MGU Mathematics Series — Calculus
Part V: Applications of Integration, Differential Equations, and Series
Slope Fields · Separable Equations · Growth · Decay · Newton's Law of Cooling
Chapter 13 of 14
MGU Mathematics Series  ·  Calculus  ·  Library Textbook Edition

Chapter 13:
Differential Equations
and Mathematical Modeling

Solutions as functions, slope fields, separation of variables, exponential growth and decay, logistic models, Newton's Law of Cooling, and the language of change across science, engineering, and finance

■ Math Level: Calculus I ■

■ 14 Sections  •  Part V of 5 ■

Follows Chapter 12: Applications of Integration  ·  Prepares students for Differential Equations and Calculus II

Chapter 13 — Front Matter

Purpose and Overview

This chapter introduces differential equations as the language of change. A differential equation is not simply a harder algebra equation — it is a different kind of mathematical statement entirely. Rather than asking "what is the value of \( x \)?", a differential equation asks "what function satisfies this relationship between a quantity and its rate of change?"

Differential equations arise naturally in almost every quantitative field. A biologist describing a growing bacterial colony, an engineer tracking the temperature of a cooling object, an economist modeling a savings account, and a physicist analyzing a falling object are all working with differential equations, whether they name them as such or not. The goal of this chapter is to make that connection explicit, to build careful reasoning about what differential equations mean, and to develop reliable techniques for solving the most accessible class: separable first-order differential equations.

Students who have worked through Chapters 9 through 12 have already encountered antiderivatives, initial value problems, and accumulation. This chapter extends and formalizes those ideas.

Opening Question

A scientist introduces a small number of bacteria into a nutrient-rich dish and measures the population every hour. The measurements suggest that the rate at which the population grows is proportional to the population itself — larger populations grow faster. No fixed formula for the population was given. Instead, only the rule governing how it changes is known. How do you find the population function? And once you have it, what does it tell you about the future behavior of the colony? This is the central kind of question a differential equation answers.

14Sections
14Objectives
22Key Terms
15+Worked Examples

Learning Objectives

  1. 1Identify a differential equation and describe what it says in plain language.
  2. 2Distinguish between general solutions and particular solutions.
  3. 3Verify that a proposed function is a solution to a differential equation.
  4. 4Sketch and interpret slope fields for first-order differential equations.
  5. 5Use initial conditions to determine a particular solution.
  6. 6Identify and solve separable differential equations.
  7. 7Model and solve exponential growth problems, including doubling time.
  8. 8Model and solve exponential decay problems, including half-life.
  9. 9Apply Newton's Law of Cooling to temperature change problems.
  10. 10Describe the logistic growth model and interpret its qualitative behavior.
  11. 11Connect differential equations to motion, terminal velocity, and position models.
  12. 12Apply differential equations to science, finance, and engineering contexts.
  13. 13Recognize and avoid common mistakes in solving and interpreting differential equations.
  14. 14Identify what techniques await in the full Differential Equations course.
Chapter 13 — Contents

Sections at a Glance

Sections 13.1 – 13.5
What Differential Equations Say: Reading, Solutions, and Slope Fields
Definitions · Solutions as Functions · Verification · Slope Fields · Initial Conditions
Section 13.1

What a Differential Equation Is

A differential equation is an equation that involves an unknown function and one or more of its derivatives. Rather than asking for a specific number, it asks for a function whose derivative relationship matches the equation.

To make this concrete, consider a simple example. Suppose you know that the rate at which a quantity \( y \) changes with respect to time \( t \) is always equal to the quantity itself:

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

This is a differential equation. It says: "find a function \( y(t) \) such that when you differentiate it, you get the function back." The function \( y = e^t \) satisfies this, and so does \( y = 2e^t \), and more generally \( y = Ce^t \) for any constant \( C \). This family of functions is the general solution.

The Order of a Differential Equation

The order of a differential equation is the highest derivative that appears. \( \frac{dy}{dt} = y \) is first-order; \( \frac{d^2y}{dt^2} = -y \) is second-order. This chapter focuses on first-order equations.

EquationOrderWhat It Describes
\( \frac{dy}{dt} = ky \)FirstGrowth or decay proportional to current value
\( \frac{dy}{dt} = k(A - y) \)FirstRate of change proportional to distance from target
\( \frac{d^2y}{dt^2} = -\omega^2 y \)SecondSimple harmonic oscillation (springs, pendulums)
\( \frac{dP}{dt} = rP\!\left(1 - \frac{P}{K}\right) \)FirstLogistic population growth with carrying capacity
Read First, Compute Second

Before solving any differential equation, read it carefully and describe in words what it says. Developing this habit builds understanding and prevents many common errors. Ask: what does the rate depend on? Is it always growing, always decaying, or does behavior change at a certain value?

Section 13.2

Solutions as Functions

The solution to a differential equation is not a number — it is a function. This is the single most important conceptual shift in this chapter. When you solve \( 2x + 3 = 11 \), you find the number \( x = 4 \). When you solve \( \frac{dy}{dt} = 2t \), you find the function \( y(t) = t^2 + C \).

General vs. Particular Solutions

The general solution is a family of functions satisfying the equation, containing one or more arbitrary constants. For a first-order ODE, it typically contains one constant \( C \).

A particular solution is a single member of that family, selected by applying an initial condition \( y(t_0) = y_0 \).

Graph Instruction — Family of Solutions

Draw several curves of the form \( y = t^2 + C \) for \( C = -2,\, 0,\, 2,\, 4 \). Each curve is a parabola shifted vertically. Together they form the general solution to \( \frac{dy}{dt} = 2t \). No single curve is the complete answer; the entire family is. An initial condition pins down which parabola is the answer to a specific problem.

Section 13.3

Verifying Solutions

One of the most important and underused skills in differential equations is verification: checking whether a given function actually satisfies a given equation. Verification requires substituting the proposed function and its derivative into the equation and confirming that both sides are equal for all values of the variable — not just at one point.

Example 13.3.1
Verifying an Exponential Solution
  • ProblemVerify that \( y = 3e^{2t} \) is a solution to \( \frac{dy}{dt} = 2y \).
  • ThinkCompute \( \frac{dy}{dt} \) from the proposed function, then compute \( 2y \), and confirm they are equal.
  • Calculate $$\frac{dy}{dt} = \frac{d}{dt}\!\left(3e^{2t}\right) = 6e^{2t}$$ $$2y = 2 \cdot 3e^{2t} = 6e^{2t}$$
  • CheckBoth expressions equal \( 6e^{2t} \) for all \( t \). ✓
Answer: \( y = 3e^{2t} \) is a solution to \( \frac{dy}{dt} = 2y \).
Example 13.3.2
A Function That Is Not a Solution
  • ProblemDetermine whether \( y = t^2 \) satisfies \( \frac{dy}{dt} = y \).
  • Calculate\( \frac{dy}{dt} = 2t \) and \( y = t^2 \).
  • CheckIs \( 2t = t^2 \) for all \( t \)? No — it holds only at \( t = 0 \) and \( t = 2 \), not as an identity.
Answer: \( y = t^2 \) is not a solution. It fails to satisfy the equation for all values of \( t \).
Example 13.3.3
Verifying a Trigonometric Solution
  • ProblemVerify that \( y = \sin(3t) \) satisfies \( \frac{d^2y}{dt^2} = -9y \).
  • Calculate $$\frac{dy}{dt} = 3\cos(3t), \qquad \frac{d^2y}{dt^2} = -9\sin(3t)$$ $$-9y = -9\sin(3t)$$
  • CheckBoth sides equal \( -9\sin(3t) \). ✓
Answer: \( y = \sin(3t) \) is a solution to this second-order equation.
Section 13.4

Slope Fields

When a differential equation cannot be solved analytically, or when you want a visual picture of all possible solutions before computing, a slope field (also called a direction field) is an invaluable tool. A first-order equation of the form \( \frac{dy}{dt} = f(t, y) \) specifies the slope of the solution curve at every point \( (t, y) \) in the plane. A slope field visualizes this by drawing a short segment at each grid point with the indicated slope.

Graph Instruction — Slope Field for \( \frac{dy}{dt} = y \)

Choose a grid with \( t \in \{-2,-1,0,1,2\} \) and \( y \in \{-2,-1,0,1,2\} \). At each point \( (t,y) \), draw a short segment with slope \( = y \). Along \( y = 0 \), all segments are horizontal. Along \( y = 1 \), all segments have slope 1. Along \( y = -1 \), all segments have slope \( -1 \). The resulting field shows exponential curves flowing upward above the axis and downward below it, consistent with the general solution \( y = Ce^t \).

Any solution curve must be tangent to the slope field at every point it passes through. Tracing a path through the field following the indicated slopes sketches a particular solution.

Equilibrium Solutions

An equilibrium solution is a constant function \( y = c \) where the slope field shows horizontal segments along the entire line — meaning \( f(t, c) = 0 \) for all \( t \). The logistic equation \( \frac{dP}{dt} = rP\!\left(1 - \frac{P}{K}\right) \) has equilibria at \( P = 0 \) (unstable) and \( P = K \) (stable).

Example 13.4.1
Interpreting a Slope Field for \( \frac{dy}{dt} = 1 - y \)
  • SituationA slope field is drawn. At each point \( (t, y) \), the segment has slope \( 1 - y \). Along \( y = 1 \), all slopes are zero. Above \( y = 1 \), slopes are negative. Below \( y = 1 \), slopes are positive.
  • ThinkWhat does this tell us about solution behavior?
  • AnalyzeSolutions starting above \( y = 1 \) decrease toward it. Solutions starting below increase toward it. The equilibrium at \( y = 1 \) is stable — all nearby solutions are attracted to it.
  • VerifyThe general solution is \( y = 1 + Ce^{-t} \). As \( t \to \infty \), \( Ce^{-t} \to 0 \), so all solutions approach \( y = 1 \). ✓
The slope field reveals qualitative behavior — stable equilibrium at \( y = 1 \) — before any algebraic computation.
Section 13.5

Initial Conditions

The general solution to a first-order ODE contains one arbitrary constant \( C \). An initial condition specifies the value of the function at one particular input — written \( y(t_0) = y_0 \) — and selects a unique particular solution from the general family. Together, a differential equation and initial condition form an initial value problem (IVP):

$$\frac{dy}{dt} = f(t, y), \qquad y(t_0) = y_0$$
Example 13.5.1
Solving an Initial Value Problem
  • ProblemSolve \( \frac{dy}{dt} = 3t^2 \) with \( y(0) = 4 \).
  • ThinkIntegrate both sides to find the general solution, then apply the initial condition.
  • Calculate $$y = \int 3t^2\,dt = t^3 + C$$ Apply \( y(0) = 4 \): \( \quad 4 = 0 + C \implies C = 4 \)
  • CheckDifferentiate \( y = t^3 + 4 \): gives \( 3t^2 \). ✓ At \( t = 0 \): \( y = 4 \). ✓
Particular solution: \( y = t^3 + 4 \)
Example 13.5.2
Initial Condition in a Growth Model
  • ProblemA bacterial colony satisfies \( \frac{dP}{dt} = 0.5P \) with initial population \( P(0) = 200 \) (in thousands).
  • CalculateGeneral solution: \( P = Ce^{0.5t} \). Apply \( P(0) = 200 \): \( C = 200 \).
  • Result\( P(t) = 200e^{0.5t} \). After 2 hours: \( P(2) = 200e^1 \approx 543 \) thousand cells.
Particular solution: \( P(t) = 200e^{0.5t} \)
Section 13.6
Separation of Variables
The Central Technique for Solving Separable First-Order Differential Equations
Section 13.6

Separable Differential Equations

The most important solution technique in this chapter is separation of variables. It applies to differential equations where the variables \( y \) and \( t \) can be arranged on opposite sides of the equation. A first-order ODE is separable if it can be written in the form:

$$\frac{dy}{dt} = g(t) \cdot h(y) \qquad \Longleftrightarrow \qquad \frac{1}{h(y)}\,dy = g(t)\,dt$$
Separation of Variables — Procedure

1. Rearrange so all \( y \) terms and \( dy \) are on one side, all \( t \) terms and \( dt \) are on the other.

2. Integrate both sides.

3. Solve for \( y \) if possible.

4. Apply the initial condition if given to find \( C \).

5. Verify the result satisfies the original equation.

Example 13.6.1
Basic Separation of Variables
  • ProblemSolve \( \frac{dy}{dt} = 3y \).
  • Separate\( \dfrac{1}{y}\,dy = 3\,dt \)
  • Integrate $$\int \frac{1}{y}\,dy = \int 3\,dt \implies \ln|y| = 3t + C_1$$
  • SolveExponentiate: \( |y| = e^{C_1}e^{3t} \). Let \( C = \pm e^{C_1} \), so \( y = Ce^{3t} \).
  • Check\( \frac{dy}{dt} = 3Ce^{3t} = 3y \). ✓
General solution: \( y = Ce^{3t} \)
Example 13.6.2
Separable with Both Variables — Curves of Integration
  • ProblemSolve \( \frac{dy}{dt} = \frac{t}{y} \).
  • Separate\( y\,dy = t\,dt \)
  • Integrate $$\frac{y^2}{2} = \frac{t^2}{2} + C_1 \implies y^2 = t^2 + C$$
  • CheckDifferentiate implicitly: \( 2y\,\frac{dy}{dt} = 2t \implies \frac{dy}{dt} = \frac{t}{y} \). ✓
General solution: \( y^2 = t^2 + C \) (a family of hyperbolas)
Example 13.6.3
Separable with Initial Condition
  • ProblemSolve \( \frac{dy}{dx} = xy \) with \( y(0) = 5 \).
  • Separate\( \dfrac{1}{y}\,dy = x\,dx \)
  • Integrate $$\ln|y| = \frac{x^2}{2} + C_1 \implies y = Ce^{x^2/2}$$
  • Apply IC\( 5 = Ce^0 = C \), so \( C = 5 \).
  • Check\( \frac{dy}{dx} = 5 \cdot x \cdot e^{x^2/2} = x \cdot y \). ✓ At \( x = 0 \): \( y = 5 \). ✓
Particular solution: \( y = 5e^{x^2/2} \)
Not All First-Order Equations Are Separable

The equation \( \frac{dy}{dt} = y + t \) cannot be separated — the right side does not factor into a product of a function of \( t \) alone times a function of \( y \) alone. Forcing separation on a non-separable equation produces incorrect results. Always check whether the equation factors cleanly before proceeding.

Sections 13.7 – 13.12
Modeling with Differential Equations
Exponential Growth · Exponential Decay · Logistic Growth · Newton's Law of Cooling · Motion · Science and Finance
Section 13.7

Exponential Growth Models

Exponential growth arises when the rate of change of a quantity is proportional to its current value. The differential equation and its solution are among the most important in all of applied mathematics:

$$\frac{dP}{dt} = kP \quad (k > 0) \qquad \Longrightarrow \qquad P(t) = P_0 e^{kt}$$

The doubling time — the time for the quantity to double — depends only on \( k \):

$$t_d = \frac{\ln 2}{k}$$
Example 13.7.1
Bacterial Population Growth
  • ProblemA colony starts at \( P_0 = 500 \) with growth constant \( k = 0.693 \) per hour. Find \( P(3) \) and the doubling time.
  • Calculate $$P(3) = 500e^{0.693 \cdot 3} = 500e^{2.079} \approx 500 \cdot 8 = 4000$$ $$t_d = \frac{\ln 2}{0.693} = \frac{0.693}{0.693} = 1 \text{ hour}$$
  • CheckAfter 1 hr: 1000. After 2 hr: 2000. After 3 hr: 4000. Three doublings from 500. ✓
  • InterpretThe population doubles every hour. After 3 hours it has doubled three times.
After 3 hours: approximately 4000 cells. Doubling time: 1 hour.
Example 13.7.2
Finding the Growth Constant from Data
  • ProblemAn investment starts at \$1000. After 5 years the balance is \$1349. Find the continuous growth rate and doubling time.
  • Represent\( 1349 = 1000e^{5k} \)
  • Solve $$e^{5k} = 1.349 \implies 5k = \ln(1.349) \approx 0.2993 \implies k \approx 0.0599 \text{ per year}$$ $$t_d = \frac{\ln 2}{0.0599} \approx 11.6 \text{ years}$$
  • InterpretThe account earns approximately 5.99% continuous annual rate. The balance doubles every 11.6 years.
\( k \approx 0.0599 \) per year; doubling time \( \approx 11.6 \) years.
Section 13.8

Exponential Decay Models

Exponential decay arises when the rate of change is negative and proportional to the current value:

$$\frac{dQ}{dt} = -kQ \quad (k > 0) \qquad \Longrightarrow \qquad Q(t) = Q_0 e^{-kt}$$

The half-life depends only on the decay constant:

$$t_{1/2} = \frac{\ln 2}{k}$$
Example 13.8.1
Radioactive Decay — Carbon-14 Dating
  • ProblemCarbon-14 has a half-life of 5730 years. A bone fragment contains 25% of its original carbon-14. How old is it?
  • Find \( k \) $$k = \frac{\ln 2}{5730} \approx 0.0001209 \text{ per year}$$
  • Solve for \( t \) $$0.25Q_0 = Q_0 e^{-kt} \implies \ln(0.25) = -kt \implies t = \frac{\ln 4}{k} \approx \frac{1.3863}{0.0001209} \approx 11{,}460 \text{ years}$$
  • CheckTwo half-lives: \( 2 \times 5730 = 11{,}460 \). 25% is exactly two halvings from 100%. ✓
  • InterpretThe fragment is approximately 11,460 years old.
Age ≈ 11,460 years (two half-lives)
Example 13.8.2
Drug Concentration in the Bloodstream
  • ProblemInitial concentration: 80 mg/L. After 4 hours: 60 mg/L. Find the concentration after 10 hours.
  • Find \( k \) $$60 = 80e^{-4k} \implies e^{-4k} = 0.75 \implies k = \frac{-\ln(0.75)}{4} \approx 0.0719 \text{ per hour}$$
  • Evaluate $$C(10) = 80e^{-0.0719 \cdot 10} \approx 80 \cdot 0.487 \approx 38.9 \text{ mg/L}$$
  • InterpretAfter 10 hours, approximately 38.9 mg/L remains. This model is used in pharmacokinetics to determine dosing intervals.
Concentration after 10 hours ≈ 38.9 mg/L
Section 13.9

Logistic Growth as an Introduction

Exponential growth assumes unlimited resources. The logistic growth model accounts for the real constraint that no population can grow without bound. As a population approaches the environment's capacity, growth slows and eventually stops.

$$\frac{dP}{dt} = rP\!\left(1 - \frac{P}{K}\right)$$

Here \( r \) is the intrinsic growth rate and \( K \) is the carrying capacity. When \( P \ll K \), the factor \( (1 - P/K) \approx 1 \) and growth is nearly exponential. As \( P \to K \), the factor approaches 0 and growth stops.

Graph Instruction — S-Shaped Logistic Curve

Draw an S-shaped (sigmoidal) curve beginning with slow growth near \( P = 0 \), steepening to maximum rate at \( P = K/2 \), then leveling off as \( P \to K \). Draw a dashed horizontal line at \( P = K \) (carrying capacity) and mark the inflection point at \( P = K/2 \) where the growth rate is maximum. Compare to an exponential curve that would continue rising without bound.

Example 13.9.1
Reading a Logistic Model
  • ProblemA fish population satisfies the logistic model with \( r = 0.4 \) and \( K = 5000 \). Current population: 1000 fish. What is the current growth rate?
  • Calculate $$\frac{dP}{dt} = 0.4 \cdot 1000 \cdot \left(1 - \frac{1000}{5000}\right) = 400 \cdot 0.8 = 320 \text{ fish per unit time}$$
  • CompareAt \( P = 4500 \): \( \frac{dP}{dt} = 0.4 \cdot 4500 \cdot (1 - 0.9) = 180 \) fish per unit time — much slower.
  • InterpretGrowth slows as the population approaches 5000. The rate at 4500 fish is barely more than half of the rate at 1000 fish.
Current growth rate: 320 fish per unit time
MGU Library Connection
  • MGU Differential Equations course — full derivation and solution of the logistic equation
  • MGU Biology and Ecology modeling pages — logistic model, Lotka-Volterra predator-prey systems
  • MGU Data Science Foundations — logistic regression and its relation to the logistic function
Section 13.10

Newton's Law of Cooling as an Introduction

Newton's Law of Cooling states that the rate of change of an object's temperature is proportional to the difference between its current temperature and the surrounding (ambient) temperature \( T_s \):

$$\frac{dT}{dt} = -k(T - T_s), \quad k > 0$$

The substitution \( u = T - T_s \) converts this into the exponential decay equation \( \frac{du}{dt} = -ku \), giving the general solution:

$$T(t) = T_s + (T_0 - T_s)e^{-kt}$$
Example 13.10.1
A Cooling Coffee Cup
  • ProblemCoffee starts at 90°C in a 20°C room. After 5 minutes: 74°C. Find \( k \) and the temperature after 15 minutes.
  • Model\( T(t) = 20 + 70e^{-kt} \)
  • Find \( k \) $$74 = 20 + 70e^{-5k} \implies 54 = 70e^{-5k} \implies k = \frac{\ln(70/54)}{5} \approx 0.0518 \text{ per minute}$$
  • Evaluate $$T(15) = 20 + 70e^{-0.0518 \cdot 15} \approx 20 + 70 \cdot 0.460 \approx 52.2°C$$
  • InterpretAfter 15 minutes the coffee is about 52°C. It approaches room temperature (20°C) asymptotically but never drops below it.
After 15 minutes: ≈ 52.2°C
Example 13.10.2
A Forensic Application — Time of Death
  • ProblemA body is found in a 20°C room. Current temperature: 30°C. One hour later: 28°C. Normal body temperature: 37°C. Estimate time of death.
  • Model\( T(t) = 20 + 17e^{-kt} \), where \( t = 0 \) is time of death.
  • Find \( k \) Let \( t_1 \) = time of discovery. From \( T(t_1) = 30 \): \( e^{-kt_1} = \frac{10}{17} \). From \( T(t_1+1) = 28 \): $$8 = 10e^{-k} \implies e^{-k} = 0.8 \implies k = -\ln(0.8) \approx 0.2231 \text{ per hour}$$
  • Find \( t_1 \) $$e^{-0.2231 t_1} = \frac{10}{17} \implies t_1 = \frac{\ln(17/10)}{0.2231} \approx \frac{0.5306}{0.2231} \approx 2.38 \text{ hours}$$
  • InterpretApproximately 2.38 hours elapsed between death and discovery. Note: Newton's Law is a simplified model; actual forensic estimates carry substantial uncertainty.
Time of death ≈ 2.38 hours before discovery
Section 13.11

Motion Models

Differential equations provide the natural language for motion. The relationships between position, velocity, and acceleration are rate relationships:

$$v(t) = s'(t) = \frac{ds}{dt}, \qquad a(t) = v'(t) = \frac{dv}{dt}$$

In free fall with air resistance proportional to velocity, Newton's second law gives:

$$m\frac{dv}{dt} = mg - bv$$

Terminal velocity occurs when \( \frac{dv}{dt} = 0 \), meaning drag equals gravity: \( v_{\text{term}} = \frac{mg}{b} \).

Example 13.11.1
Terminal Velocity and Velocity Model
  • ProblemA car starts from rest. Its velocity satisfies \( \frac{dv}{dt} = 4 - 0.2v \), \( v(0) = 0 \). Find the terminal velocity and \( v(t) \).
  • Terminal velocity $$0 = 4 - 0.2v_{\text{term}} \implies v_{\text{term}} = 20 \text{ m/s}$$
  • Solve Separate: \( \dfrac{dv}{4 - 0.2v} = dt \). Integrate (substitution \( u = 4 - 0.2v \)): $$-5\ln|4 - 0.2v| = t + C_1 \implies v(t) = 20\!\left(1 - e^{-t/5}\right)$$
  • CheckAt \( t = 0 \): \( v = 0 \). ✓ As \( t \to \infty \): \( v \to 20 \). ✓ Differentiate: \( \frac{dv}{dt} = 4e^{-t/5} = 4 - 0.2v \). ✓
  • InterpretThe car accelerates from rest and asymptotically approaches 20 m/s. Acceleration decreases as velocity increases.
\( v(t) = 20\!\left(1 - e^{-t/5}\right) \); terminal velocity = 20 m/s
Section 13.12

Differential Equations in Science and Finance

Continuous Compounding

If money \( A(t) \) grows at continuous rate \( r \), then \( \frac{dA}{dt} = rA \), giving \( A(t) = A_0 e^{rt} \). This is the mathematical foundation of continuous compounding.

Example 13.12.1
Continuous Compound Interest
  • ProblemAn account begins with \$5000 at 6% continuous annual interest. Balance after 10 years?
  • Calculate $$A(10) = 5000e^{0.06 \cdot 10} = 5000e^{0.6} \approx 5000 \cdot 1.8221 \approx \$9{,}111$$
  • InterpretThe balance nearly doubles in 10 years due to continuous compounding.
Balance after 10 years ≈ \$9,111

Mixing Tank Problems

Example 13.12.2
Setting Up a Mixing Problem
  • SituationA 100-gallon tank holds 50 lbs of dissolved salt. Fresh water enters at 3 gal/min; the well-mixed solution exits at 3 gal/min. Let \( A(t) \) = pounds of salt at time \( t \).
  • ThinkRate of change = rate in − rate out. Fresh water brings no salt in. Salt flows out at the tank's current concentration times the flow rate.
  • Represent $$\frac{dA}{dt} = 0 - \frac{A}{100} \cdot 3 = -\frac{3A}{100}$$
  • Solve This is exponential decay: \( A(t) = 50e^{-0.03t} \).
  • InterpretThe salt washes out continuously. As \( t \to \infty \), \( A \to 0 \): the tank eventually holds pure water.
\( A(t) = 50e^{-0.03t} \) pounds of salt
Sections 13.13 – 13.14
Common Mistakes and the Road Ahead
Error Analysis · Preparing for the Full Differential Equations Course
Section 13.13

Common Differential Equation Mistakes

Mistake 1 — Forgetting the Constant of Integration
Solve \( \frac{dy}{dt} = 2t \): answer is \( y = t^2 \) Correct: y = t² + C. Without the constant, only the single solution through the origin is captured. Always include +C and determine it from an initial condition if one is given.
Mistake 2 — Forcing Separation on a Non-Separable Equation
Separate \( \frac{dy}{dt} = y + t \) as \( \frac{dy}{y} = (1 + t/y)\,dt \) Correct: \( y + t \) does not factor into g(t)·h(y). This equation is not separable. First-order linear methods (integrating factor) are needed — covered in the Differential Equations course.
Mistake 3 — Applying Growth Formula with Wrong Sign
Exponential decay: \( Q(t) = Q_0 e^{kt} \) with \( k > 0 \) Correct: For decay, the exponent must be negative: Q(t) = Q₀e^(−kt) with k > 0. The positive sign gives growth; the negative sign gives decay.
Mistake 4 — Forgetting to Apply the Initial Condition
Report general solution \( y = Ce^{3t} \) when \( y(0) = 4 \) was given Correct: Apply y(0) = 4 to get C = 4, then report the particular solution y = 4e^(3t). An unapplied initial condition leaves the answer incomplete.
Mistake 5 — Confusing Decay Constant and Half-Life
"Half-life is 10 years, so k = 10 per year" Correct: k = ln 2 / t₁/₂ = ln 2 / 10 ≈ 0.0693 per year. The decay constant and the half-life have an inverse relationship — they are not equal.
Mistake 6 — Misreading Newton's Law of Cooling
\( \frac{dT}{dt} = -kT \) for cooling toward \( T_s = 25°C \) Correct: \( \frac{dT}{dt} = -k(T - T_s) \). The first equation drives T toward 0; the correct equation drives T toward the ambient temperature T_s. The difference matters enormously in applications.
Mistake 7 — Applying Exponential Growth Near Carrying Capacity
Using \( P(t) = P_0 e^{rt} \) when \( P \approx K \) Correct: Exponential growth approximates logistic growth only when P ≪ K. Near the carrying capacity, the logistic brake is strong and exponential growth dramatically overestimates the population.
Mistake 8 — Omitting Units on the Rate Constant
"The growth constant is k ≈ 0.0693" (no units given) Correct: Always state the units — k ≈ 0.0693 per hour, or 0.0693 year⁻¹, etc. A rate constant without units is meaningless and will produce wrong answers if units are mixed.
Mistake 9 — Skipping the Verification Step
Finding y and not checking whether it satisfies the original equation Correct: After finding a solution, differentiate it and substitute back into the equation to confirm both sides match identically. Algebraic slips during separation can produce functions that do not satisfy the equation.
Mistake 10 — Treating General and Particular Solutions as Interchangeable
"The solution is Ce^(3t)" when a specific initial condition was provided Correct: When an initial condition is given, the problem asks for a particular solution, not the general family. The general solution is mathematically complete but contextually incomplete until C is determined.
Section 13.14

Preparing for Advanced Differential Equations

This chapter has developed the most accessible tools: reading equations, verifying solutions, drawing slope fields, applying initial conditions, and solving separable equations. These are powerful tools that account for many of the most important models in applied mathematics. The world of differential equations, however, is much larger.

TopicKey IdeaWhere Covered
First-order linear equations\( \frac{dy}{dt} + P(t)y = Q(t) \) — solved with an integrating factorMGU Differential Equations, Chapter 2
Second-order linear equations\( y'' + by = 0 \) — solutions involve \(\sin\) and \(\cos\); require two initial conditionsMGU Differential Equations, Chapters 3–4
Systems of ODEsPredator-prey, electrical circuits, multi-compartment pharmacokineticsMGU Differential Equations, Chapter 7
Euler's MethodNumerical approximation when no closed-form solution existsMGU Scientific Computing, Chapter 5
Partial differential equationsHeat equation, wave equation, Laplace equation — functions of multiple variablesMGU Advanced Engineering Mathematics
MGU Library Connection
  • MGU Differential Equations — integrating factors, second-order linear equations, series solutions, Laplace transforms, systems, numerical methods
  • MGU Physics — Newton's second law, harmonic oscillation, circuit equations
  • MGU Electrical Engineering — RC, RL, and RLC circuits described by first- and second-order ODEs
  • MGU Scientific Computing — Euler's method, Runge-Kutta methods, stiff equations
  • MGU Machine Learning Mathematics — gradient flow, continuous-time neural network models, neural ODEs
Chapter 13 — Back Matter

Chapter Summary

Key Terms Review

differential equation
An equation involving an unknown function and one or more of its derivatives.
ordinary differential equation
A differential equation involving a function of one variable and its derivatives with respect to that variable.
order
The order of a differential equation equals the order of the highest derivative that appears in it.
general solution
The family of all functions satisfying a differential equation, containing one or more arbitrary constants.
particular solution
A specific member of the general solution family, determined by applying an initial condition.
initial condition
A specified value of the solution function at a particular input: \( y(t_0) = y_0 \).
initial value problem
A differential equation together with one or more initial conditions, which determine a unique particular solution.
slope field
A visual representation showing the slope specified by a differential equation at each point in the plane. Also called a direction field.
equilibrium solution
A constant function \( y = c \) satisfying the differential equation, occurring where the rate of change is zero.
separable equation
A first-order ODE that can be written as \( \frac{dy}{dt} = g(t) \cdot h(y) \), enabling separation of variables.
separation of variables
A solution method: place all \( y \) terms on one side, all \( t \) terms on the other, then integrate both sides.
exponential growth
Growth governed by \( \frac{dP}{dt} = kP \) with \( k > 0 \), producing \( P(t) = P_0 e^{kt} \).
exponential decay
Decay governed by \( \frac{dQ}{dt} = -kQ \) with \( k > 0 \), producing \( Q(t) = Q_0 e^{-kt} \).
doubling time
The time for an exponentially growing quantity to double: \( t_d = \frac{\ln 2}{k} \).
half-life
The time for an exponentially decaying quantity to halve: \( t_{1/2} = \frac{\ln 2}{k} \).
logistic growth
Growth governed by \( \frac{dP}{dt} = rP(1 - P/K) \), producing S-shaped curves that level off at the carrying capacity.
carrying capacity
The maximum sustainable population \( K \) in the logistic model. A stable equilibrium solution.
Newton's Law of Cooling
The rate of temperature change is proportional to the difference between the object's temperature and the ambient temperature: \( \frac{dT}{dt} = -k(T - T_s) \).
terminal velocity
The limiting speed reached when air resistance exactly balances gravity, giving zero net acceleration: \( v_{\text{term}} = mg/b \).
growth constant
The proportionality constant \( k > 0 \) in an exponential growth model, with units of (time)⁻¹.
decay constant
The proportionality constant \( k > 0 \) in an exponential decay model, related to half-life by \( k = \ln 2 / t_{1/2} \).
verification
Confirming a proposed solution by substituting it and its derivative into the original equation and checking both sides are equal for all values.

Solution Verification Practice

For each differential equation and proposed solution, determine whether the function is a solution. Show all derivative computations.

  1. Equation: \( \frac{dy}{dt} = 4y \). Proposed: \( y = e^{4t} \).
  2. Equation: \( \frac{dy}{dt} = 4y \). Proposed: \( y = 2e^{4t} \).
  3. Equation: \( \frac{dy}{dx} = 3x^2 \). Proposed: \( y = x^3 + 7 \).
  4. Equation: \( \frac{d^2y}{dt^2} + y = 0 \). Proposed: \( y = \cos t \).
  5. Equation: \( \frac{d^2y}{dt^2} + y = 0 \). Proposed: \( y = e^t \).
  6. Equation: \( \frac{dy}{dt} = 2 - y \). Proposed: \( y = 2 - 3e^{-t} \).
  7. Equation: \( \frac{dy}{dt} = y^2 \). Proposed: \( y = \frac{1}{1-t} \).
  8. Equation: \( \frac{dT}{dt} = -0.1(T - 25) \). Proposed: \( T = 25 + 15e^{-0.1t} \). What is \( T(0) \)?

Slope Field Practice

  1. For \( \frac{dy}{dt} = -y \): (a) What is the slope at \( (0, 2) \)? (b) What is the slope at \( (3, -1) \)? (c) Along which line are all slopes zero? (d) What is the equilibrium solution? (e) Describe the behavior of a solution starting at \( y(0) = 3 \).
  2. For \( \frac{dy}{dt} = 1 - y \): (a) Where are slope segments horizontal? (b) Where are slopes positive? (c) Where are slopes negative? (d) Describe the long-term behavior starting at \( y(0) = 0 \) and starting at \( y(0) = 4 \).
  3. Sketch (or describe in words) the slope field for \( \frac{dy}{dt} = y(1 - y) \) on \( t \in [0,4] \), \( y \in [-0.5, 2] \). Identify both equilibrium solutions and describe the behavior of solutions between and outside the equilibria.
  4. A slope field shows horizontal segments along \( y = 0 \) and \( y = 3 \). Solutions between these lines form S-shaped curves increasing toward \( y = 3 \). Which equation fits? Explain. (a) \( \frac{dy}{dt} = y \) (b) \( \frac{dy}{dt} = y(3-y) \) (c) \( \frac{dy}{dt} = 3 - y \) (d) \( \frac{dy}{dt} = y - 3 \).

Separable Equation Practice

Find the general solution. Apply the initial condition if given.

Part AGeneral Solutions
  1. \( \dfrac{dy}{dt} = 5y \)
  2. \( \dfrac{dy}{dt} = -0.2y \)
  3. \( \dfrac{dy}{dt} = \dfrac{t}{y} \)
  4. \( y\,\dfrac{dy}{dx} = x + 1 \)
Part BWith Initial Conditions
  1. \( \dfrac{dy}{dx} = 4x^3,\; y(0)=1 \)
  2. \( \dfrac{dy}{dt} = \dfrac{t}{y},\; y(0)=3 \)
  3. \( \dfrac{dP}{dt} = 0.03P,\; P(0)=10000 \)
  4. \( \dfrac{dQ}{dt} = -0.15Q,\; Q(0)=200 \)
  5. \( \dfrac{dy}{dt} = (2t+1)(y-3),\; y(0)=5 \)
  6. \( \dfrac{dy}{dt} = y^2,\; y(0)=\tfrac{1}{2} \)

Growth and Decay Applications

  1. Population Growth. A city of 250,000 grows at continuous rate 1.8% per year. (a) Write the differential equation and solution. (b) Population after 10 years. (c) When does it reach 350,000?
  2. Bacterial Culture. A culture starts with 400 cells and doubles every 3 hours. (a) Find \( k \). (b) Cells after 9 hours. (c) When does population reach 10,000?
  3. Radioactive Decay. Iodine-131 has a half-life of 8 days. A patient receives 150 mg. (a) Find \( k \). (b) Amount remaining after 16 days. (c) When does the dose fall below 10 mg?
  4. Investment. Account starts at \$8000, earning 4.5% continuous annual interest. (a) Balance after 20 years. (b) Doubling time.
  5. Carbon-14 Dating. An artifact retains 60% of its original carbon-14. Half-life: 5730 years. Estimate the artifact's age.
  6. Newton's Law of Cooling. A metal rod at 200°C is placed in a 22°C room. After 10 minutes: 140°C. (a) Find \( k \). (b) Temperature after 30 minutes. (c) When does it reach 50°C?
  7. Drug Decay. \( \frac{dC}{dt} = -0.08C \), \( C(0) = 120 \) mg/L. (a) Half-life. (b) Concentration after 6 hours. (c) When does it drop below 20 mg/L?
  8. Logistic Population. A deer herd satisfies \( \frac{dP}{dt} = 0.6P(1 - P/800) \). Current population: 200. (a) Current growth rate. (b) Growth rate at population 400. (c) At what population is growth rate maximum? (d) Long-run behavior.

Modeling Problems

  1. Mixing Tank. A 500-gallon tank holds pure water. Brine (2 lbs salt/gallon) enters at 4 gal/min; well-mixed solution exits at 4 gal/min. (a) Why does tank volume stay constant? (b) Write \( \frac{dA}{dt} \). (c) Find the equilibrium salt amount and interpret it. (d) Write the particular solution with \( A(0) = 0 \).
  2. Spread of Information. A post begins spreading in a community of 20,000. The rate of new viewers is proportional to the number who haven't seen it. (a) Write the differential equation for \( N(t) \). (b) Identify the carrying capacity. (c) How does this differ from simple exponential growth?
  3. Terminal Velocity. A skydiver of mass 75 kg falls with drag coefficient \( b = 15 \) kg/s and \( g = 9.8 \) m/s². (a) Write the velocity differential equation. (b) Find terminal velocity. (c) Interpret physically.
  4. Continuous Reinvestment. A company's revenue satisfies \( \frac{dR}{dt} = 0.12R \), \( R(0) = 3.5 \) million dollars. (a) Find \( R(t) \). (b) Revenue after 5 years. (c) Doubling time. (d) What real-world factors might eventually prevent continued exponential growth?
  5. Forensics. A body is found in a 18°C room. At 9:00 PM: body temperature 31°C. At 10:00 PM: 29°C. Normal body temperature: 37°C. Estimate the time of death.

Error Analysis

Each problem shows a student's work. Identify the error, explain why it is wrong, and provide the correct solution.

Problem 1
Solve \( \frac{dy}{dt} = 2t + 3 \), \( y(0) = 1 \): Student writes \( y = 2t + 3 \). At \( t = 0 \): \( y = 3 \neq 1 \). "The equation has no solution."

Identify the error (the student differentiated rather than integrated) and find the correct particular solution.

Problem 2
Solve \( \frac{dy}{dt} = 3y \): Student writes \( y = \frac{3y^2}{2} \).

Identify the algebraic error and solve correctly using separation of variables.

Problem 3
Half-life is 10 years, so \( k = 10 \) per year.

Explain the misconception and compute the correct decay constant.

Problem 4
Newton's Law of Cooling in a 25°C room: \( \frac{dT}{dt} = -kT \).

Explain what is physically wrong with this equation and write the correct one.

Problem 5
Verify whether \( y = t^3 \) satisfies \( \frac{dy}{dt} = 3y \). Student: "\( \frac{dy}{dt} = 3t^2 \) and \( 3y = 3t^3 \). Both have a factor of 3 — close enough."

Identify the flaw in this reasoning and give the correct conclusion.

Chapter 13 Checkpoint

Chapter 13 Checkpoint
Complete without assistance — assesses all major skills and concepts from Chapter 13
Part A — Concept Questions
  1. In your own words, explain the difference between the general solution and a particular solution to a differential equation.
  2. A slope field shows that all segments are horizontal along the line \( y = 5 \). What does this tell you about the differential equation and its solutions?
  3. Explain what an initial condition does in solving a differential equation.
  4. Give one real-world situation modeled by exponential growth and one modeled by Newton's Law of Cooling.
Part B — Verification
  1. Verify that \( y = 4e^{-2t} \) is a solution to \( \frac{dy}{dt} = -2y \).
  2. Determine whether \( y = t^2 + 3t + 1 \) satisfies \( \frac{dy}{dt} = y \). Show your work and explain your conclusion.
Part C — Separable Equations
  1. Solve \( \frac{dy}{dt} = 6y \) with \( y(0) = 2 \).
  2. Solve \( \frac{dy}{dx} = \frac{2x}{y} \) with \( y(0) = 4 \).
  3. Solve \( \frac{dQ}{dt} = -0.05Q \) with \( Q(0) = 800 \). Find the half-life.
Part D — Applications
  1. A population of 1000 bacteria triples every 2 hours. (a) Find \( k \). (b) Write the particular solution \( P(t) \). (c) How many bacteria after 6 hours?
  2. A cup of tea starts at 95°C in a 22°C room. After 8 minutes: 72°C. (a) Find \( k \). (b) Temperature after 20 minutes. (c) When does temperature reach 40°C?
  3. A radioactive element has decay constant \( k = 0.00012 \) per year. (a) Find the half-life. (b) If 500 grams are present initially, how much remains after 10,000 years?
Part E — Slope Fields
  1. For \( \frac{dy}{dt} = 2(3 - y) \): (a) Identify the equilibrium solution. (b) Describe slopes when \( y < 3 \) and when \( y > 3 \). (c) Describe the long-term behavior of all solutions.
Part F — Modeling
  1. A lake has a pollution level of 80 units that decreases at rate \( k = 0.03 \) per year. (a) Write the differential equation and solution. (b) Pollution level after 15 years. (c) When does it drop below 10 units? (d) What model assumption is being made, and what real-world factors could violate it?
Looking Ahead
Chapter 14: Sequences, Series, and Calculus Readiness for Higher Mathematics

Chapter 13 introduced differential equations as the language of change. Chapter 14 will introduce a different kind of infinite process: sequences and series. You will explore what it means for an infinite list of numbers to converge, how to sum an infinite geometric series, and how functions can be represented as infinite polynomials. Taylor polynomials will give you a preview of one of the most powerful ideas in all of mathematics — approximating any smooth function as a polynomial to any desired precision. Chapter 14 serves as the bridge between Calculus I and the rich landscape of Calculus II, where these ideas are developed in full.

MGU Library Connection
  • MGU Differential Equations — integrating factors, second-order linear equations, Laplace transforms, series solutions, systems of ODEs, numerical methods
  • MGU Calculus II — integration techniques, parametric and polar curves, sequences and series in depth
  • MGU Physics — Newton's second law, harmonic oscillation, thermodynamics and cooling
  • MGU Calculus Appendix O — Differential Equations Preview Reference
  • MGU Calculus Appendix F — Derivative Rules Formula Sheet (differentiation for verification)
  • MGU Dictionary entries: differential equation, separable equation, slope field, exponential growth, half-life, logistic growth, Newton's Law of Cooling