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

These notes condense sections 1.08k and 1.08l of the OCR A Level Mathematics A specification (Version 3.1, August 2026, for first assessment in 2018). Both are Stage 2, so they are A Level only. For full explanations and worked examples, use the study guide. Test yourself with the practice questions.

The course page is OCR A Level Mathematics, the printable list is the checklist, and the free diagnostics show which topics to revise first. The integration techniques you need are in the Pure Mathematics revision notes.

Key terms

Term	Meaning
First order differential equation	Contains dy/dx (or dN/dt) but no higher derivative
General solution	Contains an arbitrary constant; a family of curves
Initial condition	A known value, such as y = 3 when x = 1
Particular solution	The one member of the family that fits the initial condition
Separable	Can be written dy/dx = f(x)g(y)

Direct integration (1.08k)

If dy/dx = f(x), then y = ∫ f(x) dx + c. Find c from the initial condition.

Worked reminder. dy/dx = 6x − 5/x² with y = 4 when x = 1. Then y = 3x² + 5/x + c. At x = 1: 4 = 3 + 5 + c, so c = −4. Particular solution: y = 3x² + 5/x − 4.

Separating variables (1.08k)

Method in steps:

1. Factorise the right-hand side if you can: dy/dx = f(x)g(y).
2. Rearrange to get the y terms with dy and the x terms with dx: (1/g(y)) dy = f(x) dx.
3. Integrate each side, adding a single constant to one of them.
4. Put the given values of x and y into the result to find that constant.
5. Make y the subject where the question asks, or where it is simple.

Form	Separated	Typical result
dy/dx = ky	∫ 1/y dy = ∫ k dx	y = Ae^(kx)
dy/dx = y² f(x)	∫ y⁻² dy = ∫ f(x) dx	−1/y = ∫ f(x) dx
dy/dx = f(x)/y	∫ y dy = ∫ f(x) dx	y²/2 = ∫ f(x) dx
dy/dx = k(y − a)	∫ 1/(y − a) dy = ∫ k dx	y = a + Ae^(kx)
dy/dx = e^(x + y)	∫ e^(−y) dy = ∫ e^x dx	e^(−y) result, then take logs

Worked reminder (common factor). dy/dx = 2xy + 6x with y = 1 when x = 0. Factorise: dy/dx = 2x(y + 3). Then ∫ 1/(y + 3) dy = ∫ 2x dx, so ln|y + 3| = x² + c. Hence y + 3 = Ae^(x²). At x = 0: 4 = A. So y = 4e^(x²) − 3.

Worked reminder (powers of y). dy/dx = 6x²y² with y = 1/2 when x = 0. Then ∫ y⁻² dy = ∫ 6x² dx gives −1/y = 2x³ + c. At x = 0: −2 = c. So 1/y = 2 − 2x³ and y = 1/(2 − 2x³).

From logarithm to exponential

ln|y − a| = g(x) + c becomes y − a = Ae^(g(x)), where A = ±eᶜ. Apply the initial condition to A.

• Never write y − a = e^(g(x)) + c.
• If the question tells you y > a, you may write ln(y − a) with no modulus.
• A constant solution y = a is covered by A = 0.

Exponential growth and decay (1.08l)

Words	Equation	Solution
Rate of increase proportional to the amount	dN/dt = kN, k > 0	N = N₀e^(kt)
Rate of decrease proportional to the amount	dN/dt = −kN, k > 0	N = N₀e^(−kt)
Rate proportional to the difference from a fixed value a	dT/dt = −k(T − a)	T = a + Ae^(−kt)

• Half-life: solve N = N₀/2, so t = ln 2/k.
• Doubling time: solve N = 2N₀, so t = ln 2/k.
• N₀ is the value at t = 0.
• Use two data points to find k: divide one equation by the other, or take logs.

Worked reminder. A population satisfies dN/dt = 0.04N, with N = 2000 at t = 0. Then N = 2000e^(0.04t). It reaches 5000 when e^(0.04t) = 2.5, so t = ln 2.5/0.04 = 22.9 years.

Cooling and the constant in the exponent

Worked reminder. A metal block at 90 °C is placed in a room at 15 °C. Its temperature obeys dT/dt = −k(T − 15). After 6 minutes it is 60 °C.

1. Separate and integrate: ln(T − 15) = −kt + c, so T = 15 + Ae^(−kt).
2. At t = 0, T = 90, so A = 75.
3. At t = 6: 45 = 75e^(−6k), so e^(−6k) = 0.6 and k = −ln 0.6/6 = 0.0851.
4. Time to reach 30 °C: 15 = 75e^(−kt), so t = ln 5/k = 18.9 minutes.

Always keep k exact (or to at least 4 significant figures) until the last line, and round only the final answer.

Setting up an equation from words

Statement	Equation
The rate of growth of x is proportional to x	dx/dt = kx
The rate of decrease of m is proportional to m²	dm/dt = −km²
The gradient at (x, y) is twice y divided by x	dy/dx = 2y/x
The rate of cooling is proportional to the excess temperature over 15	dT/dt = −k(T − 15)

Define every symbol with its units, state whether k is positive, and add the initial condition before you solve.

Kinematics links (1.08l)

• v = ds/dt and a = dv/dt.
• If a depends on v, separate in v and t. If a depends on s, the equation may need a different method, and it is not covered here.
• A speed that approaches a limit has a terminal velocity: set the acceleration to zero and solve.

Worked reminder. dv/dt = 12 − 3v with v = 2 when t = 0. Then ∫ 1/(12 − 3v) dv = ∫ dt gives −(1/3) ln|12 − 3v| = t + c. So 12 − 3v = Ae^(−3t). At t = 0: A = 6. Then v = 4 − 2e^(−3t). As t → ∞, v → 4, so the terminal velocity is 4. Check by setting 12 − 3v = 0: v = 4.

Interpreting a solution

Question to ask	What to write
What is the value at t = 0?	The starting amount, temperature or speed
What happens as t → ∞?	Limiting value, unlimited growth, or decay towards 0
Does it make sense for all t?	Check for negative or impossible values, or division by zero
What does the model ignore?	Limited resources, changing surroundings, air resistance

Limitations to quote with a reason:

• Exponential growth continues for ever, but real populations are limited by space or food.
• Exponential decay never reaches zero, but real quantities are made of whole units.
• Newton cooling assumes steady surroundings and uniform temperature.
• A formula valid for an interval may fail outside it (a denominator can become zero).

Must-know distinctions

• General or particular? General has c or A; particular has a number.
• dy/dt = ky or dy/dt = k(y − a)? The first tends to 0 (or grows without limit); the second tends to a.
• “Increases” or “decreases”? Decide the sign of the right-hand side from the words, not from k.
• Solution or check? Differentiate your answer and put it in the equation if time allows.

Quick self-test

1. Find the general solution of dy/dx = 8x³ − 2.
2. Solve dy/dx = 4y, given y = 3 when x = 0.
3. Solve dy/dx = x/y², given y = 1 when x = 0.
4. Solve dy/dx = x(y − 1), given y = 5 when x = 0.
5. Solve dy/dx = e^(2x − y), given y = 0 when x = 0. Give y in terms of x.
6. A quantity satisfies dm/dt = −0.05m. Find the time for it to halve.
7. A population satisfies dP/dt = kP, with P = 400 at t = 0 and P = 600 at t = 2. Find k.
8. A hot drink obeys dT/dt = −0.1(T − 20). State what T approaches.
9. A particle has dv/dt = 5 − v and v = 0 at t = 0. Find v in terms of t.
10. Give one limitation of exponential growth as a model for a population.

Answers

1. y = 2x⁴ − 2x + c
2. y = 3e^(4x)
3. ∫ y² dy = ∫ x dx, so y³/3 = x²/2 + c; at (0, 1), c = 1/3. y = ∛(1 + 3x²/2).
4. ln|y − 1| = x²/2 + c; y − 1 = Ae^(x²/2); A = 4. y = 1 + 4e^(x²/2).
5. e^y dy = e^(2x) dx, so e^y = e^(2x)/2 + c; at (0, 0), c = 1/2. y = ln((e^(2x) + 1)/2).
6. t = ln 2/0.05 = 13.9
7. 600 = 400e^(2k), so k = ln 1.5/2 = 0.203
8. 20 (the surroundings)
9. −ln|5 − v| = t + c; 5 − v = 5e^(−t); v = 5 − 5e^(−t)
10. It gives unlimited growth; real populations are limited by food or space.

Where marks are usually lost

• Putting a constant on both sides, or none at all.
• Missing the common factor, so the variables never separate.
• Writing y − a = e^(g(x)) + c instead of y − a = Ae^(g(x)).
• Substituting the initial condition before integrating.
• Getting the sign of k wrong for decay.
• Giving a limitation that is vague (“not realistic”) with no reason tied to the context.
• Leaving a solution with a domain problem unmentioned, such as a denominator that reaches zero.
• Stopping at the solution when the question asks for a time, a limit or a distance.

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.