Solution to Homogeneous Differential Equation

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Theorem

Let:

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

be a homogeneous differential equation‎.


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 separable differential equation.


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.


Sources

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