# Solution to Homogeneous Differential Equation

## Theorem

Let:

$\map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$

It can be solved by making the substitution $z = \dfrac y x$.

Its solution is:

$\ds \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$

where:

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

## Proof

From the original equation‎, we see:

$\dfrac {\d y} {\d x} = \map f {x, y} = -\dfrac {\map M {x, y} } {\map N {x, y} }$

From Quotient of Homogeneous Functions‎ it follows that $\map f {x, y}$ is homogeneous of degree zero.

Thus:

$\map f {t x, t y} = t^0 \map f {x, y} = \map f {x, y}$

Set $t = \dfrac 1 x$ in this equation‎:

 $\ds \map f {x, y}$ $=$ $\ds \map f {\paren {\frac 1 x} x, \paren {\frac 1 x} y}$ $\ds$ $=$ $\ds \map f {1, \frac y x}$ $\ds$ $=$ $\ds \map f {1, z}$

where $z = \dfrac y x$.

Then:

 $\ds z$ $=$ $\ds \frac y x$ $\ds \leadsto \ \$ $\ds y$ $=$ $\ds z x$ $\ds \leadsto \ \$ $\ds \frac {\d y} {\d x}$ $=$ $\ds z + x \frac {\d z} {\d x}$ Product Rule for Derivatives $\ds \leadsto \ \$ $\ds z + x \frac {\d z} {\d x}$ $=$ $\ds \map f {1, z}$ $\ds \leadsto \ \$ $\ds \int \frac {\d z} {\map f {1, z} - z}$ $=$ $\ds \int \frac {\d x} x$

This is seen to be a differential equation with separable variables.

On performing the required integrations and simplifying as necessary, the final step is to substitute $\dfrac y x$ back for $z$.

$\blacksquare$

## Historical Note

This method of Solution to Homogeneous Differential Equation was described by Johann Bernoulli between the years $\text {1694}$ – $\text {1697}$.

He applied this technique to problems on orthogonal trajectories.