Newton's second law is a differential equation: $F = ma$ means force equals mass times the second derivative of position. Every radioactive decay law, every population growth model, every pendulum swing, and every electrical circuit is governed by a differential equation. While standard equations ask you to find a number that satisfies a condition, differential equations ask you to find a function whose derivatives satisfy a condition. They are the language of change — and in a universe where everything is changing, they are the most important equations in science.
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A population of bacteria doubles every hour. If $P(t)$ is the population at time $t$, write a differential equation that captures this fact. What does the equation say about the rate of change of $P$?
Core Content
A differential equation is an equation that involves a function and one or more of its derivatives.
Examples:
The order of a differential equation is the order of the highest derivative that appears.
The general solution of a first-order DE contains one arbitrary constant (like $+C$ in integration). A particular solution is found by using an initial condition to determine the constant.
Example: Solve $\frac{dy}{dx} = 2x$ with $y(0) = 3$.
General solution: y = x^2 + C
y(0) = 3: 0 + C = 3, so C = 3
Particular solution: y = x^2 + 3
To check whether $y = f(x)$ is a solution to a differential equation, substitute $y$ and its derivatives into the equation and confirm both sides are equal.
Example: Verify that $y = e^{2x}$ is a solution to $\frac{dy}{dx} = 2y$.
LHS: dy/dx = 2e^{2x}
RHS: 2y = 2e^{2x}
LHS = RHS ✓
Example: Verify that $y = A\cos x + B\sin x$ is a solution to $\frac{d^2y}{dx^2} + y = 0$.
y' = -A sin x + B cos x
y'' = -A cos x - B sin x = -y
y'' + y = -y + y = 0 ✓
Solve $\frac{dy}{dx} = 3x^2$ with $y(1) = 4$.
The particular solution.
General solution: y = ∫3x^2 dx = x^3 + C
y(1) = 4: 1^3 + C = 4
C = 3
Particular solution: y = x^3 + 3
$y = x^3 + 3$.
Solve $\frac{dy}{dx} = e^x$ with $y(0) = 2$.
Answer:
$y = \int e^x \, dx = e^x + C$. $y(0) = 2$: $e^0 + C = 2 \Rightarrow C = 1$.
Particular solution: $y = e^x + 1$.
Equation involving a function and its derivatives
Highest derivative appearing in the equation
Contains arbitrary constants
Uses initial conditions to find constants
The differential equation for bacteria doubling every hour is $\frac{dP}{dt} = kP$ where $k = \ln 2 \approx 0.693$ per hour. This says the rate of change of population is proportional to the current population — the more bacteria there are, the faster they reproduce. The solution is $P(t) = P_0 e^{kt} = P_0 \cdot 2^t$, which confirms that the population doubles every hour. Differential equations capture the underlying mechanism (proportional growth) rather than just describing the final pattern (doubling). This is their power: they explain why things change, not just how they change.
5 random questions from a replayable lesson bank — feedback shown immediately
Short Answer
8. Solve $\frac{dy}{dx} = 6x^2 - 4x + 1$ with $y(1) = 3$. Show all working. 3 MARKS
9. Verify that $y = 2e^{3x}$ satisfies $\frac{dy}{dx} = 3y$. Then find the particular solution to $\frac{dy}{dx} = 3y$ with $y(0) = 5$. 4 MARKS
10. A cup of coffee at 80°C is placed in a room at 20°C. Newton's Law of Cooling gives $\frac{dT}{dt} = -k(T - 20)$. (a) Explain what this equation means physically. (b) What happens to $\frac{dT}{dt}$ as $T$ approaches 20°C? (c) Sketch a graph of $T$ versus $t$ and explain its shape. 4 MARKS
1. (a) First order, (b) Second order, (c) First order (the power on the derivative doesn't affect the order).
2. $y = e^{-x}$, $y' = -e^{-x}$. $y' + y = -e^{-x} + e^{-x} = 0$ ✓.
3. $y = x^4 + C$. $y(0) = 5 \Rightarrow C = 5$. So $y = x^4 + 5$.
1. Each integration introduces one constant. First order needs one integration → one constant → one condition. Second order needs two integrations → two constants → two conditions.
2. $\frac{dP}{dt} = -kP$ where $k > 0$.
3. $g$ is gravitational acceleration (increases velocity downward). $-kv$ is air resistance (opposes motion, proportional to velocity). At terminal velocity, $g = kv$ so acceleration is zero.
Q8 (3 marks): $y = \int (6x^2 - 4x + 1) \, dx = 2x^3 - 2x^2 + x + C$ [1]. $y(1) = 3$: $2 - 2 + 1 + C = 3 \Rightarrow C = 2$ [1]. $y = 2x^3 - 2x^2 + x + 2$ [1].
Q9 (4 marks): $y = 2e^{3x}$, $\frac{dy}{dx} = 6e^{3x}$ [0.5]. $3y = 6e^{3x}$ [0.5]. LHS = RHS ✓ [0.5]. General solution: $y = Ae^{3x}$ [1]. $y(0) = 5$: $A = 5$ [0.5]. Particular solution: $y = 5e^{3x}$ [1].
Q10 (4 marks): (a) The rate of temperature change is proportional to the difference between coffee temperature and room temperature. The negative sign means the coffee cools [1.5]. (b) As $T \to 20$, $\frac{dT}{dt} \to 0$ — cooling slows and eventually stops [1]. (c) Graph starts at $(0, 80)$, decreases exponentially toward $T = 20$ as $t \to \infty$. It is concave up, with the steepest slope at the start [1.5].
Climb platforms by identifying DE orders, verifying solutions, and solving with initial conditions. Pool: lesson 11.
Tick when you've finished all activities and checked your answers.