Differential Equations with Exponential Growth and Decay Interactive Widget

One constant k controls both halving and doubling

Whether a population explodes or a substance decays, the mathematics is identical — only the sign of k changes. The same size, ln(2)/|k|, gives you the half-life for decay (k < 0) and the doubling time for growth (k > 0). Change k and watch where the curve doubles or halves.

Exponential solution
k = +0.50
Set k. Find where y reaches 2 (or 0.5).
Growth: Doubling time = 1.39 units
Rate constant k
−1.00 0.00 +1.00
y at x = 1
1.65
Doubling time
1.39

To solve dy/dx = ky, separate variables to get dy/y = k dx, then integrate both sides. The left side becomes ln|y| and the right becomes kx + C. Exponentiating both sides gives |y| = e^(kx+C) = e^C · e^(kx). Since e^C is just another constant, we write y = Ae^(kx) where A = ±e^C represents the initial value at x = 0. This is the general solution to all exponential growth and decay problems — one elegant formula capturing everything from bacterial colonies to radioactive atoms.

Know This

The half-life for decay, t½ = ln(2)/|k| (k < 0), and the doubling time for growth, ln(2)/k (k > 0), are the same formula in size — both equal ln(2)/|k| — revealing that exponential processes have a characteristic timescale determined entirely by the rate constant.