Question: 1 + [x/y - sin(y)]y’ = 0

We firstly simplify the equation into

dx + [x/y - sin(y)] dy= 0

Then we have M and N

M(x,y) = 1

N(x,y) = [x/y - sin(y)]

Then, we find the derivative of M with respect to y and N with respect to x

My = 0

Nx = 1/y

Since My is not equal to Nx, it is not exact.

Thus, we need to multiply a factor 𝓾 that satisfies the equation

R1 = [ (My - Nx)/ M] = [(0-1/y)] = -1/y

𝓾 = e^{-∫R1dy} = e^{-∫(-1/y)dy} = e^{lny} = y

We multiply the 𝓾 on both sides of the equation to find an exact equation

𝓾 dx + 𝓾[x/y - sin(y)] dy= 0

y dx + y [x/y - sin(y)] dy= 0

Then we have our new M’ and N’

M’(x,y) = y

N’(x,y) = y[x/y - sin(y)] = x - ysin(y)

Thus, there exist a function 𝒞(x,y) such that

𝒞x = M’

𝒞y = N’

By Integrating M’ with respect to x

𝒞x = M’

𝒞 = ∫ M’ dx = ∫ y dx = xy + h(y)

By differentiating with respect to y and equating to 𝒞y = N’

We get x + h’(y) = x - ysin(y)

Therefore, h’(y) = - ysin(y)

By integrating on both sides

h(y) =∫ - ysin(y) dy = ycos(y) - sin(y)

Now, we have

𝒞 = xy + ycos(y) - sin(y)

Thus, the solutions of differential equation are given implicitly by

xy + ycos(y) - sin(y) = C