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Bacteria multiplying in a dish, money growing at a continuous rate and a cup of tea cooling on a desk look like three different chapters. Mathematically they are one equation: the rate of change is proportional to the amount present. This post solves that equation once, shows how to get the constant k from data, works one example of each kind and lists the mistakes that cost marks.
Where is this in the syllabus?
The JEE (Main) 2026 syllabus lists "the solution of differential equation by the method of separation of variables" in Unit 9. The NCERT Class 12 chapter Differential Equations (reprint 2026-27) covers variables separable equations in section 9.4.1, with a continuous-interest problem as Example 9 and a bacteria growth problem in its exercises.
Cooling is different. Newton's law of cooling is in section 10.10 of NCERT Class 11 Thermal Properties of Matter, and the JEE (Advanced) 2026 brochure names it. The JEE (Main) 2026 physics syllabus lists heat transfer by conduction, convection and radiation but does not name the law. Check the current year's documents.
How do you solve dy/dt = ky?
Separate the variables and integrate:
dy/y = k dt
ln y = kt + C
y = e^C e^(kt). At t = 0, y = y₀, so e^C = y₀.
So y = y₀e^(kt). If k > 0 the quantity grows; if k < 0 it decays.
Two results follow directly. The time to double is ln 2 ÷ k, and for decay the time to halve is ln 2 ÷ |k|. Neither depends on y₀: a culture of 100 cells and one of a million double in the same time.
How do you find k from data?
Usually you are not given k. You are given two readings, and k comes from their ratio:
k = ln(y₂/y₁) ÷ (t₂ − t₁).
Often you do not even need the number. If a quantity grows by a factor r in time T, it grows by r² in time 2T, because e^(2kT) = (e^(kT))².
Worked example 1: a bacteria culture
Sample numbers: a culture grows from 5000 to 8000 bacteria in 3 hours, at a rate proportional to the number present. How many are there after 6 hours, and what is the doubling time?
In 3 hours the count is multiplied by 8000 ÷ 5000 = 1.6.
In 6 hours it is multiplied by 1.6² = 2.56, so the count is 5000 × 2.56 = 12 800. No logarithms needed.
k = ln 1.6 ÷ 3 ≈ 0.4700 ÷ 3 ≈ 0.157 per hour.
Doubling time = ln 2 ÷ k = 3 ln 2 ÷ ln 1.6 ≈ 3 × 0.6931 ÷ 0.4700 ≈ 4.42 hours.
Worked example 2: continuous interest, from NCERT
NCERT's Example 9: principal grows continuously at 5% a year. When does Rs 1000 double?
dP/dt = (5/100)P = P/20, so P = 1000e^(t/20).
Set 2000 = 1000e^(t/20). Then t/20 = ln 2, so t = 20 ln 2 ≈ 20 × 0.6931 ≈ 13.9 years.
Worked example 3: cooling, with a shift
NCERT's section 10.10 writes Newton's law as dT/dt proportional to −(T − Ts), where Ts is the temperature of the surroundings, and integrates it to T = Ts + C′e^(−Kt). The quantity that decays exponentially is the excess temperature y = T − Ts, not T itself. NCERT also notes that the law is a good approximation for small temperature differences.
Sample numbers: a beaker of water cools from 85 °C to 65 °C in 4 minutes in a room at 25 °C. Find its temperature after 8 minutes and the time to reach 35 °C.
The excess falls from 60 to 40 in 4 minutes, a factor of 2/3.
After 8 minutes the excess is 60 × (2/3)² ≈ 26.7, so T ≈ 51.7 °C.
To reach 35 °C the excess must fall from 60 to 10, a factor of 1/6. So t = 4 × ln 6 ÷ ln 1.5 ≈ 4 × 1.792 ÷ 0.405 ≈ 17.7 minutes.
Our post on Newton's law of cooling compares this exact method with the average-temperature shortcut that many notes teach.
Which mistakes cost marks?
Writing T = T₀e^(−kt) for cooling. The room temperature must be subtracted first.
Using a percentage as k, as the callout shows.
Losing the constant. Put in the initial condition before solving for time.
Mixing log bases. The solution uses natural logarithms, ln.
Using the model outside its range. Exponential growth is an approximation for limited times; NCERT's cooling law is for small temperature differences.
For the radioactive version of the same equation, see our post on radioactive decay and half-life.
When the exponential curve itself is hard to picture, the Fermi Simulation Lab has interactive simulations for the ideas that are easier to see than to read.
A practice routine
Derive y = y₀e^(kt) from dy/dt = ky without notes, including the constant.
Solve NCERT Exercise 9.3, question 22 (the bacteria problem), then check it with the ratio trick.
Write three cooling problems with your own sample numbers and solve them with the excess temperature.
For each answer, estimate first with doubling or halving times, then compute.
Today, find the doubling time of something that grows 25% in 5 years, and check that four doublings take four times as long.
