Exponential growth and decay

Example

Let $k$ and $y_0$ be real numbers. The initial value problem $y'(t) = k\, y(t)$, $y(0) = y_0$ has the unique solution $$y(t) = y_0\, e^{k t}, \qquad t \in \mathbb{R}.$$ If $k > 0$ and $y_0 > 0$, the solution grows: this is exponential growth. If $k < 0$ and $y_0 > 0$, the solution decreases to $0$: this is exponential decay.

Remarks

For $k > 0$, the doubling time is $\ln 2 / k$: we have $y(t + \ln 2 / k) = 2\, y(t)$ for each $t$. For $k < 0$, the half-life is $\ln 2 / |k|$. Radioactive decay follows this law. Uniqueness: if $y$ is a solution, then the derivative of $y(t) e^{-k t}$ is $0$, so $y(t) e^{-kt}$ is the constant $y_0$.

Explanations

L1 human

The equation says that the rate of change is proportional to the present amount. A population with no limits, or money with continuous interest, grows like this. The larger the amount, the faster it grows, so the growth gets faster and faster.