Formulating differential equations from rates of change

By the end of this lesson

By the end you can write a differential equation from a sentence about a rate of change, decide whether the equation needs a minus sign, and use given values to work out the constant $k$ .

A rate of change is a derivative

A rate of change measures how fast one quantity changes when another quantity changes.

In calculus, a rate of change is a derivative. If $y$ depends on $x$ , the symbol

$\frac{dy}{dx}$

means "how fast $y$ changes for each unit that $x$ increases". Its units are the units of $y$ divided by the units of $x$ .

For example, suppose $V$ is the volume of water in a tank in litres, and $t$ is the time in seconds. Then

$\frac{dV}{dt}$

is measured in litres per second. It is positive while the tank fills, and negative while the tank drains.

Key to the symbols

Symbol	What it means
$y,\ V,\ P,\ \theta$	the quantity that is changing
$x,\ t$	the quantity it changes with ( $t$ is usually time)
$\frac{dy}{dx}$	the rate of change of $y$ with respect to $x$
$k$	the constant of proportionality: a fixed number for the problem

Always say what your letters mean: "Let $y$ be the mass in grams at time $t$ seconds." The examiner marks the equation, and an undefined letter makes it impossible to follow.

From words to symbols

A differential equation is an equation that contains a derivative. The ones in this topic are first order: they contain $\frac{dy}{dx}$ (or $\frac{dV}{dt}$ and so on) but no second derivative.

The syllabus wording is "formulate a simple statement involving a rate of change as a differential equation". In practice it means: read the sentence, spot the derivative, write the equation.

Translation table

Words	Symbols
rate of change of $y$ with respect to $x$	$\frac{dy}{dx}$
rate of change of $V$ with respect to time $t$	$\frac{dV}{dt}$
is proportional to $y$	$= ky$
is inversely proportional to $r$	$= \frac{k}{r}$
is proportional to the square of $r$	$= kr^2$
is proportional to the square root of $h$	$= k\sqrt{h}$
increases at a rate proportional to $y$	$\frac{dy}{dt} = ky$
decreases at a rate proportional to $y$	$\frac{dy}{dt} = -ky$

In words: proportional means "equals a fixed number times", and inversely proportional means "equals a fixed number divided by".

Worked example: A first equation from a sentence

The rate of change of $y$ with respect to $x$ is proportional to $y$ . Write a differential equation connecting $y$ and $x$ .

Have a go first?

Quick check

One-line answerQuestion 1 · Recall

One-line answerQuestion 2 · Apply

Answer the 2 questions above to continue.