Question

If $\frac{dy}{dx}=\frac{y}{x}\left(\frac{x\,cos \frac{y}{x}+y\,sin \frac{y}{x}}{y\,sin \frac{y}{x}-x\,cos \frac{y}{x}}\right)$, then

• A. $x\,cos\, \frac{y}{x}=cy$
• B. $x\,sec\, \frac{y}{x}=cy$
• C. $cos\, \frac{y}{x}=cxy$
• D. $sec\, \frac{y}{x}=cxy$

Answer 1

Answer: (D)

Substitute $y=vx$
$\Rightarrow \frac{dy}{dx}=\frac{xdv}{dx}+v$
Now, given equation becomes
$\frac{xdv}{dx}+v=v\left(\frac{cos\,v+v\,sin\,v}{v\,sin\,v-cos\,v}\right)$
$\Rightarrow \frac{xdv}{dx}=\frac{2v\,cos\,v}{v\,sin\,v-cos\,v}$
$\Rightarrow \frac{2dx}{x}+\frac{\left(cos\,v-v\,sin\,v\right)dv}{v\,cos\,v}=0$
$\Rightarrow 2\,ln\,x+ln\left(v\,cos\,v\right)-ln\,c'$
$\Rightarrow x^{2}\left(\frac{y}{x}cos \frac{y}{x}\right)=c'$
$\Rightarrow xy\,cos\, \frac{y}{x}=c'$
$\Rightarrow sec\, \frac{y}{x}=cxy$.