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 = elny = 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
