OxfordAQA International AS and A-level Mathematics: Forming and solving first-order differential equations (9660) – Revision Notes

Quick recall for section P2.8 of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2. It belongs to International A-level (A2) Unit P2 and is not examined at International AS. For full explanations use the study guide. Then try the practice questions. Also see the course hub, the printable checklist and the free 10-minute diagnostics.

Words to equations

Wording	Equation
Rate of change of y with respect to x	dy/dx
Proportional to y	dy/dx = ky
Inversely proportional to y	dy/dx = k/y
Decreasing, proportional to the amount	dm/dt = −km, k > 0
Temperature T above surroundings S, cooling	dT/dt = −k(T − S), k > 0
Rate in minus rate out	dx/dt = (in) − (out)

Rules:

• Introduce k and say what it is (k > 0 or k < 0).
• Match the sign to the words: “increases” gives positive, “decreases” gives negative.
• “Rate of change of volume” is dV/dt. If the question gives r, use dV/dt = (dV/dr)(dr/dt).

Method for separable equations

1. dy/dx = f(x) g(y)
2. (1/g(y)) dy = f(x) dx
3. integrate each side, adding a single +c
4. use the initial condition for c
5. rearrange for y if asked

Check that variables really separate. dy/dx = x + y does not, because the right side cannot be written as a product of a function of x and a function of y. dy/dx = x e^y does, and so does dy/dx = (x + 1)/y.

Key results

Equation	Solution	Notes
dy/dt = ky	y = Ae^(kt)	A = y at t = 0
dy/dt = −ky	y = Ae^(−kt)	Never reaches 0
dT/dt = −k(T − S)	T − S = Be^(−kt)	T tends to S
dy/dx = k/y	y² = 2kx + c	Reciprocal on the y side
dh/dt = −k√h	√h = √h₀ − kt/2	Reaches 0 in finite time
dx/dt = a − bx	x = a/b + Be^(−bt)	Limit a/b

Doubling time and half-life are both (ln 2)/k. They do not depend on the starting amount.

Useful integrals for this topic

∫ 1/y dy = ln y + c                    (y > 0)
∫ y^n dy = y^(n+1)/(n+1) + c           (n ≠ −1)
∫ e^(ay) dy = e^(ay)/a + c
∫ 1/(ax + b) dx = (1/a) ln(ax + b) + c
∫ 1/(py + q) dy = (1/p) ln(py + q) + c

For logistic-type equations (a product of two linear factors in y), split with partial fractions first.

Combining logarithms to tidy the answer

ln y = 3 ln x + c
write c = ln A
ln y = ln(Ax³)
y = Ax³

Take care with signs: the constant A can only be positive if you started with ln y and y > 0.

Worked reminders

Growth. A fish stock has dS/dt = 0.06S and S = 500 at t = 0. Then S = 500e^(0.06t). It doubles when e^(0.06t) = 2, so t = (ln 2)/0.06 = 11.6 (3 s.f.).

Separable. Solve dy/dx = 2x/(y + 1) with y = 1 when x = 0, for y > −1.

(y + 1) dy = 2x dx
(y + 1)²/2 = x² + c
x = 0, y = 1:  2 = c
(y + 1)² = 2x² + 4
y = −1 + √(2x² + 4)

Cooling. A body at 70 °C in surroundings at 18 °C reaches 46 °C after 10 minutes. Then T − 18 = 52e^(−kt), so 28 = 52e^(−10k), k = (1/10) ln(52/28) = 0.0619. At t = 20, T = 18 + 52e^(−20k) = 33.1 °C (3 s.f.).

Method in steps for a context problem

1. Name the quantity and the variable (for example N after t days).
2. Write the rate sentence as dN/dt = ... with a constant k.
3. Separate and integrate. One constant.
4. Use the starting condition (usually t = 0) to find the constant.
5. Use the second piece of data to find k. Keep it exact.
6. Answer the question in words, with units.

Always check the answer against the story. A decaying amount must fall. A cooling body must stay above the room temperature.

Forming from geometry: a reminder

An oil slick spreads as a circle so that its area A increases at a rate proportional to its circumference. With radius r, A = πr² and circumference = 2πr.

dA/dt = k × 2πr            (k > 0)
dA/dr = 2πr
dr/dt = 2πkr ÷ 2πr = k

The radius grows at a constant rate. The same method gives dr/dt for spheres, circles and cylinders: write the quantity formula, differentiate it with respect to the size variable, then divide.

Checking a solution by substitution

To show that y = 5 − 2e^(−4x) satisfies dy/dx = 4(5 − y):

dy/dx = 8e^(−4x)
5 − y = 2e^(−4x), so 4(5 − y) = 8e^(−4x)

Both sides match, so the equation is satisfied. A “show that” question needs this line-by-line comparison, not just a statement.

Reading the shape of a solution

• Exponential growth y = Ae^(kt): the curve rises faster as y grows. The gradient is proportional to the height.
• Exponential decay: the curve falls towards the t-axis and flattens.
• Cooling: the curve falls towards the line T = S, which is a horizontal asymptote.
• Infusion with removal: the curve rises towards the limit a/b, which is the value of x where rate in equals rate out (dx/dt = 0).

Setting the rate equal to zero finds the limiting value without solving the equation.

Must-know distinctions

• General solution has +c. Particular solution has c found.
• Growth model (exponential) has no upper limit. Cooling and infusion models have a limiting value.
• Exponential decay never reaches zero. Square-root draining does, at a finite time.
• dm/dt = −km with k > 0 and dm/dt = km with k < 0 describe the same decay. Do not mix them.
• Rate of change of V and rate of change of r are different quantities. Link them with the chain rule.

Quick self-test

1. Write an equation for “the rate of increase of N is proportional to the square of N”.
2. Solve dy/dx = 2y given y = 3 when x = 0. Find y when x = (ln 2)/2.
3. A radioactive mass decays at a rate proportional to the mass, with dm/dt = −0.05m. Find the time for the mass to fall to 40% of its starting value.
4. Solve dy/dx = 2x/(y + 1), given y = 1 when x = 0 and y > −1. Find y when x = 2.
5. Solve dy/dx = 3e^y with y = 0 when x = 0, and find y when x = 0.2.
6. Bacteria grow at a rate proportional to their number, dN/dt = 0.03N. How long for the number to become five times its starting value?
7. State the limit of x as t → ∞ for dx/dt = 6 − 0.3x with x = 0 at t = 0.
8. A drink at 70 °C is in a room at 18 °C. After 10 minutes it is 46 °C. Find its temperature after 20 minutes in total.
9. Water drains with dh/dt = −0.2√h. If h = 9 initially, find the time to empty.
10. Find the general solution of x dy/dx = 3(y + 2) for x > 0, y > −2.

Answers

1. dN/dt = kN², with k > 0.
2. y = 3e^(2x). At x = (ln 2)/2, y = 3 × 2 = 6.
3. m = m₀e^(−0.05t). 0.4 = e^(−0.05t), t = (ln 2.5)/0.05 = 18.3 (3 s.f.).
4. y = −1 + √(2x² + 4). At x = 2, y = −1 + √12 = 2.46 (3 s.f.).
5. e^(−y) dy = 3 dx, so −e^(−y) = 3x + c. At (0, 0): c = −1, so e^(−y) = 1 − 3x and y = −ln(1 − 3x). At x = 0.2, y = −ln 0.4 = 0.916 (3 s.f.).
6. e^(0.03t) = 5, t = (ln 5)/0.03 = 53.6 (3 s.f.).
7. The limit is 20.
8. T = 33.1 °C (3 s.f.), from the worked reminder.
9. 2√h = −0.2t + 6, so √h = 3 − 0.1t. Empty at t = 30.
10. ln(y + 2) = 3 ln x + c, so y = Ax³ − 2.

Where marks are usually lost

• Not separating the variables first, so y and x are integrated together.
• Missing the constant of integration or adding it after the initial condition has been applied.
• Rewriting ln y = kt + c as y = kt + c.
• Leaving k unknown when the question gives enough data to find it.
• Rounding k early, then quoting answers that do not match to 3 significant figures.
• Wrong sign of k in a decay or cooling model, giving growth instead of decay.
• Applying the initial condition to the wrong quantity (for example using T = 85 in place of T − 20 = 65).
• Not answering the question asked: the time, the amount, or the limiting value, with units.
• Treating dV/dt and dr/dt as the same thing.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2 (International AS exams from May/June 2018, A-level from May/June 2019), published by OxfordAQA. These notes cover section P2.8 of Unit P2.