Q: 1+((x/y)-siny)y’= 0
Let M(x,y)=1 , N(x,y)= ((x/y)-siny)
Then, ∂/∂y{M(x,y)}=0 , ∂/∂x{N(x,y)}=1/y
Notice that
(N_x – M_y)/M = 1/y
It contains y only, so
dμ/dy = [(N_x – M_y)/M] μ = μ/y
μ=y
Multiplying original equation by μ(y),we get
y + (x-ysin(y))y’= 0
Now,this equation is exact, since
M_y = N_x
Therefore,
∃φ(x,y) s.t. φ_x = M = y
φ=∫ydx= xy + h(y)
φ_y= x +h’(y)
Also, φ_y=N=x-ysin(y)
So h’(y)=ysin(y)
h(y)= ∫ysin(y) dy=ycos(y)-sin(y)+c
Thus
φ=xy+ ycos(y)-sin(y)=C
