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1)

A curve passes through the point   (1,π6) let the slope of the curve at each point  (x,y) be   yx+sec(yx),x>0.  , then , the equation  of the curve is 


A) sin(yx)=logx+12

B) cosec(yx)=logx+2

C) sec(2yx)=logx+2

D) cos(2yx)=logx+12

Answer:

Option A

Explanation:

Concept involved

 solving of homogeneous   differential equation  i.e,  substitute  yx=v

                         y=vx

     dydx=v+xdvdx

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

   dydx=yx+sec(yx)

 put      yx=v

     v+xdvdx=v+sec(v)

      xdvdx=sec(v)

     dvsecv=dxx

      cosvdv=dxx

     sinv=logx+logc

    sin(yx)=log(cx)

 as it passes through  (1,π6)

      sin(π6)=logc

  logc=12

         sin(yx)=logx+12