Question

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

• A. $xcos\frac{y}{x}=cy$
• B. $xsec\frac{y}{x}=cy$
• 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
$\frac{xdv}{dx}+v=v\left(\frac{cosv+vsinv}{vsinv-cosv}\right)$
$\Rightarrow \frac{xdv}{dx}=\frac{2vcosv}{vsinv-cosv}$
$\Rightarrow \frac{2dx}{x}+\frac{\left(cosv-vsinv\right)dv}{vcosv}=0$
$\Rightarrow 2lnx+ln\left(vcosv\right)-ln{c}^{\prime }$
$\Rightarrow x^2\left(\frac{y}{x}cos\frac{y}{x}\right)=c^{\prime }$
$\Rightarrow xycos\frac{y}{x}=c^{\prime }$
$\Rightarrow sec\frac{y}{x}=cxy$.