OCR A Level Mathematics A: Differential equations: separable equations and their interpretation (H240)

This study guide follows the OCR A Level Mathematics A specification (Version 3.1, August 2026, for first assessment in 2018). It covers sections 1.08k and 1.08l of the Pure Mathematics content: solving first order differential equations with separable variables, and interpreting the solution in context. The specification lists both under Stage 2, so they are A Level only and are not part of the AS qualification. Pure Mathematics content is assessed in all three written papers.

Use it with the revision notes and the practice questions. The course page is OCR A Level Mathematics and the printable list is the checklist. To find your weak spots first, try the free diagnostics. The general ideas of integration are in the Pure Mathematics study guide; this page assumes you can already integrate the standard functions and does not teach the integration techniques themselves.

What you must be able to do

Section	Outcome	Stage
1.08k	Find the general solution of a simple first order differential equation by direct integration or by separating variables, including cases that need a common factor taken out	A Level only
1.08k	Use an initial condition to find a particular solution	A Level only
1.08l	Interpret a solution in the context of a problem, including exponential growth and decay and modelling	A Level only
1.08l	Identify limitations of a solution, including links to kinematics	A Level only

What a differential equation is

A differential equation links a function to its derivatives. A first order equation contains only the first derivative, such as dy/dx or dN/dt.

Solving one means finding the function. Integrating produces an arbitrary constant, so you first get a general solution, which describes a whole family of curves. An initial condition (or boundary condition), such as “y = 3 when x = 1”, picks out one curve. That curve is the particular solution.

Direct integration

If the right-hand side contains only x, integrate straight away:

dy/dx = f(x) gives y = ∫ f(x) dx

Worked example 1. Find the general solution of dy/dx = 4x + 9/x³, then the particular solution through (1, 3).

dy/dx = 4x + 9x⁻³
y = 2x² + 9x⁻²/(−2) + c = 2x² − 9/(2x²) + c

Substitute x = 1, y = 3: 3 = 2 − 9/2 + c, so c = 11/2.

Particular solution: y = 2x² − 9/(2x²) + 11/2.

Separating variables

An equation is separable when it can be written dy/dx = f(x)g(y): a function of x multiplied by a function of y. Then move every y to one side and every x to the other:

(1/g(y)) dy = f(x) dx
∫ 1/g(y) dy = ∫ f(x) dx

Treat dy/dx as a ratio only as a shorthand. What you are really doing is integrating both sides with respect to x. Put one constant of integration on one side.

Method in steps:

1. Write the equation as dy/dx = f(x)g(y). Take out any common factor first.
2. Rearrange so all the y terms and dy are on the left and all the x terms and dx are on the right.
3. Integrate both sides and add + c on one side.
4. Use the given values of x and y to find c.
5. Rearrange to make y the subject if the question asks for it, or if it is simple to do.

Worked example 2. Solve dy/dx = x/(3y²), given y = 2 when x = 0.

3y² dy = x dx
∫ 3y² dy = ∫ x dx
y³ = x²/2 + c

When x = 0, y = 2: 8 = c.

y³ = 8 + x²/2, so y = ∛(8 + x²/2).

Worked example 3 (common factor). Solve dy/dx = 3xy − 12x, given y = 7 when x = 0.

The two terms share a factor 3x, so factorise first: dy/dx = 3x(y − 4).

∫ 1/(y − 4) dy = ∫ 3x dx
ln|y − 4| = 3x²/2 + c
y − 4 = Ae^(3x²/2)          (A = ±eᶜ)

When x = 0, y = 7: 3 = A. So y = 4 + 3e^(3x²/2).

Why the modulus and the constant A? Integrating 1/(y − 4) gives ln|y − 4|. Removing the logarithm gives |y − 4| = eᶜ × e^(3x²/2). Writing A = ±eᶜ absorbs the sign and the constant together. If the question tells you y > 4, you may write ln(y − 4) directly.

Worked example 4 (trigonometric). Solve dy/dx = y² sin x, given y = 1 when x = π/2.

∫ y⁻² dy = ∫ sin x dx
−1/y = −cos x + c

When x = π/2, y = 1: −1 = 0 + c, so c = −1. Then 1/y = cos x + 1, and y = 1/(1 + cos x).

Notice that this solution is undefined when cos x = −1, that is at x = π and every odd multiple of π. The solution is only meaningful on the interval −π < x < π that contains the starting point.

Exponential growth and decay (1.08l)

If a quantity changes at a rate proportional to its size, then dy/dt = ky for a constant k.

∫ 1/y dy = ∫ k dt
ln y = kt + c        (y > 0)
y = Ae^(kt)

When t = 0, y = A, so A is the initial value. If k > 0 you have exponential growth. If k < 0 you have exponential decay; it is usual to write dy/dt = −ky with k > 0 and read the sign from the equation.

Meaning of the words matters. “The rate of decrease is proportional to the mass” gives dm/dt = −km with k > 0. The minus sign records that m is falling.

Worked example 5 (decay). The mass m grams of a radioactive sample satisfies dm/dt = −0.02m, where t is in years, and m = 50 when t = 0. Find the half-life.

ln m = −0.02t + c,  so  m = 50e^(−0.02t)

Half-life: 25 = 50e^(−0.02t), so e^(−0.02t) = 1/2, giving t = ln 2/0.02 = 34.7 years (3 s.f.).

Interpretation: after 100 years m = 50e⁻² = 6.77 g. The model says m gets closer and closer to 0 but never reaches it.

Worked example 6 (growth and its limit). A bacterial colony has P cells at time t hours with dP/dt = kP. At t = 0, P = 500, and at t = 3, P = 1300.

P = 500e^(kt)
1300 = 500e^(3k)  →  e^(3k) = 2.6  →  k = ln 2.6/3 = 0.3185

Time to reach 5000: 5000 = 500e^(kt), so t = ln 10/k = 7.23 hours.

Limitation: at t = 8 the model gives P ≈ 6391, and the numbers keep growing without limit. A real colony runs out of food or space, so the model is only reasonable while the colony is small. Also P should be a whole number, so a decimal answer is only an approximation.

Modelling with a constant on the right

Many rates depend on a difference, so the right-hand side has the form k(y − a). The method is the same. Newton’s law of cooling says the rate of change of an object’s temperature T is proportional to the difference between T and the surroundings.

Worked example 7. A pie at 180 °C is put in a room at a steady 22 °C. Its temperature satisfies dT/dt = −k(T − 22), where t is in minutes and k > 0. After 10 minutes the temperature is 130 °C. Find T in terms of t, and the time to reach 60 °C.

∫ 1/(T − 22) dT = ∫ −k dt
ln(T − 22) = −kt + c        (T > 22)
T − 22 = Ae^(−kt)

At t = 0, T = 180: A = 158. At t = 10, T = 130: 108 = 158e^(−10k), so k = ln(158/108)/10 = 0.0380.

T = 22 + 158e^(−0.0380t).

For T = 60: 38 = 158e^(−kt), so t = ln(158/38)/k = 37.5 minutes.

Interpretation: as t → ∞, T → 22, so the pie approaches room temperature and never falls below it. Limitations: the model assumes the room stays at 22 °C and that the pie has one uniform temperature.

Kinematics

Velocity and acceleration are related by a = dv/dt, and v = ds/dt. If the acceleration depends on v, you get a separable equation. For example, resistance can make the acceleration depend on the speed.

Worked example 8. A boat’s speed v m s⁻¹ satisfies dv/dt = 6 − 2v, and v = 0 when t = 0. Find v in terms of t, describe the motion, and find the distance travelled in the first 2 seconds.

∫ 1/(6 − 2v) dv = ∫ 1 dt
−½ ln|6 − 2v| = t + c
ln|6 − 2v| = −2t + c′
6 − 2v = Ae^(−2t)

At t = 0, v = 0: A = 6, so 2v = 6 − 6e^(−2t) and v = 3 − 3e^(−2t).

Description: the boat starts from rest and speeds up, but more and more slowly. As t → ∞, e^(−2t) → 0 and v → 3, so the speed approaches 3 m s⁻¹ and never reaches it. The acceleration 6 − 2v is zero at v = 3, which is why the speed levels off.

Distance: s = ∫₀² (3 − 3e^(−2t)) dt = [3t + 1.5e^(−2t)] from 0 to 2 = 6 + 1.5e⁻⁴ − 1.5 = 4.53 m (3 s.f.).

Interpreting the solution and its limitations

Whenever a question says “describe”, “comment” or “interpret”, turn the formula into a sentence about the situation. Check these in turn:

• Initial value. Put t = 0 and say what it means.
• Long-term behaviour. What happens as t → ∞? Is there a limiting value (an asymptote), unlimited growth, or decay towards zero?
• Sign and size of the rate. Is the quantity increasing or decreasing, and is that sensible?
• Domain. Is the solution defined for all the values the context allows? Negative lengths, masses or populations signal that the model has failed.
• Assumptions. Say what the model ignores, such as a limit on resources, a changing room temperature or air resistance.

Be specific. “The model is not realistic” earns nothing; “the model gives the population growing without limit, but food supply would restrict it” is a point.

Common errors

• Two constants. Writing + c on both sides. One is enough.
• Splitting the integral wrongly. Integrating a term in y with respect to x, or leaving a y on the right-hand side.
• Forgetting the common factor. In dy/dx = 3xy − 12x you cannot separate until you factorise.
• Losing the constant when taking exponentials. ln|y − 4| = 3x²/2 + c becomes y − 4 = Ae^(3x²/2). Writing y − 4 = e^(3x²/2) + c is wrong.
• Applying the initial condition too early. Substitute it after integrating, not into the differential equation.
• Sign of k. Read the words. “Decreases at a rate proportional to” needs a minus sign.
• Dividing by zero. Dividing by g(y) assumes g(y) is not zero. A constant solution, such as y = 4 in example 3, can be lost this way. It is usually covered by allowing A = 0.
• Not answering the question asked. If it asks for the half-life or the distance, finish the job after finding the solution.

Practise on the practice questions once you can reproduce examples 2, 3, 7 and 8 without looking.

Official syllabus

OCR, A Level Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018. Sections 1.08k and 1.08l.