This topic is on Paper 3 only. If you are sitting AS Level with Paper 2, you do not need it.
🎯What you need to be able to do
- Form a differential equation from a description of a rate of change.
- Solve a first-order equation by separating the variables.
- Use partial fractions to carry out the integration when required.
- Find the constant of integration from an initial condition, and give the particular solution.
- Rearrange a solution to make the dependent variable the subject.
- Interpret the solution in context, including its long-term behaviour.
📚The mathematics
Forming the equation
This step is examinable in its own right, and it is pure translation. The phrase “the rate of
change of \(P\) with respect to time” is \( \dfrac{\mathrm{d}P}{\mathrm{d}t} \), and
“proportional to” introduces a constant \(k\):
proportional to \(P\): \( \dfrac{\mathrm{d}P}{\mathrm{d}t} = kP \)
proportional to \( \sqrt{P} \): \( \dfrac{\mathrm{d}P}{\mathrm{d}t} = k\sqrt{P} \)
inversely proportional to \(P\): \( \dfrac{\mathrm{d}P}{\mathrm{d}t} = \dfrac{k}{P} \)
Read the direction carefully. A quantity that is decreasing needs a negative sign, so
“the rate of decrease is proportional to \(P\)” is
\( \dfrac{\mathrm{d}P}{\mathrm{d}t} = -kP \) with \( k > 0 \). And a rate proportional to the
difference from some fixed value, as in cooling, is
\( \dfrac{\mathrm{d}\theta}{\mathrm{d}t} = -k(\theta - \theta_0) \) — the exponential term
describes the gap, not the quantity itself.
Separating the variables
An equation of the form \( \dfrac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{f}(x)\,\mathrm{g}(y) \) is
separable. Gather all the \(y\) terms on one side with \( \mathrm{d}y \), all the \(x\) terms on the
other with \( \mathrm{d}x \), and integrate both sides:
\[ \int \frac{1}{\mathrm{g}(y)}\,\mathrm{d}y = \int \mathrm{f}(x)\,\mathrm{d}x \]
One constant only. Both sides produce a constant, but they combine into a single
arbitrary constant on one side. Writing \( c_1 \) and \( c_2 \) and carrying both is not wrong, but
it is clutter; put \( +c \) on the \(x\) side and move on.
Handling the integration
The \(y\) integral is often the awkward one. Three patterns cover most questions:
- \( \displaystyle\int \frac{1}{y}\,\mathrm{d}y = \ln|y| + c \) — the exponential growth
and decay case.
- \( \displaystyle\int \frac{1}{y(a - y)}\,\mathrm{d}y \) — needs
partial fractions first. This is the logistic model, and it is a
Paper 3 favourite.
- \( \displaystyle\int \frac{1}{\sqrt{y}}\,\mathrm{d}y = 2\sqrt{y} + c \) — a power, not a
logarithm.
Applying the initial condition
The general solution contains the arbitrary constant; the particular
solution has it determined by a given condition, typically the value at \( t = 0 \).
Substitute the condition as soon as you have integrated — before rearranging — because
the equation is usually simplest at that moment.
Combine the exponential constants properly. From \( \ln|y| = kt + c \),
exponentiating gives \( y = \mathrm{e}^{kt + c} = \mathrm{e}^{c}\mathrm{e}^{kt} \). Since
\( \mathrm{e}^{c} \) is just another positive constant, write it as a single letter \(A\):
\( y = A\mathrm{e}^{kt} \). Carrying \( \mathrm{e}^{c} \) through the rest of the question is
correct but painful, and writing \( y = \mathrm{e}^{kt} + c \) — adding rather than
multiplying — is simply wrong and very common.
Interpreting the answer
Questions almost always end by asking what the solution means. Two standard requests:
- A value at a given time, or the time at which a value is reached —
substitute and solve, often taking logarithms.
- The long-term behaviour, as \( t \to \infty \). For \( y = A\mathrm{e}^{-kt} + B \)
with \( k > 0 \), the exponential vanishes and \( y \to B \). Say what that limit
means: the temperature settles at room temperature, the population stabilises at the
carrying capacity, the concentration approaches zero.
✏️Worked example
A tank contains \(V\) litres of liquid at time \(t\) minutes. The liquid drains so that the rate
of decrease of \(V\) is proportional to \( \sqrt{V} \). Initially \( V = 400 \), and after
10 minutes \( V = 225 \).
(a) Form a differential equation and solve it to find \(V\) in terms of \(t\).
(b) Find the time at which the tank is empty.
(a) The rate of decrease is proportional to \( \sqrt{V} \), so
\[ \frac{\mathrm{d}V}{\mathrm{d}t} = -k\sqrt{V}, \qquad k > 0 \]
Separating the variables:
\[ \int \frac{1}{\sqrt{V}}\,\mathrm{d}V = \int -k\,\mathrm{d}t
\;\Longrightarrow\; 2\sqrt{V} = -kt + c \]
Apply the initial condition \( V = 400 \) when \( t = 0 \): \( 2(20) = c \), so \( c = 40 \).
Then use \( V = 225 \) at \( t = 10 \): \( 2(15) = -10k + 40 \), so \( 30 = 40 - 10k \), giving
\( 10k = 10 \) and \( k = 1 \).
So \( 2\sqrt{V} = 40 - t \), and making \(V\) the subject:
\[ \sqrt{V} = \frac{40 - t}{2} \;\Longrightarrow\; V = \frac{(40-t)^{2}}{4} \]
(b) The tank is empty when \( V = 0 \), which requires \( 40 - t = 0 \), so
\( t = 40 \) minutes.
Check it. Test both given data points against the solution. At \( t = 0 \):
\( V = \dfrac{1600}{4} = 400 \) ✓. At \( t = 10 \): \( V = \dfrac{900}{4} = 225 \) ✓. Both
conditions are satisfied, so the constants are right. Note also that the model only makes sense
for \( 0 \le t \le 40 \) — beyond that the formula gives a positive \(V\) again, which is
physically meaningless, and saying so is a legitimate interpretation point.
The minus sign, and the order of operations. Omitting the minus in
\( \dfrac{\mathrm{d}V}{\mathrm{d}t} = -k\sqrt{V} \) produces a tank that fills up, and every
subsequent answer is wrong in a way the arithmetic cannot reveal. Read whether the quantity is
increasing or decreasing before writing anything. And substitute the initial condition
immediately after integrating, while the equation is still
\( 2\sqrt{V} = -kt + c \) — rearranging first makes finding \(c\) much messier.
📝Practise
Work through these, then reveal the answer. Each question targets a different objective from the list above.
1. Solve \( \dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^{2}y \) given that \( y = 2 \) when \( x = 0 \).
Separate: \( \displaystyle\int \dfrac{1}{y}\,\mathrm{d}y = \int 3x^{2}\,\mathrm{d}x \), so \( \ln|y| = x^{3} + c \). Exponentiating, \( y = A\mathrm{e}^{x^{3}} \) where \( A = \mathrm{e}^{c} \). Applying \( y(0) = 2 \): \( 2 = A\mathrm{e}^{0} = A \). So \( y = 2\mathrm{e}^{x^{3}} \).
2. A population \(P\) grows so that \( \dfrac{\mathrm{d}P}{\mathrm{d}t} = 0.04P \). If \( P = 5000 \) at \( t = 0 \), find \(P\) when \( t = 25 \), and the time for the population to double.
Separating gives \( \ln P = 0.04t + c \), so \( P = A\mathrm{e}^{0.04t} \) with \( A = 5000 \). At \( t = 25 \): \( P = 5000\mathrm{e}^{1} = 13\,591 \). For doubling, \( 10000 = 5000\mathrm{e}^{0.04t} \), so \( \mathrm{e}^{0.04t} = 2 \) and \( 0.04t = \ln 2 \), giving \( t = \dfrac{0.6931}{0.04} = 17.3 \) years. Note the doubling time does not depend on the starting population — a defining feature of exponential growth.
3. Solve \( \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{y\cos x}{2} \) given \( y = 4 \) when \( x = 0 \).
Separate: \( \displaystyle\int \dfrac{1}{y}\,\mathrm{d}y = \int \dfrac{\cos x}{2}\,\mathrm{d}x \), giving \( \ln|y| = \tfrac12\sin x + c \). So \( y = A\mathrm{e}^{\frac12\sin x} \). At \( x = 0 \), \( \sin 0 = 0 \), so \( 4 = A \). Hence \( y = 4\mathrm{e}^{\frac12\sin x} \). Since \( \sin x \) is bounded, \(y\) oscillates between \( 4\mathrm{e}^{-0.5} = 2.43 \) and \( 4\mathrm{e}^{0.5} = 6.59 \) rather than growing without limit.
4. A body cools so that \( \dfrac{\mathrm{d}\theta}{\mathrm{d}t} = -k(\theta - 20) \). Initially \( \theta = 90 \), and after 12 minutes \( \theta = 60 \). Find \( \theta \) in terms of \(t\), and the long-term temperature.
Separate: \( \displaystyle\int \dfrac{1}{\theta - 20}\,\mathrm{d}\theta = \int -k\,\mathrm{d}t \), so \( \ln|\theta - 20| = -kt + c \) and \( \theta - 20 = A\mathrm{e}^{-kt} \). At \( t = 0 \): \( 70 = A \). At \( t = 12 \): \( 40 = 70\mathrm{e}^{-12k} \), so \( \mathrm{e}^{-12k} = \dfrac{4}{7} \), giving \( -12k = \ln(0.5714) = -0.5596 \) and \( k = 0.04664 \). Hence \( \theta = 20 + 70\mathrm{e}^{-0.0466t} \). As \( t \to \infty \) the exponential tends to zero, so \( \theta \to 20 \) — the body cools to room temperature, which is exactly what the model should predict.
5. Solve \( \dfrac{\mathrm{d}y}{\mathrm{d}t} = \dfrac{y(10 - y)}{50} \) given \( y = 2 \) when \( t = 0 \).
Separate: \( \displaystyle\int \dfrac{1}{y(10-y)}\,\mathrm{d}y = \int \dfrac{1}{50}\,\mathrm{d}t \). Partial fractions: \( \dfrac{1}{y(10-y)} \equiv \dfrac{A}{y} + \dfrac{B}{10-y} \) gives \( 1 = A(10-y) + By \); at \( y = 0 \), \( A = \tfrac{1}{10} \); at \( y = 10 \), \( B = \tfrac{1}{10} \). So \( \tfrac{1}{10}\left[\ln|y| - \ln|10-y|\right] = \tfrac{t}{50} + c \), that is \( \ln\left|\dfrac{y}{10-y}\right| = \dfrac{t}{5} + c' \). At \( t = 0 \), \( y = 2 \): \( \ln\left(\tfrac{2}{8}\right) = c' \), so \( c' = \ln 0.25 \). Hence \( \dfrac{y}{10-y} = 0.25\mathrm{e}^{t/5} \). Note the sign on the second logarithm: differentiating \( \ln|10-y| \) with respect to \(y\) brings down a \( -1 \), which is why the two logarithms subtract.
6. The rate at which a chemical dissolves is proportional to the amount \(m\) still undissolved. Write a differential equation, and explain what happens to \(m\) in the long run.
The amount undissolved decreases, so \( \dfrac{\mathrm{d}m}{\mathrm{d}t} = -km \) with \( k > 0 \). Solving gives \( m = A\mathrm{e}^{-kt} \), where \(A\) is the initial amount. As \( t \to \infty \), \( \mathrm{e}^{-kt} \to 0 \), so \( m \to 0 \): the chemical dissolves completely, but only asymptotically — the model says it never quite finishes, which is a limitation worth noting when interpreting it.
🔗Go deeper — other people’s work
These are external resources, not mine. If one stops working, tell me and
everything above it on this page still stands.
- 3Blue1Brown — Differential equations, for what these equations actually describe
- Desmos — plot a family of solution curves by varying the constant of integration
- Khan Academy — separable differential equations