Question

The solution of the differential equatio $ysin(x/y)dx=(xsin(x/y)-y)dy$ satisfying $y(\pi /4)=1$ is

• A. $cos\frac{x}{y}=-lo{g}_{e}y+\frac{1}{\sqrt{2}}$
• 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. $cos\frac{x}{y}=-lo{g}_{e}x-\frac{1}{\sqrt{2}}$

Answer 1

Answer: (A)

Given differential equation is
$y\sin \left(\frac{x}{y}\right)dx=\left\{x\sin \left(\frac{x}{y}\right)-y\right\}dy$
$\Rightarrow \frac{dx}{dy}=\frac{x\sin \left(\frac{x}{y}\right)-y}{y\sin \left(\frac{x}{y}\right)}=\frac{x}{y}-\frac{1}{\sin \left(\frac{x}{y}\right)}$ ... (i)
On putting $v=\frac{x}{y}\Rightarrow x=v$
$\Rightarrow \frac{dx}{dy}=v\cdot 1+y\frac{dv}{dy}$ in Eq. (i), 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)
$\Rightarrow \cos v=\log y+C$
$\Rightarrow \cos \left(\frac{x}{y}\right)=\log y+C...(i)$
Given at $x=\frac{\pi }{4},y=1$ then from Eq. (i)
$\Rightarrow \cos \left(\frac{\pi }{4}\right)=\log (1)+C$
$\Rightarrow \left(C=\frac{1}{\sqrt{2}}\right)$
On putting the value of $C$ in Eq. (i), 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.