QCE Specialist Mathematics - Unit 4 - Rates of change and differential equations

Differential Equations | QCE Specialist Mathematics

Learn QCE Specialist differential equations, separation of variables, slope fields, logistic growth, cooling and decay models.

Updated 2026-05-18 - 5 min read

A differential equation gives information about a function through its derivative. Instead of starting with $y$ and finding $\frac{dy}{dx}$, you are given a relationship involving $\frac{dy}{dx}$ and asked to work back toward $y$.

General and particular solutions

The equation

$ \frac{dy}{dx}=x^2+1 $

can be solved by integrating:

$ y=\frac{x^3}{3}+x+c. $

This is a general solution because it contains the constant $c$. If an initial condition is provided, such as $y(0)=4$, then $c=4$ and you have a particular solution.

The syllabus includes first-order equations of the forms:

$ \frac{dy}{dx}=f(x),\quad \frac{dy}{dx}=g(y),\quad \frac{dy}{dx}=f(x)g(y). $

The first form is solved by integrating with respect to $x$. The other two forms usually need separation of variables.

Separation of variables

Some differential equations can be separated into all $y$ terms on one side and all $x$ terms on the other.

If

$ \frac{dy}{dx}=f(x)g(y), $

then write:

$ \frac1{g(y)}\,dy=f(x)\,dx. $

Then integrate both sides.

That last sentence matters. When you divide by $y$, you temporarily exclude $y=0$. If the original differential equation allows $y=0$, include it as a separate solution or explain why the initial condition rules it out.

For a differential equation of the form $\frac{dy}{dx}=g(y)$, you can still separate:

$ \frac{dy}{g(y)}=dx. $

For example, if:

$ \frac{dy}{dx}=1+y^2, $

then:

$ \frac{1}{1+y^2}\,dy=dx. $

Integrating gives:

$ \tan^{-1}y=x+c. $

If the initial condition is $y(0)=1$, then:

$ \frac{\pi}{4}=c, $

so:

$ \tan^{-1}y=x+\frac{\pi}{4}. $

On a suitable interval, this can be written as $y=\tan\left(x+\frac{\pi}{4}\right)$. Sometimes making $y$ the subject is not helpful or not possible in simple functions; the implicit solution can still be valid.

Slope fields

A slope field shows small line segments whose gradients come from the differential equation. A solution curve should follow the local direction of those segments.

Slope fields are useful because they let you understand the behaviour of solutions without solving the equation exactly.

For an equation $\frac{dy}{dx}=F(x,y)$, the slope at each point depends on the value of $F$ there. If the right-hand side depends only on $y$, then points with the same $y$-coordinate have the same slope. If it depends only on $x$, then vertical columns share the same slope.

To construct a slope field by hand, substitute grid points into the derivative rule. If:

$ \frac{dy}{dx}=\frac{x}{y}, $

then the point $(2,1)$ has slope $2$, while the point $(2,-1)$ has slope $-2$. Points with $y=0$ are undefined, so the field has no ordinary line segment there.

An initial condition such as $y(1)=2$ selects one solution curve from the whole family. Draw the curve through $(1,2)$ so that it follows the nearby slope marks without crossing other solution curves in a first-order well-behaved field.

The direction field can also reveal geometric structure. For example, $\frac{dy}{dx}=-\frac{x}{y}$ has tangent segments that follow circles centred at the origin, because separating and integrating gives

$ y\,dy=-x\,dx \quad\Longrightarrow\quad x^2+y^2=C. $

Even if you did not solve it first, a circular slope pattern suggests that the solution curves stay on level sets of a distance-from-origin expression.

Harder separations and domains

Some separated equations need another substitution during integration. For example:

$ \frac{dy}{dx}=\frac{y\ln x}{x},\quad x>0,\ y>0. $

Separate:

$ \frac1y\,dy=\frac{\ln x}{x}\,dx. $

The right side uses $u=\ln x$, $du=\frac1x\,dx$, so:

$ \ln y=\frac12(\ln x)^2+c. $

Because $y>0$, no absolute value is needed for $\ln y$ in this context. If the domain was not specified, think carefully about signs and intervals before simplifying.

Common models

Exponential growth or decay has the form:

$ \frac{dP}{dt}=kP. $

Its solution is:

$ P=P_0e^{kt}. $

Logistic growth limits the population by including a carrying capacity $L$:

$ \frac{dP}{dt}=kP\left(1-\frac{P}{L}\right). $

Newton's law of cooling models temperature moving toward the surrounding temperature $T_s$:

$ \frac{dT}{dt}=-k(T-T_s). $

Radioactive decay has the same basic shape as exponential decay.

For Newton's law of cooling, the temperature approaches the surrounding temperature rather than decreasing forever:

$ \frac{dT}{dt}=-k(T-T_s). $

For radioactive decay:

$ \frac{dN}{dt}=-kN. $

For logistic growth, the factor $1-\frac{P}{L}$ slows growth as $P$ approaches the carrying capacity $L$.

A logistic equation can be solved by separation:

$ \frac{dP}{dt}=kP\left(1-\frac{P}{L}\right) =\frac{k}{L}P(L-P). $

Then:

$ \frac{L}{P(L-P)}\,dP=k\,dt. $

Using partial fractions:

$ \frac{L}{P(L-P)}=\frac1P+\frac1{L-P}. $

After integrating:

$ \ln\left|\frac{P}{L-P}\right|=kt+c. $

This form shows the S-shaped behaviour: growth is small when $P$ is small, fastest near the middle, and slows as $P$ approaches $L$.

Mixing-tank models use the same rate-in minus rate-out logic:

$ \frac{dQ}{dt}=\text{solute in per time}-\text{solute out per time}. $

If volume changes, the concentration in the tank is $\frac{Q(t)}{V(t)}$, so the outgoing solute rate is outflow multiplied by $\frac{Q(t)}{V(t)}$. That is the step students often miss.

Continuous compounding with deposits can be modelled by:

$ \frac{dA}{dt}=rA+D, $

where $D$ is the continuous deposit rate. Newton cooling, radioactive decay and finance models all share the same structure: define the quantity, write the rate law, attach the initial condition, then interpret the solution.

Modelling with technology

QCAA allows models to be solved with and without technology. The setup should still show the differential equation, variables, parameter meanings and initial condition. Technology can solve or graph the equation, but your interpretation should say what the solution means in the context.

Quick check

Sources

• QCAA Specialist Mathematics subject page
• QCAA Specialist Mathematics 2025 syllabus
• OpenStax Calculus Volume 2: Basics of Differential Equations
• OpenStax Calculus Volume 2: The Logistic Equation