# Separation of Variables

## Theorem

Suppose a first order ordinary differential equation can be expressible in this form:

- $\dfrac {\d y} {\d x} = \map g x \map h y$

Then the equation is said to **have separable variables**, or **be separable**.

Its general solution is found by solving the integration:

- $\ds \int \frac {\d y} {\map h y} = \int \map g x \rd x + C$

### General Result

Suppose a first order ordinary differential equation can be expressible in this form:

- $\map {g_1} x \map {h_1} y + \map {g_2} x \map {h_2} y \dfrac {\d y} {\d x} = 0$

Then the equation is said to **have separable variables**, or **be separable**.

Its general solution is found by solving the integration:

- $\ds \int \frac {\map {g_1} x} {\map {g_2} x} \rd x + \int \frac {\map {h_2} y} {\map {h_1} y} \rd y = C$

## Proof

Dividing both sides by $\map h y$, we get:

- $\dfrac 1 {\map h y} \dfrac {\d y} {\d x} = \map g x$

Integrating both sides with respect to $x$, we get:

- $\ds \int \frac 1 {\map h y} \frac {\d y} {\d x} \rd x = \int \map g x \rd x$

which, from Integration by Substitution, reduces to the result.

The arbitrary constant $C$ appears during the integration process.

$\blacksquare$

## Also presented as

Some sources present this as an equation in the form:

- $\dfrac {\d y} {\d x} = \dfrac {\map g x} {\map h y}$

or:

- $\map h y \dfrac {\d y} {\d x} = \map g x$

whose general solution is found by solving the integration:

- $\ds \int \map h y \rd y = \int \map g x \rd x + C$

Other sources have:

- $\map g x + \map h y \dfrac {\d y} {\d x} = 0$

whose general solution is found by solving the integration:

- $\ds \int \map g x \rd x = -\int \map h y \rd y + C$

## Mnemonic Device

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As derivatives are not fractions, the following is a mnemonic device only.

This is an an abuse of notation that is likely to make some Calculus professors upset.

But it's useful.

\(\ds \frac {\d y} {\d x}\) | \(=\) | \(\ds \map g x \map h y\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds \d y\) | \(=\) | \(\ds \map g x \map h y \rd x\) | solve for $\d y$ | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds \frac 1 {\map h y} \rd y\) | \(=\) | \(\ds \map g x \rd x\) | collecting like terms on each side |

## Examples

### Arbitrary Example 1

Consider the first order ODE:

- $(1): \quad \map {\dfrac \d {\d x} } {\map f x} = 3 x$

where we are given that $\map f 1 = 2$.

The particular solution to $(1)$ is:

- $\map f x = \dfrac {3 x^2 + 1} 2$

## Historical Note

The method of **Separation of Variables** was described by Johann Bernoulli between the years $\text {1694}$ – $\text {1697}$.

## Sources

- 1972: George F. Simmons:
*Differential Equations*... (previous) ... (next): $\S 2.7$: Homogeneous Equations - 1998: David Nelson:
*The Penguin Dictionary of Mathematics*(2nd ed.) ... (previous) ... (next):**differential equation** - 2008: David Nelson:
*The Penguin Dictionary of Mathematics*(4th ed.) ... (previous) ... (next):**differential equation** - 2014: Christopher Clapham and James Nicholson:
*The Concise Oxford Dictionary of Mathematics*(5th ed.) ... (previous) ... (next):**separable first-order differential equation** - 2021: Richard Earl and James Nicholson:
*The Concise Oxford Dictionary of Mathematics*(6th ed.) ... (previous) ... (next):**separable first-order differential equation**