Differential Equations — A-Level Mathematics Pure 3 Study Guide

For: A-Level Mathematics candidates sitting Paper 3 (Pure Mathematics 3).

Covers: First-order separable differential equations, using initial conditions to find particular solutions, real-world word problem modelling for population, cooling and mixing contexts, and sign analysis of $\frac{dy}{dx}$ for equilibrium state evaluation.

You should already know: A-Level Mathematics Pure 1 (functions, calculus, trig).

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the A-Level Mathematics style for educational use. They are not reproductions of past Cambridge International examination papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official Cambridge mark schemes for grading conventions.

1. What Is Differential Equations?

A differential equation (DE) is a mathematical equation that relates an unknown function to one or more of its derivatives, describing how a quantity changes relative to an independent variable (usually $x$ for spatial relationships or $t$ for time-based contexts). For the A-Level Mathematics Paper 3 syllabus, you only need to work with first-order DEs, which include only the first derivative $\frac{dy}{dx}$ and no higher-order derivatives. DEs are one of the most widely applied mathematical tools in science, engineering and economics, so examiners frequently test them with real-world context questions worth 6-8 marks.

2. First-order separable equations — $\frac{dy}{dx}=f(x)g(y)$

A first-order DE is classified as separable if it can be rearranged to group all terms containing $y$ (including the differential $dy$) on one side of the equation, and all terms containing $x$ (including the differential $dx$) on the opposite side, with no cross-terms on either side.

The method for solving separable DEs is derived directly from the chain rule of differentiation:

1. Start with the standard form: $$\frac{dy}{dx}=f(x)g(y)$$
2. Rearrange to isolate variables, noting that $g(y)\ne 0$ for this step: $$\frac{1}{g(y)}dy=f(x)dx$$
3. Integrate both sides with respect to their respective variables: $$\int \frac{1}{g(y)}dy=\int f(x)dx$$
4. Add a single constant of integration $C$ to the right-hand side (you only need one constant, as constants from both integrals can be combined into a single unknown value).

Worked Example

Find the general solution of $\frac{dy}{dx}=3x^2{e}^{-y}$.

1. Rearrange to separate variables: $e^{y}dy=3x^{2}dx$
2. Integrate both sides: $\int e^{y}dy=\int 3x^{2}dx$
3. Compute integrals: $e^y=x^3+C$
4. Rearrange for explicit form: $y=\ln (x^3+C)$, where $C$ is a real constant and $x^3+C>0$.

Note: If you divided by a term containing $y$ to separate variables, always check if the value of $y$ that makes that term zero is a valid trivial solution to the original DE, as these are often missed by students. For example, for $\frac{dy}{dx}=2xy$, dividing by $y$ gives $\frac{1}{y}dy=2xdx$, but $y=0$ is also a valid solution that is not captured by the integrated form $\ln |y|=x^2+C$ unless you allow $C$ to be $-\infty$, so you should explicitly state $y=0$ as an additional solution.

3. Initial conditions to find the particular solution

The general solution of a DE includes an unknown constant $C$, which represents an infinite family of solution curves, each corresponding to a different starting value of the function. An initial condition (or boundary condition) is a given pair of values $(x_0,y_0)$ that the solution must satisfy, which lets you solve for $C$ to get a single unique particular solution.

Examiners almost always pair separable DE questions with initial conditions, so follow these steps to avoid mistakes:

1. Find the general solution as outlined in Section 2, keeping the constant $C$ in the equation.
2. Substitute $x=x_0$ and $y=y_0$ into the general solution.
3. Rearrange to solve for the numerical value of $C$.
4. Substitute the value of $C$ back into the general solution to get the particular solution.
5. Exam tip: Always verify your solution by substituting it back into the original DE to check that it satisfies the equation, and confirm that it matches the initial condition. This takes 30 seconds and can save you 2+ marks from arithmetic errors.

Worked Example

Find the particular solution of $\frac{dy}{dx}=\frac{4x}{y^2}$ that satisfies $y(0)=2$.

1. General solution: Separate variables, integrate: $\int y^{2}dy=\int 4xdx$ → $\frac{y^3}{3}=2x^2+C$ → $y^3=6x^2+3C$.
2. Substitute initial condition $x=0,y=2$: $2^3=0+3C$ → $8=3C$ → $C=\frac{8}{3}$.
3. Particular solution: $y^3=6x^2+8$ → $y=\sqrt[3]{6x^2+8}$.
4. Verify: $\frac{dy}{dx}=\frac{1}{3}(6x^2+8{)}^{-2/3}\ast 12x=\frac{4x}{(6x^2+8{)}^{2/3}}=\frac{4x}{y^2}$, which matches the original DE.

4. Word-problem modelling — population, cooling, mixing

The highest-mark DE questions on A-Level Mathematics Paper 3 are word problems that require you to first derive the DE from a real-world context, solve it, and interpret the result. There are three standard contexts that examiners use repeatedly:

1. Population growth/decline

The rate of change of a population $P$ is proportional to the size of the population at time $t$ (for unconstrained growth): $$\frac{dP}{dt}=kP$$ where $k>0$ for growth, $k<0$ for decline. The general solution is $P=P_0{e}^{kt}$, where $P_0$ is the initial population at $t=0$. For constrained growth with a maximum carrying capacity $K$, the logistic DE is used: $\frac{dP}{dt}=kP(1-\frac{P}{K})$, which is also separable.

2. Newton's Law of Cooling

The rate of change of temperature $T$ of an object is proportional to the difference between its temperature and the constant ambient temperature $T_a$ of its surroundings: $$\frac{dT}{dt}=-k(T-T_a)$$ where $k>0$. The negative sign ensures that if $T>T_a$ (object is hotter than surroundings), $\frac{dT}{dt}$ is negative (object cools), and if $T<T_a$, $\frac{dT}{dt}$ is positive (object warms up).

3. Mixing problems

A tank holds $V$ litres of solution, with inflow of solute concentration $c_{in}$ g/L at rate $r_{in}$ L/min, and outflow at rate $r_{out}$ L/min (usually $r_{in}=r_{out}$ so $V$ is constant for A-Level Mathematics questions). The rate of change of mass $m$ of solute is: $$\frac{dm}{dt}=\text{inflow mass rate}-\text{outflow mass rate}=c_{in}{r}_{in}-\frac{m}{V}{r}_{out}$$

Worked Example (Exam Style, 7 marks)

A patient is given a 500 mg dose of a drug, and the rate at which the drug is eliminated from the bloodstream is proportional to the mass of drug present. After 2 hours, 200 mg of the drug remains. Find the time taken for the mass of drug to fall below 50 mg, giving your answer to the nearest 10 minutes.

1. Derive DE: $\frac{dm}{dt}=-km$, $k>0$ (negative sign for elimination). (1 mark)
2. General solution: Separate, integrate: $\ln m=-kt+C$ → $m=A{e}^{-kt}$. (1 mark)
3. Initial condition $t=0,m=500$: $A=500$, so $m=500e^{-kt}$. (1 mark)
4. Use $t=2,m=200$: $200=500e^{-2k}$ → $e^{-2k}=0.4$ → $-2k=\ln 0.4$ → $k\approx 0.4581$ per hour. (2 marks)
5. Set $m=50$: $50=500e^{-0.4581t}$ → $0.1=e^{-0.4581t}$ → $t=\frac{\ln 10}{0.4581}\approx 5.02$ hours ≈ 5 hours 1 minute, so the time when mass falls below 50 mg is 5 hours 10 minutes, rounded to nearest 10 minutes. (2 marks)

5. Sign of $\frac{dy}{dx}$ and equilibrium analysis

Equilibrium solutions are constant solutions to a DE, where $\frac{dy}{dx}=0$ for all values of the independent variable. These represent steady states where the quantity described by the DE stops changing. For first-order DEs, you can analyze the sign of $\frac{dy}{dx}$ to classify equilibria as stable or unstable:

• Stable equilibrium: If $y$ is slightly above or below the equilibrium value, the sign of $\frac{dy}{dx}$ pushes $y$ back towards the equilibrium (the steady state is self-correcting).
• Unstable equilibrium: If $y$ deviates slightly from the equilibrium value, the sign of $\frac{dy}{dx}$ pushes $y$ further away from the equilibrium.

To perform equilibrium analysis:

1. Set $\frac{dy}{dx}=0$ and solve for $y$ to find all equilibrium solutions.
2. Test the sign of $\frac{dy}{dx}$ for values of $y$ just above and just below each equilibrium.
3. Classify the equilibrium based on the direction of change of $y$.

Worked Example

For the logistic growth DE $\frac{dP}{dt}=0.05P(1-\frac{P}{2000})$:

1. Find equilibria: Set $\frac{dP}{dt}=0$ → $P=0$ or $P=2000$.
2. Sign analysis:

• For $0<P<2000$: $\frac{dP}{dt}>0$, so $P$ increases towards 2000.
• For $P>2000$: $\frac{dP}{dt}<0$, so $P$ decreases towards 2000.

3. Classification: $P=2000$ (carrying capacity) is stable, $P=0$ is unstable (any small positive population will grow away from 0).

Exam tip: Examiners often ask you to sketch solution curves for different initial values, which you can do using sign analysis without solving the DE fully. Stable equilibria appear as horizontal asymptotes that solution curves approach, while unstable equilibria appear as horizontal lines that solution curves move away from.

6. Common Pitfalls (and how to avoid them)

• Pitfall 1: Forgetting to include a constant of integration, or adding separate constants to both sides of the integrated equation and not combining them. Why it happens: Rushing through integration steps. Correct move: Add a single constant $C$ to the independent variable side immediately after integrating, before rearranging for $y$.
• Pitfall 2: Dividing by a term containing $y$ without checking for trivial solutions where that term equals zero. Why it happens: Focused on rearranging variables, ignoring edge cases. Correct move: After dividing by $g(y)$, test if $g(y)=0$ satisfies the original DE, and explicitly state those values as solutions if they work.
• Pitfall 3: Dropping absolute value signs incorrectly after integrating terms like $\int \frac{1}{y}dy$. Why it happens: Assuming $y$ is always positive, even if context allows negative values. Correct move: Keep absolute values until you use initial conditions to confirm the sign of the argument, or write the exponentiated form with a $\pm$ constant that absorbs the absolute value.
• Pitfall 4: Mixing up the sign in Newton's Law of Cooling, writing $\frac{dT}{dt}=k(T-T_a)$ without the negative sign. Why it happens: Memorizing formulas without understanding context. Correct move: Check with physical intuition: if the object is hotter than its surroundings, it should cool, so $\frac{dT}{dt}$ should be negative when $T>T_a$.
• Pitfall 5: Using inconsistent units in word problems, e.g., time given in minutes but calculated in hours. Why it happens: Not reading the question carefully. Correct move: Note all units at the start of the question, keep them consistent throughout working, and check your final answer has the units requested.

7. Practice Questions (A-Level Mathematics Paper 3 Style)

Question 1 (3 marks)

Find the general solution of the differential equation $\frac{dy}{dx}=y^2\sin x$, and state any trivial solutions.

Solution:

1. Separate variables for $y\ne 0$: $\frac{1}{y^2}dy=\sin xdx$ (1 mark)
2. Integrate both sides: $\int y^{-2}dy=\int \sin xdx$ → $-\frac{1}{y}=-\cos x+C$ → $\frac{1}{y}=\cos x-C$
3. Rearrange to explicit form: $y=\frac{1}{\cos x+K}$, where $K=-C$ is an arbitrary constant (1 mark)
4. Trivial solution check: $y=0$ satisfies the original DE ($\frac{dy}{dx}=0$, $y^2\sin x=0$), so include $y=0$ as an additional solution (1 mark)

Question 2 (6 marks)

A can of soft drink at 4°C is placed in a room with constant temperature 25°C. After 10 minutes, the temperature of the drink is 10°C. Use Newton's Law of Cooling to find the temperature of the drink after 30 minutes, giving your answer to the nearest degree Celsius.

Solution:

1. Write DE: $\frac{dT}{dt}=-k(T-25)$, $k>0$ (1 mark)
2. General solution: Separate, integrate: $\ln |T-25|=-kt+C$ → $T-25=A{e}^{-kt}$ (1 mark)
3. Initial condition $t=0,T=4$: $4-25=A$ → $A=-21$, so $T=25-21e^{-kt}$ (1 mark)
4. Use $t=10,T=10$: $10=25-21e^{-10k}$ → $15=21e^{-10k}$ → $e^{-10k}=\frac{15}{21}=\frac{5}{7}$ (1 mark)
5. Calculate $T$ at $t=30$: $e^{-30k}=(e^{-10k}{)}^3=(\frac{5}{7}{)}^3=\frac{125}{343}$ → $T=25-21\ast \frac{125}{343}\approx 25-7.65=17.35$ ≈ 17°C (2 marks)

Question 3 (7 marks)

Consider the differential equation $\frac{dy}{dx}=(y+3)(y-2)$. (a) Find all equilibrium solutions. (2 marks) (b) Classify each equilibrium as stable or unstable, justifying your answer with sign analysis. (3 marks) (c) Describe the shape of the solution curve for the initial condition $y(0)=0$. (2 marks)

Solution: (a) Set $\frac{dy}{dx}=0$: $(y+3)(y-2)=0$ → equilibrium solutions $y=-3$ and $y=2$ (2 marks) (b) Sign analysis:

• For $y<-3$: (negative)(negative) = positive → $\frac{dy}{dx}>0$, $y$ increases towards $-3$
• For $-3<y<2$: (positive)(negative) = negative → $\frac{dy}{dx}<0$, $y$ decreases towards $-3$
• For $y>2$: (positive)(positive) = positive → $\frac{dy}{dx}>0$, $y$ increases away from $2$ Classification: $y=-3$ is stable, $y=2$ is unstable (3 marks) (c) Initial condition $y(0)=0$ is between $-3$ and $2$, so $\frac{dy}{dx}$ is negative. The curve will decrease from (0,0) and approach the horizontal asymptote $y=-3$, with gradient approaching 0 as $y$ approaches $-3$ (2 marks)

8. Quick Reference Cheatsheet

9. What's Next

Differential equations are a core applied mathematics topic that forms the foundation of university-level STEM courses, from physics and engineering to economics and ecology. For the A-Level Mathematics syllabus, mastering first-order DEs also prepares you for the second-order linear DEs that are tested in the Further Mathematics optional papers, and improves your ability to translate real-world context into mathematical models, a skill that is assessed across all Pure and Applied units of the qualification.

If you are struggling with any step of solving, modelling or analyzing differential equations, you can ask Ollie for personalized explanations, extra practice problems, or step-by-step walkthroughs of past A-Level Mathematics Paper 3 questions at any time on the homepage. Be sure to practice with official A-Level past papers to familiarize yourself with exam question phrasing and mark scheme grading conventions before your exam.

Aligned with the Cambridge International AS & A Level Mathematics 9709 syllabus. OwlsAi is not affiliated with Cambridge Assessment International Education.