Cambridge International AS & A Level Mathematics 9709 (2026-2027) · Paper 3: Pure Mathematics 3 · official syllabus
DIFFERENTIAL EQUATIONS LAB
Learn it by playing it
Four short missions on turning rates of change into equations, separating variables, pinning the constant and fitting growth and cooling models. In each one you read the graph, predict the rule, and adjust your answer to match the model - when a mission glows green, read why it worked, then take the next one.
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Mission 1 of 4
Pin the constant
Integrating a rate of change always leaves a constant, and one point on the curve pins it down. The curve obeys dy/dx = 2x and passes through (0, 3). The teal curve is y = x^2 + c with your constant; white is the reference curve; the teal ring is the point (0, 3). Slide the constant until the curve matches the ring.
Before you slide: integrating dy/dx = 2x gives...
Teal: your curve y = x^2 + c. White: reference curve y = x^2 + 3. Ring: (0, 3).
Your constant 1.0 against the true 3.0 - put x = 0 and y = 3 into y = x^2 + c.
Study notes
Played the lab above? These notes are the reference version of what you just felt - use them to check exact formulas and the exam wording.
Turning a rate of change into an equation 3.8
A differential equation is any statement linking a quantity to its rate of change: 'the rate is proportional to the size' becomes dN/dt = kN.
Proportional to x means k times x for some constant k; inversely proportional means k/x; 'falls' or 'decreases' puts a minus sign in front.
Worked example: a colony growing in proportion to its size obeys dN/dt = kN; coffee cooling in proportion to the gap to a 20-degree room obeys d(theta)/dt = -k(theta - 20).
The words carry the whole model: name the variables, translate the proportionality, and only then start solving.
Separation of variables 3.8
To solve a first-order separable equation, rearrange so every y sits with dy and every x with dx, then integrate each side on its own variable.
Worked example: y dy/dx = x separates to y dy = x dx, and integrating gives y^2/2 = x^2/2 + C, tidied to y^2 = x^2 + c.
One arbitrary constant is enough - two constants from two integrations always combine into a single c.
Worked example: dy/dx = 2x integrates directly to y = x^2 + c; through (0, 3) the constant is c = 3 exactly, and the curve checks out at (2, 7).
The constant and the initial condition 3.8
Integration gives a family of curves; one given point - the initial or boundary condition - picks the single member you need.
After finding a general solution, substitute the given condition to determine its arbitrary constant. You can also use a definite-integral method with limits taken from the condition.
Worked example: y^2 = x^2 + c through (0, 4) gives c = 16 exactly, so y = sqrt(x^2 + 16), and y(3) = sqrt(25) = 5 exactly.
Take the positive or negative root from the context: here the curve passes through y = 4, so the positive root is the right one.
Exponential growth and decay models 3.8
For N > 0, dN/dt = kN separates to dN/N = k dt, giving ln N = kt + c and so N = N0 e^(kt): growth for k > 0, decay for k < 0. Dividing by N excludes the equilibrium solution N = 0; check it separately if the context permits it.
Worked example: N(0) = 50 and N(5) = 50e^2 exactly give e^(5k) = e^2, so k = 2/5 = 0.40 exactly. The count 50e^2 is approximately 369.45 (2 d.p.); do not substitute that rounded count when finding an exact rate.
Newton's law of cooling shifts the same shape: theta = theta_ambient + (theta0 - theta_ambient)e^(-kt), decaying toward the surroundings, never below them.
Worked example: theta = 20 + 70e^(-kt) with theta(5) = 20 + 70/e exactly gives e^(-5k) = e^(-1), so k = 1/5 = 0.20 exactly. The temperature is approximately 45.75 (2 d.p.).
Exam craft 3.8
Form, solve, fit, interpret: write the model explicitly, show separation and integration, use the condition, then interpret the answer in context. Follow the question's instructions rather than assuming a fixed mark pattern.
Keep exact forms (ln 2, e^2, sqrt(25)) until the final line, then round as the question asks - early rounding drifts the fitted constant.
A quick substitution of the fitted constants back into the given condition is a free accuracy check: y(3) = 5 and theta(5) = 20 + 70/e (approximately 45.75) both confirm their models here.
If the answer must be interpreted ('when does the coffee reach 30 degrees?'), solve the equation for t with ln and state the units.
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