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:

\(\ds \dfrac {\partial M} {\partial y}\)

\(=\)

\(\ds -\frac 1 {x - y}\)

\(\ds \dfrac {\partial N} {\partial x}\)

\(=\)

\(\ds -\frac 1 {x - y}\)

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:

\(\ds \dfrac {\partial f} {\partial x}\)

\(=\)

\(\ds \map M {x, y}\)

\(\ds \dfrac {\partial f} {\partial y}\)

\(=\)

\(\ds \map N {x, y}\)

Hence:

\(\ds f\)

\(=\)

\(\ds \map M {x, y} \rd x + \map g y\)

\(\ds \)

\(=\)

\(\ds \int \left({1 + \map \ln {x - y} }\right) \rd x + \map g y\)

\(\ds \)

\(=\)

\(\ds x + \paren {x - y} \map \ln {x - y} - \paren {x - y} + \map g y\)

Primitive of $\ln x$

and:

\(\ds f\)

\(=\)

\(\ds \int \map N {x, y} \rd y + \map h x\)

\(\ds \)

\(=\)

\(\ds \int \left({-\map \ln {x - y} }\right) \rd y + \map h x\)

\(\ds \)

\(=\)

\(\ds \paren {x - y} \map \ln {x - y} - \paren {x - y} + \map h x\)

Primitive of $\ln x$

Thus:

\(\ds \map f {x, y}\)

\(=\)

\(\ds x + \paren {x - y} \map \ln {x - y} - \paren {x - y}\)

\(\ds \)

\(=\)

\(\ds \paren {x - y} \map \ln {x - y} + y\)

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

\(\ds \paren {x - y} \map \ln {x - y} + y\)

\(=\)

\(\ds C\)

\(\ds \leadsto \ \ \)

\(\ds \paren {x - y} \map \ln {x - y}\)

\(=\)

\(\ds C - y\)

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): Miscellaneous Problems for Chapter $2$: Problem $18$