First Order ODE/(x + (2 over y)) dy + y dx = 0

Theorem

The first order ordinary differential equation:

$(1): \quad \paren {x + \dfrac 2 y} \rd y + y \rd x = 0$

is an exact differential equation with solution:

$x y + 2 \ln y = C$

This can also be presented as:

$\dfrac {\d y} {\d x} + \dfrac y {x + \dfrac 2 y} = 0$

Proof

Let:

$\map M {x, y} = y$

$\map N {x, y} = x + \dfrac 2 y$

Then:

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

\(=\)

\(\ds 1\)

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

\(=\)

\(\ds 1\)

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:

\(\ds f\)

\(=\)

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

\(\ds \)

\(=\)

\(\ds \int y \rd x + \map g y\)

\(\ds \)

\(=\)

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

and:

\(\ds f\)

\(=\)

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

\(\ds \)

\(=\)

\(\ds \int \paren {x + \dfrac 2 y} \rd y + \map h x\)

\(\ds \)

\(=\)

\(\ds x y + 2 \ln y + \map h x\)

Thus:

$\map f {x, y} = x y + 2 \ln y$

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

$x y + 2 \ln y = C$

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 2.8$: Exact Equations: Problem $1$