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. $\sin \frac{x}{y}=lo{g}_{e}y+\frac{1}{\sqrt{2}}$
• C. $\sin \frac{x}{y}=lo{g}_{e}x-\frac{1}{\sqrt{2}}$
• D. None of these

Answer 1

Answer: (D)

None of these

Given differential equation is
$y\sin \left(\frac{x}{y}\right)dx=\{x\sin \left(\frac{x}{y}\right)-y\}dy$
$\Rightarrow \frac{dx}{dy}=\frac{x\sin \left(\frac{x}{y}\right)-y}{y\sin \left(\frac{x}{y}\right)}=\frac{x}{y}-\frac{1}{\sin \left(\frac{x}{y}\right)}...\left(i\right)$
On putting $v=\frac{x}{y}$ $\Rightarrow$ $x=vy$
$\Rightarrow \frac{dx}{dy}=v.1+y.\frac{dv}{dy}$ in eq $\left(i\right)$, we get
$v+y\frac{dv}{dy}=v-\frac{1}{\sin v}$
$\Rightarrow y\frac{dv}{dy}=-\frac{1}{\sin v}$
$\Rightarrow -\int \sin vdv=\int \frac{dy}{y}$ (on integrating)
$\Rightarrow \cos v=\log y+C$
$\Rightarrow \cos \left(\frac{x}{y}\right)=\log y+C...\left(ii\right)$
Given at $x=\frac{\pi }{4},y=1$, then from eq. $\left(i\right)$
$\Rightarrow \cos \frac{\pi }{4}=\log 1+C$
$\Rightarrow (C=\frac{1}{\sqrt{2}})$
On putting the value of $C$ in Eq. $\left(i\right)$, we get
$\cos \left(\frac{x}{y}\right)={\log }_{e}y+\frac{1}{\sqrt{2}}$
which is the required solution. So no option is correct