# 5.9: Growth and Decay

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We’ve seen a differential equation pop up several times already, and it is the most common and simplest of all differential equations:

$$G'(t) = k G(t)$$

When $$k$$ is positive, this is saying that $$G$$ is growing at a rate proportional to the value of the function at any given point. As we’ve seen, population tends to follow this rule, but several other things do as well. When $$k$$ is negative, this is saying that $$G'$$ is decreasing at a rate proportional to its value, and this is true for several things as well. These are called the growth and decay equations respectively.

And there is a simple solution to the differential equation $$G'(t) = kG(t)$$. It is $$G(t) = Ae^{k t}$$. Let’s see some examples

## Decay

A radioactive isotope decays at a rate of $$0.003$$ times its mass in grams per day. Initially, a sample contains $$40$$ grams of the isotope at $$t = 0$$.
1. How much of the isotope will there be left at $$t = 365$$?
2. At what time will there be $$1$$ gram left?

Let $$I(t)$$ be the mass of isotope in grams at time $$t$$. Thus, $$I(0) = 40$$, since we start with $$40$$ grams. Since the isotope decays at a rate of $$0.003$$ times its current mass, we see that $$I'(t) = -0.003 \cdot I(t)$$. The negative is in there because it is a decay rate — the amount of isotope is going down. We know the solution to a differential equation like this is

$$I(t) = A e^{-0.003 t}$$

Since $$I(0) = 40$$, we also have $$I(0) = A e^{-0.003(0)} = A e^0 = A$$, and hence $$A = 40$$. Our equation for the mass of the isotope is now $$I(t) = 40 e^{-0.003 t}$$

From here, we can now tell how much isotope will be left at $$t = 365$$. We plug in $$t = 365$$ and have

$$I(365) = 40 e^{-0.003 (365)} \approx \boxed{13.38 \text{ grams}}.$$

This solves part (1).

To find out when there will be $$1$$ gram left, we solve

\begin{align*} 1 & = 40 e^{-0.003t} \\ \frac{1}{40} & = e^{-0.003 t} \\ \ln \left( \frac{1}{40} \right) & = -0.003t \\ \frac{1}{-0.003} \ln \left( \frac{1}{40} \right) & = t \end{align*}

Simplifying this, we see $$t \approx \boxed{1229.62\text{days}}$$ or a little over $$3$$ years. This solves part (2).

This page titled 5.9: Growth and Decay is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Tyler Seacrest via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.