If xsiny=sin(p+y),p∈R,show that sinpdy/dx+sin^2y=0?

1 Answer
Feb 12, 2018

Kindly refer to a Proof in the Explanation.

Explanation:

Given that, #xsiny=sin(p+y), p in RR#.

#:. x=sin(p+y)/siny=(sinpcosy+sinycosp)/siny#,

#=(sinpcosy)/siny+(sinycosp)/siny#.

# rArr x=sinpcoty+cosp, (p" const.)"#

Diff.ing w.r.t. #y#, we get,

#dx/dy=-csc^2ysinp+0=-sinp/sin^2y#.

#:. dy/dx=1/(dx/dy)=-sin^2y/sinp, or," what is the same as,"#

# sinpdy/dx+sin^2y=0#, as desired!