Question

use method of exactness to solve the DE:
(e^xsiny-2ysinx)dx+(e^xcosy +2cosx)dy=0

Answer 1

the eq is exact as
partial derivative of P(x,y) w.r.t y=partial derivative of Q(x,y) w.r.t x
R(x,y)=int( e^xsiny-2ysinx)dx
=e^xsiny+2ycosx +h(y)
partial derivative of R w.r.t y=e^xcosy+2cosx+h'(y) = Q(x,y)

Answer 2

which shows that h'(y)=0