Exponential Growth and Decay: dx/dt = kx

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Exponential Growth and Decay: $\frac{dx}{dt} = kx$

Quick Reference

Try This First

A sealed flask holds $100$ grams of a substance that breaks down over time. Suppose that during any short interval the amount lost is always one tenth of whatever is present at the start of that interval.

Before reading further, fill in a short table by hand. Start at $100$ grams. For each step, subtract one tenth of the current amount.

That pattern, where the rate of decrease is set by how much is present right now, is the model. The function that matches the table exactly is exponential.

Quantity-First Framing

State the behavior in words before any symbol appears. As the amount present gets larger, the amount lost per unit of time also gets larger, in direct proportion. As the amount present gets smaller, the loss per unit of time gets smaller in the same proportion. The output (the rate of change) tracks the input (the current amount) by a fixed multiplier. That fixed multiplier is the constant $k$ in the model.

This is a covariation idea taken to its continuous limit: not just the direction of change and not just the average change over a step, but the instantaneous rate at each instant tied to the value at that instant.

Prerequisite Hub

This is a starting node for the differential equations course, so it assumes only the calculus background carried in from Calculus II. There are no in-course prerequisites that come before it.

Builds on (cross-course, from MATH162 Calculus II):

• exponential-growth-decay-model (MATH162): the calculus-side first contact with quantities whose rate is proportional to their size.
• newtons-law-cooling (MATH162): the cooling application that is a shifted version of the same model.

In-course prerequisites: none. This is a starting node for the differential equations course.

Unlocks (what this node feeds into next):

• de-introduction: the formal introduction to differential equations, which uses this equation as its first concrete object.
• de-modeling: building differential equations from a described situation, where “rate proportional to amount” is the first modeling assumption.
• separable-applications: the separation-of-variables method, for which this equation is the cleanest first application.

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        D["Intro to<br/>Differential Equations"]
        E["Modeling with<br/>Differential Equations"]
        F["Separable<br/>Applications"]
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Official Definitions

The decay equation (Logan model)

Many processes in nature are decay processes described by $x' = -rx$, where $r > 0$ is the decay rate and $x = x(t)$, the state, represents the quantity of interest. In words, the rate of change of the quantity is proportional to the amount present, and $-r$ is the proportionality constant.

Source: Logan, 3rd ed., §1.1.2 Growth-Decay Models, p. 6, Eq. (1.4).

Reading the symbols carefully helps. The notation $x'$ means $\frac{dx}{dt}$, the rate of change of the state with respect to time. The right side $-rx$ says that this rate equals a fixed number $-r$ times the current value $x$. Because $r$ is positive, the product $-rx$ is negative whenever $x$ is positive, so the quantity decreases. The general growth-decay form with constant $k$ is written

\[ \frac{dx}{dt} = kx, \]

where $k = -r$ gives decay and a positive $k$ gives growth.

The growth and decay solution family

The decay equation $x' = -rx$ has infinitely many solutions $x(t) = C e^{-rt}$, one for each constant $C$. Growth (with $r$ replaced by a positive rate constant) gives $x(t) = C e^{rt}$. An initial condition $x(0) = x_0$ selects the one solution with $C = x_0$.

Source: Logan, 3rd ed., §1.1.2, p. 7.

This statement carries two ideas. First, the equation alone does not pin down a single function; it describes a whole family indexed by the constant $C$. Second, one extra piece of information, the value at the starting time, picks out exactly one member of the family. The constant $C$ is the initial amount, supplied by the starting condition.

Worked Examples

Each example follows a predict-then-check rhythm: predict the behavior first, then compute, then compare.

Example 1: Verifying a member of the solution family

Claim to test. The function $x(t) = C e^{-rt}$ solves $x' = -rx$ for every constant $C$.

Predict. Before differentiating, ask what the derivative of an exponential of the form $C e^{-rt}$ should look like. The chain rule brings down the inner factor $-r$, so the derivative should be a constant multiple of the original function. The model demands that the multiple be exactly $-r$.

Check. Differentiate, using the chain rule on the inner function $-rt$ (whose derivative is $-r$):

\[ \frac{dx}{dt} = C \cdot e^{-rt} \cdot (-r) = -r\left(C e^{-rt}\right) = -rx. \]

Compare. The result is $-r$ times the original function, which is exactly the right side of $x' = -rx$. The prediction holds, and it holds for every value of $C$, which is why the family has infinitely many members.

Example 2: Selecting one solution with an initial condition

Setup. A radioactive sample obeys $x' = -rx$ with decay rate $r = 0.05$ per year. At the starting time the sample weighs $200$ milligrams, so $x(0) = 200$. Find the one solution that matches this start, then find the amount after $10$ years.

Predict. The constant $r$ is positive, so this is decay; the amount after $10$ years should be smaller than $200$. A rough guess: a rate near five percent per year over ten years removes a noticeable but not overwhelming fraction, so expect something a little above half of $200$.

Check. Start from the family $x(t) = C e^{-rt} = C e^{-0.05 t}$. Apply the initial condition by setting $t = 0$:

\[ x(0) = C e^{0} = C = 200. \]

So $C = 200$ and the selected solution is $x(t) = 200\, e^{-0.05 t}$. Now evaluate at $t = 10$:

\[ x(10) = 200\, e^{-0.05 \cdot 10} = 200\, e^{-0.5} \approx 200 \cdot 0.6065 = 121.3 \text{ milligrams.} \]

Compare. The amount fell from $200$ to about $121$ milligrams, which is above half and below the starting value, matching the prediction. The initial condition did its job: it turned the infinite family into a single curve.

Example 3: Recovering the rate constant from two readings

Setup. A quantity follows the growth-decay model $\frac{dx}{dt} = kx$. It is measured at the starting time as $x(0) = 500$, and four time units later as $x(4) = 410$. Find $k$, and state whether this is growth or decay.

Predict. The quantity dropped from $500$ to $410$, so it is decreasing; expect $k$ to be negative (this is decay).

Check. The solution family gives $x(t) = 500\, e^{kt}$ once the initial value is used as $C = 500$. Apply the second reading:

\[ 410 = 500\, e^{4k} \quad\Longrightarrow\quad e^{4k} = \frac{410}{500} = 0.82. \]

Take the natural logarithm of both sides:

\[ 4k = \ln(0.82) \approx -0.1985 \quad\Longrightarrow\quad k \approx -0.0496 \text{ per time unit.} \]

Compare. The constant came out negative, confirming decay as predicted. Writing the rate in the decay form $x' = -rx$, the decay rate is $r = -k \approx 0.0496$ per time unit.

Common Misconceptions

Leveled Practice

Work each problem before opening the answer. A wrong first attempt here is common and shows exactly which step needs another look.

Level 1: Read the family

Problem 1. A quantity follows $x(t) = 30\, e^{-0.2 t}$. State the value at the starting time $t = 0$, and state whether the quantity is growing or decaying.

Level 2: Match the equation to a solution

Problem 2. Write the growth-decay differential equation, in the form $\frac{dx}{dt} = kx$, whose solution is $x(t) = 12\, e^{0.3 t}$. Identify $k$ and the initial value.

Level 3: Apply an initial condition

Problem 3. A quantity obeys $\frac{dx}{dt} = kx$ with $k = 0.08$. At the starting time it equals $45$. Find the selected solution, then evaluate it at $t = 5$.

Level 4: Recover the rate constant

Problem 4. A decaying sample governed by $x' = -rx$ starts at $x(0) = 80$ and is measured as $x(6) = 50$ at time $t = 6$. Find the decay rate $r$.

Level 5: Reason about the family and justify a claim

Problem 5. Two students disagree. One says the equation $\frac{dx}{dt} = kx$ has exactly one solution. The other says it has infinitely many. Explain who is correct and what extra information settles the matter. Then convince a classmate that $x(t) = 0$ for all $t$ is itself a member of the family, and find the value of $C$ that produces it.

Mastery Checklist

Foundational:

• I can state the decay equation $x' = -rx$ and say in words what each symbol means.
• I can explain why a positive $r$ produces decreasing values.
• I can read the constant $k$ in $\frac{dx}{dt} = kx$ and decide growth versus decay from its sign.

Competent:

• I can verify by differentiating that $x(t) = C e^{kt}$ solves the equation for any $C$.
• I can use an initial condition $x(0) = x_0$ to find $C$ and select one solution.
• I can evaluate the selected solution at a later time.

Proficient:

• I can recover the rate constant from two readings by taking a logarithm.
• I can explain why the family has infinitely many members and why one initial condition selects exactly one.
• I can show why the change is multiplicative rather than additive, using a specific table or pair of values.

Mental Model

Hold the model as a feedback loop tied to the current amount. The rate is never a fixed number set in advance; it is recomputed at every instant as the fixed multiplier $k$ times whatever is present right now. When the amount is large the rate is large; when the amount shrinks the rate shrinks with it. That coupling is what the single equation $\frac{dx}{dt} = kx$ records.

The solution family $x(t) = C e^{kt}$ is the set of all curves that obey that loop, and the constant $C$ is the one piece the equation cannot supply on its own: the amount you started with. Supplying $x(0) = x_0$ is what turns a family into a single history.

This one model frames the opening of the differential equations course. It is the first equation modeled from a description, the first solved by separating variables, and the first whose direction field is read. The methods that follow are practiced first on this equation precisely because the answer is already known.

Resources

Connections

Looking back (MATH162 Calculus II):

• The exponential growth and decay model met in calculus is the same equation seen again here as the first object of a differential equations course.
• Newton’s law of cooling from calculus is a shifted version of $\frac{dx}{dt} = kx$, recovered by changing variables.

Looking ahead (within MATH347):

• The introduction to differential equations uses this equation as its first worked example.
• Modeling with differential equations begins from “rate proportional to amount”, which produces this equation directly.
• The separation-of-variables method is practiced first on this cleanest separable equation.

Real-world connections:

• Radioactive decay, where the decay rate $r$ sets the half-life used in radiometric dating.
• Continuously compounded interest, where the balance grows at a fixed fraction of itself.
• The earliest, unlimited phase of a population, before resource limits bend the curve.

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