Separation of Variables

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Theorem

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

$\displaystyle \frac {\mathrm dy} {\mathrm dx} = g \left({x}\right) h \left({y}\right)$

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

Its general solution is found by solving the integration:

$\displaystyle \int \frac {\mathrm dy} {h \left({y}\right)} = \int g \left({x}\right) \ \mathrm dx + C$

Dividing both sides by $h \left({y}\right)$, we get:

$\displaystyle \frac 1 {h \left({y}\right)} \frac {\mathrm dy} {\mathrm dx} = g \left({x}\right)$

Integrating both sides WRT $x$, we get:

$\displaystyle \int \frac 1 {h \left({y}\right)} \frac {\mathrm dy} {\mathrm dx} \mathrm dx = \int g \left({x}\right) \ \mathrm dx$

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

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

$\blacksquare$

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.

$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle \frac {\mathrm dy} {\mathrm dx}$	$=$	$\displaystyle g \left({x}\right) h \left({y}\right)$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle \implies$	$\displaystyle $	$\displaystyle \mathrm dy$	$=$	$\displaystyle g \left({x}\right) h \left({y}\right) \mathrm dx$	$\displaystyle $	$\displaystyle $	$\displaystyle $	solve for $\mathrm dy$
$\displaystyle $	$\displaystyle \implies$	$\displaystyle $	$\displaystyle \frac 1 {h \left({y}\right)}\mathrm dy$	$=$	$\displaystyle g \left({x}\right) \mathrm dx$	$\displaystyle $	$\displaystyle $	$\displaystyle $	because things that have $y$ in it deserve to be together

\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \frac {\mathrm dy} {\mathrm dx}\)	\(=\)	\(\displaystyle g \left({x}\right) h \left({y}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \mathrm dy\)	\(=\)	\(\displaystyle g \left({x}\right) h \left({y}\right) \mathrm dx\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	solve for $\mathrm dy$
\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \frac 1 {h \left({y}\right)}\mathrm dy\)	\(=\)	\(\displaystyle g \left({x}\right) \mathrm dx\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	because things that have $y$ in it deserve to be together