Question

A curve passes through the point $(1,\frac{\pi }{6})$. Let the slope of the curve at each point $(x,y)$be $\frac{y}{x}+\sec \left(\frac{y}{x}\right)$, $x>0$. Then the equation of the curve is:

• A. $\sin \left(\frac{y}{x}\right)=\log x+\frac{1}{2}$
• B. $\csc \left(\frac{y}{x}\right)=\log x+2$
• C. $\sec \left(\frac{2y}{x}\right)=\log x+2$
• D. $\cos \left(\frac{2y}{x}\right)=\log x+\frac{1}{2}$

Answer 1

The correct option is A $\sin \left(\frac{y}{x}\right)=\log x+\frac{1}{2}$

$\frac{dy}{dx}=\frac{y}{x}+\sec \frac{y}{x}$

Substitute $y=vx$

$\frac{dy}{dx}=v+x\frac{dv}{dx}$
$\Rightarrow v+x\frac{dv}{dx}=v+\sec v$

$\Rightarrow \frac{dv}{secv}=\frac{dx}{x}$
$\int \cos vdv=\int \frac{dx}{x}$
$\sin v=\ln x+c$

$\sin \left(\frac{y}{x}\right)=\ln x+c$

The curve passes through $(1,\frac{\pi }{6})$

$\Rightarrow \sin \left(\frac{y}{x}\right)=\ln x+\frac{1}{2}$