Exponential Models with Differential Equations

When a quantity changes at a rate proportional to its own size, the differential equation is $\frac{dy}{dt} = ky$. Its solutions are exponential functions, and they model growth and decay in many settings.

The formula comes from separation of variables. For $y \ne 0$,

so $y = Ae^{kt}$ with $A = \pm e^C$. Setting $t = 0$ gives $A = y_0$. The constant solution $y = 0$ is the case $y_0 = 0$. When a free-response question asks for the solution, show these separation steps. The scoring guides give no points to an answer that only quotes $y_0e^{kt}$.

Dividing the equation by $y$ gives $k = \frac{dy/dt}{y}$, the rate of change as a fraction of the current amount. So $k$ has units of $1/\text{time}$. A value of $k = 0.03$ per year means the quantity is growing at $3\%$ of its current size per year at every instant.

For decay, the half-life $T$ is the time for the quantity to fall to half its value. Taking logarithms in $e^{kT} = \frac{1}{2}$ gives the half-life $T = \frac{\ln 2}{-k}$. It doesn't depend on the starting amount. For growth, the doubling time is $\frac{\ln 2}{k}$, from $e^{kT} = 2$.

In motion along a line, the same equation describes a velocity proportional to the position. Differentiating once more gives the acceleration.