# Solution to Separable Differential Equation

## Definition

Consider the separable differential equation:

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

Its general solution is found by solving the integration:

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

This technique is generally known as Separation of Variables.

### General Result

Consider the separable differential equation:

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

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.

$\blacksquare$

## Mnemonic Device

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:

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