Discussion Forum

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

Answer:

Explanation:

Concept involved

 solving of homogeneous   differential equation  i.e,  substitute  $\frac{y}{x}=v$

 $\therefore$                        y=vx

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

  here, slope of the curve  at (x, y) is

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

 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_{}^{} \cos v dv=\int_{}^{} \frac{dx}{x}$

$\Rightarrow $     $ \sin v=\log x+\log c$

$\Rightarrow$    $  \sin (\frac{y}{x})=\log (cx)$

 as it passes through  $(1,\frac{\pi}{6})$

  $\Rightarrow $    $ \sin(\frac{\pi}{6})=\log c$

  $\Rightarrow  \log c=\frac{1}{2}$

 $\therefore$        $\sin(\frac{y}{x})=\log x+\frac{1}{2}$