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

Theorem
The first order ordinary differential equation:


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

is an exact differential equation with solution:


 * $\left({x - y}\right) \ln \left({x - y}\right) = C - y$

Proof
Let $(1)$ be expressed as:
 * $\left({1 + \ln \left({x - y}\right)}\right) \mathrm d x - \ln \left({x - y}\right) \mathrm d y = 0$

Let:
 * $M \left({x, y}\right) = 1 + \ln \left({x - y}\right)$
 * $N \left({x, y}\right) = -\ln \left({x - y}\right)$

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:
 * $f \left({x, y}\right) = C$

where:

Hence:

and:

Thus:

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