Ch 8 · 3.8 Differential Equations · A Level Mathematics 9709
Topic 8 of 9 in Pure Mathematics 2 and 3 — A Level Mathematics 9709 — 57 lessons.
Introduction
"A cup of tea cools at a rate proportional to how much hotter it is than the room." "A rumour spreads at a rate proportional to the number who have heard it and the number who have not." Statements like these describe a rate of change, so they are differential equations. Solving one gives a formula you can use to predict the future. (Syllabus section 3.8, Paper 3 only.)
Key ideas — Formulating a differential equation
Translate the words into symbols:
🧑🏫 Teacher's explanation
A differential equation is simply a mathematical sentence that describes how something changes. Instead of saying 'the car is at position 5', we say 'the car's position is changing at a rate of 60 mph'. In calculus, we represent this 'rate of change' using the derivative notation, such as dy/dx or dP/dt.
📖 Definitions
🔍 Worked example — Easy
A population P grows at a rate proportional to its size. This is modeled by dP/dt = kP. Given that when t=0, P=100, and when t=5, P=200, find the value of k to 3 decimal places.
🔍 Worked example — Standard
The rate of change of volume V of a liquid leaking from a tank is proportional to the square root of the volume. This is given by dV/dt = -k√(V). Find the general solution for V in terms of t and k.
🔍 Worked example — Exam-style
The temperature θ of a cooling object follows dθ/dt = -k(θ - 20). Given that the initial temperature is 80^° C and after 10 minutes the temperature is 50^° C, find the temperature after 20 minutes.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
(a) Find the general solution for M in terms of t and k. [3] (b) Given that the initial mass is 100g and after 2 hours the mass is 60g, find the value of k to 3 decimal places. [3] [6 marks]
Key ideas — Separating the variables
A first-order equation is separable if it can be written dy/dx = f(x)g(y). Then
🧑🏫 Teacher's explanation
Imagine you are trying to untangle two different colored threads that have been twisted together. In a differential equation, the variables x and y are often 'tangled' up in the same expression. 'Separating the variables' is the mathematical process of untangling them so that every term containing y is on one side of the equals sign and every term containing x is on the other side.
📖 Definitions
🔍 Worked example — Easy
Find the general solution to dy/dx = 3x^2y.
🔍 Worked example — Standard
Solve the differential equation dy/dx = racxy given that when x=0, y=2.
🔍 Worked example — Exam-style
A population of bacteria grows at a rate proportional to the current population P. This is modeled by dP/dt = kP. If the initial population is 100 and after 2 hours the population is 400, find an expression for P in terms of t.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — Particular solutions
An initial condition (e.g. y = 2 when x = 0) fixes c and gives the particular solution. Then rearrange into the form the question asks, e.g. "y in terms of x". With logs, ln y = x^2/2 + c becomes y = Ae^x^2/2 where A = e^c.
🧑🏫 Teacher's explanation
Imagine you are watching a population of bacteria grow. A differential equation tells you how fast the population is changing at any given moment, but it doesn't tell you how many bacteria you actually have in the petri dish right now. It only gives you the 'pattern' of growth. This pattern is called the general solution, and it always contains a constant, c, because we don't know where the growth started.
📖 Definitions
🔍 Worked example — Easy
Find the particular solution to the differential equation dy/dx = 3x^2 given that y = 5 when x = 1.
🔍 Worked example — Standard
Solve the differential equation dy/dx = y/x for x > 0, given that y = 10 when x = 2.
🔍 Worked example — Exam-style
A population of bacteria P grows at a rate proportional to the population present. This is modeled by dP/dt = kP. Given that the population doubles in 3 hours, find the particular solution for P in terms of t and the initial population P_0.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — Interpreting the solution
Questions ask what happens in context: the value after a given time, the time to reach a value, or the long-term behaviour (what happens as t → ∞, e.g. e^-kt → 0). Answer in the language of the problem ("the temperature approaches 20 °C, room temperature").
🧑🏫 Teacher's explanation
A differential equation is essentially a rule that tells you how a quantity is changing at any given moment. Instead of being told 'the value is 10', you are told 'the rate at which the value changes is proportional to the value itself'. This is like being told how fast a car is accelerating rather than being told its exact position at every second.
📖 Definitions
🔍 Worked example — Easy
A population of bacteria P grows at a rate proportional to the population present. This is modeled by dP/dt = kP. Given that at t=0, P=100, find the particular solution for P in terms of t and k.
🔍 Worked example — Standard
The temperature T of an object cools according to dT/dt = -k(T - 20), where k > 0. Given T=80 when t=0, find the expression for T in terms of t and k.
🔍 Worked example — Exam-style
A solution to a differential equation is V = 50(1 - e^-0.2t). Interpret the long-term behavior of V as t → ∞.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
(a) Find an expression for M in terms of t and k. [3] (b) Find the value of k. [2] (c) Interpret the meaning of the mass as t → ∞. [1] [6 marks]
Worked examples — Example 1: a basic separable equation
Solve dy/dx = xy with y = 2 when x = 0.
🔍 Worked example — Easy
Solve dy/dx = 3x^2 given that y = 5 when x = 0.
🔍 Worked example — Standard
Solve dy/dx = xy with y = 2 when x = 0.
🔍 Worked example — Exam-style
A population P grows at a rate proportional to its size, dP/dt = kP. Given P = 100 when t = 0 and P = 400 when t = 2, find the expression for P in terms of t.
⚠️ Common errors
Worked examples — Example 2: Newton's law of cooling
Tea at 90 °C cools in a room at 20 °C according to dθ/dt = -k(θ - 20). After 10 minutes it is at 60 °C. Find when it reaches 30 °C.
🔍 Worked example — Easy
A population of bacteria grows at a rate proportional to the current population P. This is modeled by dP/dt = kP. If the initial population is 100, find the general solution for P(t).
🔍 Worked example — Standard
The temperature θ of an object follows dθ/dt = -k(θ - 20). Given that at t=0, θ=90, find the particular solution for θ in terms of t and k.
⚠️ Common errors
Worked examples — Example 3: using partial fractions
The number x (in thousands) who have heard a rumour satisfies dx/dt = 0.2x(5 - x), with x = 1 when t = 0 (t in days). Find x in terms of t.
🔍 Worked example — Easy
Solve dy/dt = 3y, given that y = 2 when t = 0.
🔍 Worked example — Standard
Solve dx/dt = 0.2x(5 - x), given x = 1 when t = 0.
🔍 Worked example — Exam-style
A population P grows such that dP/dt = kP(10-P). If P=2 when t=0 and P=5 when t=2, find the value of k.
⚠️ Common errors
Common mistakes
Practice
Answers: (1) tfrac12y^2 = x^3 + c, c = 2: y^2 = 2x^3 + 4. (2) ln y = -ln(cos x) + c, c = 0: y = sec x. (3) e^y dy = e^2x dx ⇒ e^y = tfrac12e^2x + tfrac12, so y = ln (tfrac12e^2x + tfrac12 ). (4) P = P_0e^kt, e^5k = 2 ⇒ k = ln 2/5; triple when t = ln 3/k = 5ln 3/ln 2 = 7.92 years.