A-level · Beta

Differential equations

Connect a rate law, an initial value and an exponential solution.

\frac{dy}{dt}=ky,\quad y(0)=20

Scroll the diagram sideways to see every label.

Value at t = 3: 36.44238
Instantaneous rate dy/dt: 7.28848
Rate constant k: 0.2
Initial value: 20

A constant proportional growth rate produces an increasing exponential curve.

Predict how the curve changes, then reveal the solution to check the initial condition.

Blue: exact solution of the selected idealised rate law; green: traced value. Constant k and continuous change are assumed; time units must match k. Real growth can be limited by resources. Axes rescale with parameters; displayed values are rounded to 5 decimal places.

Predict, test and explain

Step 1 of 3

Predict

For dy/dt = 0.2y with y(0) = 20, does y increase by the same amount every unit of time?

Explore and compare

Choose the investigation above. Predict what will change before you move a control.

Predict

Try it in the simulation. Change one thing at a time.

Explain

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Hints, self-check and connections

Try a self-check

Fixed values, separate from the diagram controls.

For dy/dt = 0.2y with y(0) = 20, does y increase by the same amount every unit of time?

$\frac{dy}{dt}=0.2y,\quad y(0)=20$

Hint

Separate the variables, integrate, then use the initial condition to find the constant.

Reveal answer

The derivative is proportional to the current value: dy/dt = 0.2y. The coefficient is constant, but the rate changes with y. Equal time intervals give the same multiplier, not the same addition.

$y=20e^{0.2t}$

Build an explanation

Hint 1

Separate the variables, integrate, then use the initial condition to find the constant.

Hint 2

For y > 0, dy/y = k dt gives ln y = kt + C; exponentiating gives y = Ae^(kt). Check y = 0 separately.

Hint 3

For cooling, apply the exponential decay rule to y − T. The constant solution y = T is also valid.

Worked example

This example uses fixed values, separate from the diagram controls.

Solve dy/dt = 0.2y with y(0) = 20, and find y(5).

Since the initial value is positive, separate and integrate: ∫(1/y) dy = ∫0.2 dt, so ln y = 0.2t + C.
Exponentiate: y = Ae^(0.2t). Substituting t = 0 gives A = 20.
At t = 5, y = 20e ≈ 54.37. This prediction assumes the same proportional growth law continues.

Answer: y(t) = 20e^(0.2t), so y(5) ≈ 54.37.

Connect this idea

Watch out: The derivative is proportional to the current value: dy/dt = 0.2y. The coefficient is constant, but the rate changes with y. Equal time intervals give the same multiplier, not the same addition. These are idealised models with a constant rate coefficient k. Growth does not account for limited resources. Cooling assumes a fixed ambient temperature and a uniform object temperature; an object below ambient warms towards it. When k = 0 the initial value stays constant. The zero solution, or the ambient-temperature solution, must not be lost when separating variables.

Specification and learning route

AQA H7, H8 · Edexcel Pure 8.7, Pure 8.8

A-level. This model illustrates selected content; references are not a claim of full coverage or exam-board endorsement.

Useful starting knowledge: Exponential functions; Integration of 1/y; Initial conditions.

AQA specification · Edexcel specification