An equation about a rate, not about a number
Differential equation — An equation relating a quantity to its own rate of change, whose solution is a function rather than a number.
Every equation so far has had a number as its answer. A differential equation is different: it contains a derivative, and its answer is a function. It describes how something is changing, and solving it recovers the thing itself.
This matters because rates are usually what we can observe. Nobody measures the formula for a population; they measure how fast it grows. Nobody measures the temperature curve of a cooling drink; they notice it cools faster when it is hotter. In both cases the sentence you can state is about a rate, and the function you want has to be recovered from it.
Translating the sentence into symbols
Most of the difficulty in these questions is in the first line, not the calculus. The wording follows a small number of patterns, and once you can convert them the rest is integration you already know.
The key move is that "proportional to" introduces a constant, and the sign of that constant carries the meaning: positive for growth, negative for decay. Writing dP/dt = −kP with k positive is clearer than hiding the minus inside k, and examiners prefer it.
| The sentence | The equation | Note |
|---|---|---|
| rate of increase is proportional to N | dN/dt = kN | growth; k > 0 |
| rate of decrease is proportional to N | dN/dt = −kN | decay; k still written positive |
| rate of change is inversely proportional to x | dx/dt = k/x | |
| cools at a rate proportional to the excess over 20° | dθ/dt = −k(θ − 20) | the excess, not the temperature |
| rate is proportional to the product of x and (100 − x) | dx/dt = kx(100 − x) | the logistic pattern |
| volume increases at a constant rate of 5 | dV/dt = 5 | constant, so no variable on the right |
Read what it is proportional to
Newton's law of cooling is proportional to the difference between the object and its surroundings, not to the temperature itself. Writing dθ/dt = −kθ instead of −k(θ − θ₀) is the most common modelling error in the topic, and it predicts that a drink cools all the way to absolute zero.
Separating the variables
The syllabus needs one solution method, and it works whenever the right-hand side can be written as a function of x multiplied by a function of y. Get every y with the dy and every x with the dx, then integrate both sides.
Treating dy/dx as a fraction that can be split is not strictly legitimate, but it produces the correct result and is exactly what the syllabus expects you to write.
- + c
- the arbitrary constantone is enough — combining both sides gives a single c
- general solution
- the family with c unknowna whole set of curves
- particular solution
- c found from a given conditionone specific curve
Solve dy/dx = 2xy given that y = 5 when x = 0.
- Separate: (1/y) dy = 2x dx.Every y on the left with dy, every x on the right with dx. This is possible because the right-hand side is a product of a function of x and a function of y.
- Integrate both sides: ln|y| = x² + c.The integral of 1/y is ln|y|. One constant on the right is sufficient.
- Apply the condition: ln 5 = 0 + c, so c = ln 5.Substituting the given condition immediately, before rearranging, usually keeps the algebra simpler.
- ln y − ln 5 = x², so ln(y/5) = x².Using the subtraction law for logarithms to combine before exponentiating.
- y = 5e^(x²).Exponentiating both sides. The modulus can be dropped because y = 5 is positive at the given point, so y stays positive.
y = 5e^(x²)
General and particular solutions
Integrating introduces an arbitrary constant, so a differential equation does not have one solution — it has a family of them, one for every value of c. That family is the general solution, and geometrically it is a set of curves filling the plane, none of them crossing.
A single extra piece of information — an initial condition, a boundary condition, a value at a known moment — picks out one member of that family. That is the particular solution, and it is what a modelling question always wants, because a real population has an actual size today.
The method, start to finish
- Write the differential equation from the wording, with a constant of proportionality where needed.
- Separate the variables so each side contains one letter only.
- Integrate both sides, adding a single arbitrary constant.
- Use the given condition to evaluate the constant.
- Rearrange into the form the question asks for — usually y or the modelled quantity as the subject.
- Answer the actual question, which is often a value or a limit rather than the solution itself.
Reading the answer back into the situation
The final part of a modelling question rarely asks for the solution. It asks what happens in the long run, or when some value is reached, and the marks are for interpreting rather than calculating.
The behaviour as t → ∞ is usually decided by a single exponential term. Since e^(−kt) → 0 for positive k, any term carrying that factor vanishes and whatever is left is the long-term value. Reading off that limiting value is a standard final mark.
A cup of coffee at 90 °C is left in a room at 20 °C. It cools so that dθ/dt = −k(θ − 20). After 5 minutes it is 70 °C. Find θ in terms of t, and state the temperature it approaches.
- Separate: 1/(θ − 20) dθ = −k dt.The excess over room temperature is the quantity that appears, so it is what must be isolated.
- Integrate: ln|θ − 20| = −kt + c.The left side integrates to a logarithm of the bracket, since its derivative is 1.
- At t = 0, θ = 90: ln 70 = c.The initial condition gives c directly.
- So ln((θ − 20)/70) = −kt, giving θ = 20 + 70e^(−kt).Combining the logarithms and exponentiating. The structure — room temperature plus a decaying excess — is worth recognising on sight.
- At t = 5, θ = 70: 50 = 70e^(−5k), so e^(−5k) = 5/7 and k = (1/5)ln(7/5) = 0.0673.The second condition fixes the rate constant. Note the reciprocal flip when taking the logarithm.
- θ = 20 + 70e^(−0.0673t). As t → ∞, e^(−0.0673t) → 0, so θ → 20 °C.The coffee approaches room temperature, which is the sensible answer — a good check that the model was set up correctly.
θ = 20 + 70e^(−0.0673t), approaching 20 °C
Check the answer against common sense
A solved differential equation makes a physical claim, so it can be sanity-checked. Does the population grow rather than shrink? Does the coffee approach room temperature rather than absolute zero? Does the value at t = 0 match the stated starting value? These checks cost seconds and catch sign errors that are otherwise invisible.