Question

`y'=x/y-x/(1-y)` using Separation of Variables?

Answer 1

`1/2(1-2y)^2-ln|1-2y| = 4x^2 + A`

Explanation:

We have:

`y' = x/y - x/(1-y)`

Which we can write as:

`y' = x{1/y - 1/(1-y)}`

`\ \ \ = x{((1-y)-y)/(y(1-y))}`

`\ \ \ = x{(1-2y)/(y(1-y))}`

`:. (y(1-y))/(1-2y) \ dy/dx = x`

This is a separable DE, so we can "separate the variables":

`int \ (y(1-y))/(1-2y) \ dy = int \ x \ dx` ..... [A]

To integrate the LHS, consider:

`I = int \ (y(1-y))/(1-2y) \ dy`

Along with a substitution:

`u = 1-2y => (du)/dy = -2 \ \` and `y = (1-u)/2`

Which gives us:

`I = int \ ((1-u)/2 (1-(1-u)/2))/(u)(-1/2) \ du`

`\ \ = -1/4 \ int \ ((1-u) ((2-1+u)/2))/(u) \ du`

`\ \ = 1/8 \ int \ ((u-1)(u+1))/(u) \ du`

`\ \ = 1/8 \ int \ (u^2-1)/(u) \ du`

`\ \ = 1/8 \ int \ u-1/u \ du`

`\ \ = 1/8 {1/2u^2-ln|u| }`

Using this result, we can now fully integrate [A], to get:

`1/8 {1/2(1-2y)^2-ln|1-2y| } = 1/2x^2 + C`

`:. 1/2(1-2y)^2-ln|1-2y| = 4x^2 + A`, say

Which is the Implicit General Solution.