1+(x/y-siny)y' = 0
M = 1, N = x/y-siny
My = o, Nx = 1/y
My != Nx, the equation not exact.
R1 = (My-Nx)/M = -1/y
u = e^∫R1 dy
u = y
Multiply u to both sides,
y + (x - Ysiny)y' = 0
Now, the equation exact.
there exist Φ(x, y), st.Φx = M
Φ = xy +h(y)
Φy = x + h'(y)
h'(y) = -y siny
h(y) = y cosy - siny
So, we have Φ(x,y) = xy +ycosy - Siny
Therefore, the solution of the differential equation is :
xy + y cosy - siny = C