First Order ODE/y' ln (x - y) = 1 + ln (x - y)

Theorem
The first order ordinary differential equation:


 * $(1): \quad \dfrac {\d y} {\d x} \map \ln {x - y} \d x = 1 + \map \ln {x - y}$

is an exact differential equation with solution:


 * $\paren {x - y} \map \ln {x - y} = C - y$

Proof
Let $(1)$ be expressed as:
 * $\paren {1 + \map \ln {x - y} } \rd x - \map \ln {x - y} \rd y = 0$

Let:
 * $\map M {x, y} = 1 + \map \ln {x - y}$
 * $\map N {x, y} = -\map \ln {x - y}$

Then:

Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be exact.

By Solution to Exact Differential Equation, the solution to $(1)$ is:
 * $\map f {x, y} = C$

where:

Hence:

and:

Thus:

and by Solution to Exact Differential Equation, the solution to $(1)$ is: