=-\ln y+\frac{1}{\sqrt{2}}$