Picture: sort the x's and y's onto opposite sides: A differential equation describes how fast something changes. Some can be solved by separating the variables.
Separating the variables
- If the rate can be written as a product of an x-part and a y-part, dy/dx = f(x)·g(y), you can separate the variables.
- Move everything with y (including the dy) to the left.
- Move everything with x (including the dx) to the right.
- Picture tidying a desk — all the y-things in one drawer, all the x-things in another.
Then integrate: Once they are apart, you integrate both sides. Only one constant + C is needed (it absorbs the constant from each side).
IB-style question — a draining water tank
Water drains from a tank so that its volume V (litres) after t minutes satisfies dV/dt = −2√V, with V = 400 litres at t = 0.
Find an expression for V in terms of t.
Step by step
- The right side is (a function of t = the constant −2) times (a function of V = √V), so it is separable. Put the V's on the left with dV, the t's on the right with dt.
- Integrate each side. The left is ∫ V−1/2 dV = 2√V; the right is −2t. Write one + C.
- Use the initial condition V = 400 when t = 0 to find C (√400 = 20).
- Substitute C back, then make V the subject (divide by 2, then square).
Final answer
V = (20 − t)² litres, valid until the tank empties. Check: V(0) = 400 ✓, and the tank is empty (V = 0) at t = 20 minutes — a sensible, finite draining time.
One constant, and find it straight away: Write + C just once, on the side you integrate last — not a separate constant on each side. Then apply the initial condition immediately, before rearranging, while the equation is at its simplest.
Finding C early means you rearrange clean numbers, not a trailing letter.
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Why dy/dx = ky always gives exponential change: When the rate of change is proportional to the amount present, dy/dx = ky, the solution is always exponential.
From dy/dx = ky to y = A ekx
The model
Lots of real models say the rate of change is proportional to the amount present: dy/dx = ky.
Separate
∫ (1/y) dy = ∫ k dx.
Integrate
ln|y| = kx + C.
Exponentiate
y = A ekx.
Recognising this pattern saves you re-deriving it every time.
k positive: **growth**
- Population
- Investment
k negative: **decay**
- Cooling
- Radioactive
- Drug clearance
The constant A: The constant A is just the starting value, found from the initial condition.
IB-style question — a cup of cooling coffee
A cup of coffee cools according to Newton's law of cooling: dθ/dt = −0.06(θ − 20), where θ is its temperature (°C), t is in minutes, and the room is at 20°C. The coffee starts at θ = 85°C.
(a) Solve the differential equation for θ in terms of t.
(b) Find the temperature after 10 minutes.
Step by step
- (a) Separate: the y-part is (θ − 20), so divide by it and integrate both sides.
- Integrate. The left is a 1/(linear) → logarithm; the right is −0.06t. One + C.
- Exponentiate to free θ − 20, writing A = eC, then add 20.
- Use θ = 85 at t = 0 to find A: 85 = 20 + A, so A = 65.
- (b) Substitute t = 10.
Final answer
(a) θ = 20 + 65 e−0.06t °C. (b) After 10 minutes θ ≈ 55.7°C. As t → ∞, e−0.06t → 0 so θ → 20°C — the coffee settles to room temperature, exactly as expected, which validates the model.
A second condition pins down k: Sometimes the rate constant k itself is unknown. If the question gives two data points (e.g. the value at t = 0 and at t = 5), use the first to find A and the second to find k by taking a logarithm. A GDC can solve the resulting equation for k in one step.