Growth and Decay in First-Order Differential Equations

Introduction

students, many things in the real world change over time 📈📉. A bank balance can grow because of interest, a population can shrink because of a disease, and a hot drink can cool down after being left on a table. These changes are often modeled with growth and decay differential equations, which are some of the most important examples in First-Order Equations I.

In this lesson, you will learn how to:

explain the meaning of growth and decay in differential equations,
recognize the basic form of a growth or decay model,
solve simple models using separation of variables,
connect these ideas to real-world situations,
and understand how this lesson fits into the wider study of first-order equations.

The key idea is simple: the rate of change of a quantity is often proportional to the amount that is present. When that amount is large, the change is larger; when the amount is small, the change is smaller. That pattern appears in finance, biology, chemistry, physics, and more 🌍.

The Core Idea of Growth and Decay

A growth or decay model usually starts with a quantity $y(t)$ that changes with time $t$. The simplest model says that the rate of change is proportional to the current amount:

$$\frac{dy}{dt}=ky$$

Here, $k$ is a constant.

If $k>0$, the model describes growth.
If $k<0$, the model describes decay.

This equation is called a first-order differential equation because it involves only the first derivative $\frac{dy}{dt}$. It is also a separable equation because we can rearrange it so that all $y$ terms are on one side and all $t$ terms are on the other.

Why does this make sense? Suppose a population grows by a fixed percentage each year. If there are more people now, then there are more individuals who can reproduce, so the yearly increase is larger. If there are fewer people, the increase is smaller. The same idea works for radioactive decay, where the number of undecayed atoms decreases at a rate proportional to how many are still left.

A helpful phrase is: “change proportional to amount.” That phrase captures the entire model.

Solving the Basic Exponential Model

Let’s solve the differential equation

$$\frac{dy}{dt}=ky$$

using separation of variables.

First, divide both sides by $y$ if $y\neq 0$:

$$\frac{1}{y}\frac{dy}{dt}=k$$

Then rewrite it in differential form:

$$\frac{1}{y}\,dy=k\,dt$$

Now integrate both sides:

$$\int \frac{1}{y}\,dy=\int k\,dt$$

This gives

$$\ln|y|=kt+C$$

where $C$ is a constant of integration.

Exponentiating both sides:

$$|y|=e^{kt+C}=e^C e^{kt}$$

Since $e^C$ is just another positive constant, we write

$$y=Ae^{kt}$$

where $A$ is any nonzero constant. This is the exponential growth/decay solution.

If an initial condition is given, such as $y(0)=y_0$, then

$$y_0=Ae^{k\cdot 0}=A$$

so the solution becomes

$$y(t)=y_0e^{kt}$$

This formula is one of the most important results in this topic.

Example: Population Growth

Suppose a bacteria culture starts with $200$ bacteria and grows at a continuous rate of $8\%$ per hour. Then $k=0.08$, and the model is

$$P(t)=200e^{0.08t}$$

where $P(t)$ is the population after $t$ hours.

To find the population after $5$ hours:

$$P(5)=200e^{0.08\cdot 5}=200e^{0.4}$$

This is approximately

$$P(5)\approx 298.4$$

So the population is about $298$ bacteria. Notice how the growth is not linear. It starts relatively slowly and then increases more quickly because the amount itself keeps getting larger.

Decay Models and Half-Life

Decay models use the same equation

$$\frac{dy}{dt}=ky$$

but with $k<0$.

A common example is radioactive decay. If a substance loses mass at a rate proportional to the amount remaining, then its amount decreases exponentially:

$$y(t)=y_0e^{kt},\quad k<0$$

Because $k$ is negative, the exponent becomes more negative as time increases, so the value of $y(t)$ gets smaller and smaller but never becomes negative in the ideal model.

Example: Radioactive Decay

Suppose a material has an initial mass of $80$ grams and a decay constant $k=-0.12$ per year. Then

$$m(t)=80e^{-0.12t}$$

After $10$ years:

$$m(10)=80e^{-1.2}$$

This is approximately

$$m(10)\approx 24.1$$

So about $24.1$ grams remain.

Half-Life

The half-life is the time it takes for a quantity to fall to half its original amount. For a decay model

$$y(t)=y_0e^{kt}$$

with $k<0$, the half-life $T$ satisfies

$$\frac{y_0}{2}=y_0e^{kT}$$

Divide by $y_0$:

$$\frac{1}{2}=e^{kT}$$

Take the natural logarithm:

$$\ln\left(\frac{1}{2}\right)=kT$$

$$T=\frac{\ln(1/2)}{k}$$

Since $k<0$, this gives a positive time.

For example, if $k=-0.12$,

$$T=\frac{\ln(1/2)}{-0.12}\approx 5.78$$

So the half-life is about $5.78$ years.

Interpreting the Constant $k$ and the Meaning of Units

The constant $k$ is called the rate constant. Its sign and size matter a lot.

If $k$ is positive, the quantity increases.
If $k$ is negative, the quantity decreases.
The larger the value of $|k|$, the faster the growth or decay.

The units of $k$ depend on the units of time. For example, if $t$ is measured in years, then $k$ has units of $\text{year}^{-1}$. If $t$ is measured in hours, then $k$ has units of $\text{hour}^{-1}$.

This is important because the exponent in $e^{kt}$ must be unitless. The product $kt$ should not have leftover units.

A real-world example is money in a continuously compounding account. If the principal is $P_0$ and the annual continuous interest rate is $r$, then the balance after $t$ years is

$$P(t)=P_0e^{rt}$$

This is growth with $k=r$.

For continuous decay, the same structure appears in cooling models, chemical reactions, and drug elimination in the body 🧪.

Connection to First-Order Equations I

Growth and decay are not just one isolated formula. They are a central example of the broader topic First-Order Equations I.

Here is how they fit in:

They are first-order because the highest derivative is $\frac{dy}{dt}$.
They are often separable, which means they can be solved by integration.
Their solutions are exponential functions, which are fundamental in differential equations.
They provide a foundation for more advanced models, such as logistic growth, where growth slows down because of limited resources.

In many first-order equation problems, the goal is to find a function whose derivative matches a rule about how the function changes. In growth and decay, that rule is especially clean:

$$\frac{dy}{dt}=ky$$

This simple relationship helps students learn the process of setting up a differential equation, solving it, and interpreting the result in context.

Example: Carbon Dating Style Reasoning

If a substance has a known decay constant and a measured amount remains, we can work backward to find the original amount or the age of the sample. For instance, if

$$y(t)=y_0e^{-0.03t}$$

and $y(t)$ is known at some time $t$, then we can solve for $t$ by taking logarithms:

$$t=\frac{1}{-0.03}\ln\left(\frac{y(t)}{y_0}\right)$$

This kind of reasoning is common in science and engineering.

Common Mistakes to Avoid

When working with growth and decay, students often make a few predictable mistakes:

Mixing up the sign of $k$. Positive means growth, negative means decay.
Forgetting that the solution is exponential, not linear.
Using ordinary interest formulas when the problem says continuous growth.
Treating half-life as the same as the time to disappear completely. In exponential decay, the quantity approaches $0$ but does not reach $0$ in the ideal model.
Forgetting to match units for time and rate.

A good check is to ask: does the answer make sense? For growth, does the quantity get larger over time? For decay, does it get smaller? Reasonableness is a powerful tool ✅.

Conclusion

Growth and decay models show how a quantity can change at a rate proportional to its current amount. The differential equation

$$\frac{dy}{dt}=ky$$

captures this idea and leads to the exponential solution

$$y(t)=y_0e^{kt}$$

This lesson is a key part of First-Order Equations I because it introduces separation of variables, exponential behavior, and modeling with initial conditions. These ideas appear in many applications, from bacteria growth and radioactive decay to finance and cooling processes.

If you can recognize when a situation fits the rule “change proportional to amount,” students, you are already thinking like a differential equations student.

Study Notes

Growth and decay models use the differential equation $\frac{dy}{dt}=ky$.
If $k>0$, the model describes growth; if $k<0$, it describes decay.
The solution is $y(t)=y_0e^{kt}$ when the initial condition is $y(0)=y_0$.
These equations are first-order and separable.
Half-life is the time for a quantity to become half of its initial value.
In decay, the quantity approaches $0$ over time but does not reach $0$ in the ideal model.
The constant $k$ controls the speed of growth or decay, and its units are the inverse of time.
Real-world applications include population growth, radioactive decay, continuous compound interest, and cooling processes.
This topic is a foundation for later first-order models, especially logistic growth.