Question

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

• A. $\sin \bigg ( \frac{y}{x}\bigg )=\log x+ \frac{1}{2}$
  
• B. $cosec \bigg ( \frac {y}{x}\bigg )=\log x + 2$
  
• C. $\sec \bigg ( \frac {2y}{x}\bigg )=\log x+ 2$
  
• D. $\cos \bigg ( \frac {2y}{x}\bigg )=\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
$\frac{dy}{dx}= \frac{y}{x}+ \sec \bigg (\frac {y}{x}\bigg )$
$Put \frac {y}{x}=v$
$\therefore v+x \frac{dv}{dx}=v+\sec (v)$
$\Rightarrow x \frac{dv}{dx}=\sec (v) \Rightarrow \int \limits \frac{dv}{\sec v}=\int \limits \frac{dx}{x}$
$\Rightarrow \int \limits cosvdv= \int \limits \frac {dx}{x} \Rightarrow \sin v=\log \, x+ \log \, c $
$\Rightarrow \sin \bigg (\frac{y}{x} \bigg )=\log(cx)$
As it passes through $\bigg (1, \frac{\pi}{6} \bigg ) \Rightarrow \sin \bigg (\frac{\pi}{6} \bigg )=log \, c $
$\Rightarrow \log \, c =\frac{1}{2}$
$\therefore\sin (\frac{y}{x})=\log \, x + \frac{1}{2}$