First Order ODE/(y - 1 over x) dx + (x - y) dy = 0

Theorem

The first order ODE:

$(1): \quad \paren {y - \dfrac 1 x} \rd x + \paren {x - y} \rd y = 0$

is an exact differential equation with solution:

$x y - \ln x - \dfrac {y^2} 2 + C$

Proof

Let $M$ and $N$ be defined as:

$\map M {x, y} = y - \dfrac 1 x$

$\map N {x, y} = x - y$

Then:

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

\(=\)

\(\ds 1\)

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

\(=\)

\(\ds 1\)

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:

\(\ds f\)

\(=\)

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

\(\ds \)

\(=\)

\(\ds \int \paren {y - \dfrac 1 x} \rd x + \map g y\)

\(\ds \)

\(=\)

\(\ds x y - \ln x + \map g y\)

and:

\(\ds f\)

\(=\)

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

\(\ds \)

\(=\)

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

\(\ds \)

\(=\)

\(\ds x y - \dfrac {y^2} 2 + \map h x\)

Thus:

$\map f {x, y} = x y - \ln x - \dfrac {y^2} 2$

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

$x y - \ln x - \dfrac {y^2} 2 + C$

$\blacksquare$