We have the following linear ordinary differential equation and we want to solve for y:
(x-2)y' - y = 2(x-2)^3
Can you help me find where the error is? I am getting incorrect results.
First, express the equation in the standard form of an exact differential equation:
y + 2(x-2)^3 + (-x+2)y' = 0 where M = y + 2(x-2)^3 and N = -x+2
We check if the equation is exact. It isn't, so we try to find an integrating factor u(x)
We find our u(x) and multiply our equation with it:
uy + u2(x-2)^3 + u(-x+2)y' = 0
Plug in u(x) = 1/(-x+2)^2 and we get:
1/(-x+2)^2 * y + 1/(-x+2)^2 * 2(x-2)^3 + 1/(-x+2)^2 * (-x+2)y' = 0
Clean it up and we get:
1/(-x+2)^2 * y + 2x - 4 + 1/(-x+2) * y' = 0
Let's denote our potential function with F (in some materials they use the phi symbol).
dF/dx = 1/(-x+2)^2 * y + 2x - 4
In order to find F, let's integrate both sides and add g(y), which represents the information that we are unable to retrieve from a partial derivative of x. Let's absorb the constant in it.
F = (-x+2)^(-1) * y + x^2 - 4x + g(y)
Let's find out what g(y) is.
dF/dy = (-x+2)^(-1) + g'(y) = uN
Plug in uN from earlier:
dF/dy = (-x+2)^(-1) + g'(y) = (-x+2)^(-1)
And we get g'(y) = 0 so g(y) = C
Plug g(y) into F:
F = (-x+2)^(-1) * y + x^2 - 4x + C = 0
Solve for y:
y = x^3 - 6x^2 + 8x + Cx - 2C
This is incorrect. Where did I make a mistake?
P.S. I realize there are easier choices for u available, but let's not go there. I really want to find out what is wrong with this solution. Before posting this I verified that my choice for u(x) doesn't contain variables other than x and I confirmed that the equation with u(x) is exact by comparing partials derivatives. I also tried to look for blunders by verifying just about everything with Wolfram Alpha. I can't find the error.