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

Answer 1

Answer: (A)

PLAN To solve homogeneous differential equation, i.e. substitute $\frac{y}{x}=v$
$\therefore y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
Here, slope of the curve at $(x,y)$ is
$\frac{dy}{dx}=\frac{y}{x}+\sec \left(\frac{y}{x}\right)$
$Put\frac{y}{x}=v$
$\therefore v+x\frac{dv}{dx}=v+\sec (v)$
$\Rightarrow x\frac{dv}{dx}=\sec (v)\Rightarrow \int \frac{dv}{\sec v}=\int \frac{dx}{x}$
$\Rightarrow \int cosvdv=\int \frac{dx}{x}\Rightarrow \sin v=\log x+\log c$
$\Rightarrow \sin \left(\frac{y}{x}\right)=\log (cx)$
As it passes through $(1,\frac{\pi }{6})\Rightarrow \sin \left(\frac{\pi }{6}\right)=logc$
$\Rightarrow \log c=\frac{1}{2}$
$\therefore \sin (\frac{y}{x})=\log x+\frac{1}{2}$