MGU Mathematics Series — Differential Equations
Part II: First-Order Differential Equations
Separable Equations · Growth & Decay · Cooling · Motion
Chapter 3 of 14
MGU Mathematics Series  ·  Differential Equations  ·  Library Textbook Edition

Chapter 3:
Separable Differential
Equations

The first major symbolic solution method — separating variables, integrating both sides, and recovering unknown functions from their rate laws

■ Math Level: Differential Equations ■

■ 14 Sections  •  Part II of 5 ■

Follows Chapters 1–2: Foundations and Slope Fields  ·  Prepares students for Chapter 4: Linear First-Order Equations

Chapter 3 — Front Matter

Purpose and Overview

Chapter 3 introduces the first major symbolic solution method in this course: separation of variables. After two chapters devoted to understanding what differential equations are and how to read their qualitative behavior, students are now ready to solve a wide class of first-order equations exactly.

A differential equation is called separable if it can be rewritten so that all expressions involving the dependent variable appear on one side and all expressions involving the independent variable appear on the other. Once separated, each side can be integrated independently, and the result is an equation, often solvable for the unknown function, that describes the full solution.

This chapter develops separation of variables carefully and completely. It covers the mechanics of separating variables, the meaning of constants of integration, the handling of initial conditions, the domain and validity of solutions, and a range of applications including exponential growth and decay, Newton's Law of Cooling, and simple motion. It also introduces implicit solutions and discusses what to do when solving explicitly for the dependent variable is inconvenient or impossible.

By the end of this chapter, students should understand that separating variables is not algebraic decoration. It reorganizes the relationship between rates and states so that integration, the central tool of calculus, can recover the unknown function from its rate law.

Opening Question

A population of bacteria doubles every three hours. A sample begins with 500 bacteria. How many bacteria will there be after twelve hours? After one day? You might feel ready to answer with a memorized formula, but here is the deeper question: where does that formula come from? What assumption about the bacteria produces the doubling pattern? And what happens if the growth rate depends on the population itself, or on environmental limits? The answers come from separable differential equations. The doubling formula is not assumed — it is derived from a rate relationship, then recovered by integrating.

14Sections
10Objectives
17Key Terms
8+Worked Examples

Learning Objectives

  1. 1Recognize whether a first-order differential equation is separable.
  2. 2Rewrite a separable equation so that variables appear on opposite sides.
  3. 3Integrate both sides and include the constant of integration correctly.
  4. 4Solve for the dependent variable when possible, and recognize implicit solutions when it is not.
  5. 5Apply initial conditions to determine particular solutions.
  6. 6Solve exponential growth and decay problems, including half-life and doubling time.
  7. 7Apply Newton's Law of Cooling to find temperature as a function of time.
  8. 8Set up and solve simple motion problems using separable equations.
  9. 9Determine the domain and validity of a solution.
  10. 10Identify common mistakes in separating variables and integrating.
Chapter 3 — Contents

Sections at a Glance

Sections 3.1 – 3.6
The Method: Separating, Integrating, and Solving
Recognizing Separability · Separating Variables · Integrating · Constants · Initial Value Problems
Section 3.1

What Makes an Equation Separable

A general first-order equation has the form $\frac{dy}{dx} = f(x,y)$, where $f$ may involve both $x$ and $y$ in ways that prevent direct integration. A separable equation is the special case in which $f(x,y)$ factors cleanly into a function of $x$ alone multiplied by a function of $y$ alone:

Definition — Separable Equation
$$\frac{dy}{dx} = g(x) \cdot h(y)$$

When this factoring is possible, the equation can be reorganized so that all $y$ terms and $dy$ appear on one side, and all $x$ terms and $dx$ appear on the other. Integration then becomes possible on each side separately.

How to test for separability: Ask whether the right side can be written as a product of a pure function of $x$ and a pure function of $y$. If yes, the equation is separable.

EquationSeparable?Reason / Factoring
$\dfrac{dy}{dx} = 3y$Yes$g(x)=3$, $h(y)=y$
$\dfrac{dy}{dx} = x^2 y^3$Yes$g(x)=x^2$, $h(y)=y^3$
$\dfrac{dy}{dx} = \dfrac{x}{y}$Yes$g(x)=x$, $h(y)=1/y$
$\dfrac{dy}{dt} = -ky$Yes$g(t)=-k$, $h(y)=y$
$\dfrac{dy}{dx} = x + y$NoAddition — cannot factor as $g(x)\cdot h(y)$
$\dfrac{dy}{dx} = \sin(x+y)$NoSum inside the sine cannot be factored
Tip — Manipulate Before Deciding

Some equations that do not look separable at first glance are separable after factoring or simplifying the right side. For example, $\frac{dy}{dx} = xy + x$ factors as $\frac{dy}{dx} = x(y+1)$, making it separable with $g(x)=x$ and $h(y)=y+1$. Always simplify before concluding an equation is not separable.

Section 3.2

Separating Variables

Once an equation has been identified as separable, divide both sides by $h(y)$ (assuming $h(y)\neq 0$) and multiply both sides by $dx$:

Separation Step

Starting from $\dfrac{dy}{dx} = g(x)\cdot h(y)$, divide by $h(y)$ and multiply by $dx$:

$$\frac{1}{h(y)}\,dy = g(x)\,dx$$

The left side involves only $y$ and $dy$. The right side involves only $x$ and $dx$.

Treating $dy$ and $dx$ as movable objects is a notational convenience rooted in the chain rule. For this course, the separation step is valid whenever $h(y)\neq 0$, and the result correctly sets up two separate integration problems.

Example 3.2.1
Separating $\dfrac{dy}{dx} = xy^2$
  • Identify$g(x)=x$, $h(y)=y^2$. The equation is separable.
  • SeparateDivide by $y^2$ (assuming $y\neq 0$), multiply by $dx$: $$\frac{1}{y^2}\,dy = x\,dx$$
  • ResultVariables are separated and ready to integrate.
Equilibrium Warning

The case $h(y)=0$ must always be checked separately. If $h(y_0)=0$ for some constant $y_0$, then the constant function $y=y_0$ may itself be a solution. These equilibrium solutions cannot be recovered from the formula obtained by integration. They must be noted separately.

Section 3.3

Integrating Both Sides

After separating variables, integrate both sides. The left side is integrated with respect to $y$; the right side is integrated with respect to $x$:

Integration Step
$$\int \frac{1}{h(y)}\,dy = \int g(x)\,dx$$

This step converts the differential equation — a statement about rates — into an algebraic equation relating the dependent and independent variables through their antiderivatives. This is the heart of the method.

Why only one constant? Each side produces its own constant of integration, but these are combined into a single constant $C$ on one side. Writing $C_1$ and $C_2$ and then combining as $C = C_2 - C_1$ is valid and standard.

Example 3.3.1
Integrating $\dfrac{1}{y^2}\,dy = x\,dx$
  • Integrate$$\int y^{-2}\,dy = \int x\,dx$$
  • Result$$-\frac{1}{y} = \frac{x^2}{2} + C$$
  • InterpretThis is the general solution in implicit form. Each value of $C$ gives a different solution curve.
Section 3.4

Constants of Integration

The constant of integration $C$ is one of the most important and most mishandled objects in differential equations. Different values of $C$ give different solution functions — each satisfying the original differential equation. The family of all such functions is the general solution.

Constants often appear inside logarithms or as part of exponential expressions, and it is common to rename $C$ to make the solution cleaner. If integration gives $\ln|y| = kx + C$, then exponentiating yields:

Renaming Constants After Exponentiation

$\ln|y| = kx + C \;\Longrightarrow\; |y| = e^{kx+C} = e^C \cdot e^{kx}$

Since $e^C$ is a positive constant, rename it $A$. Dropping the absolute value (allowing the sign to be absorbed by $A$):

$$y = Ae^{kx}$$

where $A$ can be any nonzero real number. Including $A=0$ corresponds to the equilibrium solution $y=0$.

Student Warning — Constant Errors

After exponentiation, $e^{kx+C} = e^C \cdot e^{kx}$, not $e^{kx} + e^C$. Exponentiation of a sum becomes a product. Absorb $e^C$ into a renamed constant $A$, and write the solution as $y = Ae^{kx}$.

Section 3.5

Solving for $y$ When Possible

After integrating, the result is usually an implicit equation. When possible, solve explicitly for $y$ as a function of $x$. Continuing from the example in Section 3.3:

Example 3.5.1
Isolating $y$ from an implicit equation
  • StartFrom Section 3.3: $\displaystyle-\frac{1}{y} = \frac{x^2}{2} + C$
  • RearrangeMultiply by $-1$ and rename $-C$ as $A$: $\displaystyle\frac{1}{y} = A - \frac{x^2}{2}$
  • Invert$\displaystyle y = \frac{1}{A - \frac{x^2}{2}} = \frac{2}{2A - x^2}$
  • SimplifyLet $B = 2A$ (still an arbitrary constant): $$y = \frac{2}{B - x^2}$$
General solution: $\displaystyle y = \dfrac{2}{B - x^2}$
Tip — When Explicit Form Is Impractical

If the implicit equation cannot be solved cleanly for $y$, accept the implicit form as the general solution. It is not incomplete — it defines the solution curves just as well. Section 3.11 discusses implicit solutions further.

Section 3.6

Initial Value Problems

The general solution contains one arbitrary constant $C$. To find a particular solution, an initial condition of the form $y(x_0) = y_0$ is applied. The pair of a differential equation and an initial condition is called an initial value problem (IVP).

Procedure — Solving an Initial Value Problem

1. Solve the differential equation to find the general solution with constant $C$.

2. Substitute the initial condition $x = x_0$, $y = y_0$.

3. Solve the resulting equation for $C$.

4. Write the particular solution by substituting $C$ back into the general solution.

Example 3.6.1
Solving an Initial Value Problem
  • ProblemSolve $\dfrac{dy}{dx} = 3y$, $\;y(0) = 4$.
  • ThinkThe equation is separable with $g(x)=3$, $h(y)=y$. Separate, integrate, then apply the initial condition.
  • Separate$\dfrac{1}{y}\,dy = 3\,dx$
  • Integrate$\ln|y| = 3x + C \;\Longrightarrow\; y = Ae^{3x}$
  • Apply IC$y(0)=4$: $\quad 4 = Ae^0 = A$, so $A=4$.
  • Answer$y = 4e^{3x}$
  • Check$y' = 12e^{3x}$. Also $3y = 12e^{3x}$. Match. $y(0)=4$. ✓
  • InterpretThe solution describes a quantity starting at $4$ when $x=0$ and growing exponentially with rate $3$ per unit of $x$.
Particular solution: $y = 4e^{3x}$
Sections 3.7 – 3.10
Growth, Decay, Cooling, and Motion
Exponential Models · Half-Life · Newton's Law of Cooling · Terminal Velocity
Section 3.7

Exponential Growth and Decay

One of the most important separable equations is $\dfrac{dy}{dt} = ky$, where $k$ is a constant. This says the rate of change of $y$ is directly proportional to the current value of $y$. When $k>0$, the quantity grows; when $k<0$, it decays.

Fundamental Exponential Growth/Decay Formula
$$y = y_0\,e^{kt}$$

where $y_0 = y(0)$ is the initial value and $k$ is the growth ($k>0$) or decay ($k<0$) constant

This formula is not memorized — it is derived. Separating $\frac{1}{y}\,dy = k\,dt$, integrating to get $\ln|y| = kt + C$, exponentiating, and applying $y(0)=y_0$ yields $y = y_0 e^{kt}$. Every instance of exponential change in science, finance, and engineering comes from this derivation.

Example 3.7.1
Bacterial Population Growth
  • ProblemA culture has $500$ bacteria at $t=0$. Growth constant $k=0.4$ per hour. Find $P(t)$ and the population after $6$ hours.
  • ThinkIVP: $\dfrac{dP}{dt}=0.4P$, $P(0)=500$. Apply the exponential formula directly.
  • Solve$P(t) = 500\,e^{0.4t}$
  • Evaluate$P(6) = 500\,e^{2.4} \approx 500 \times 11.023 \approx 5512$ bacteria
  • Check$P(0)=500$. ✓   As $t$ increases, $e^{0.4t}$ increases. ✓
  • InterpretFrom $500$ organisms, the culture grows to approximately $5{,}512$ in $6$ hours. The exponential model assumes unlimited resources. In reality, resource limits slow growth — a refinement addressed in Chapter 5 with the logistic model.
$P(t) = 500\,e^{0.4t}$. After 6 hours: $P(6) \approx 5{,}512$ bacteria.
Section 3.8

Half-Life and Doubling Time

QuantityContextFormulaDepends On
Doubling time $T_2$Exponential growth ($k>0$)$\displaystyle T_2 = \frac{\ln 2}{k}$Only on $k$, not $y_0$
Half-life $T_{1/2}$Exponential decay ($r>0$)$\displaystyle T_{1/2} = \frac{\ln 2}{r}$Only on $r$, not $y_0$

Both formulas are derived by setting $y = 2y_0$ (for doubling) or $y = \frac{1}{2}y_0$ (for half-life) in the exponential formula and solving for $t$. The initial value $y_0$ cancels, revealing that the time scale depends only on the rate constant — a remarkable property of exponential change.

Example 3.8.1
Radioactive Decay and Carbon-14
  • ProblemCarbon-14 has half-life $T_{1/2} = 5730$ years. A sample begins with $80$ g. Find the decay constant $r$, write the model, and find the mass after $2000$ years.
  • ThinkUse $r = \dfrac{\ln 2}{T_{1/2}}$, then write $Q(t) = 80e^{-rt}$, then evaluate at $t=2000$.
  • Solve $r = \dfrac{\ln 2}{5730} \approx 0.0001210 \text{ per year}$

    $Q(t) = 80\,e^{-0.0001210\,t}$

    $Q(2000) = 80\,e^{-0.2420} \approx 80 \times 0.7849 \approx 62.8\text{ g}$
  • CheckAt $t=5730$: $Q = 80e^{-\ln 2} = 80 \times 0.5 = 40\text{ g}$ — exactly half. ✓
  • InterpretAfter $2000$ years, $\approx 62.8$ g remain. Since $2000$ years is about one-third of a half-life, the sample has lost less than half its mass — consistent with $62.8 > 40$.
$r \approx 0.0001210$ per year. $Q(t) = 80e^{-0.0001210\,t}$. After 2000 years: $\approx 62.8$ g.
Section 3.9

Newton's Law of Cooling

Newton's Law of Cooling states that the rate of temperature change of an object is proportional to the difference between the object's temperature and the ambient temperature:

Newton's Law of Cooling
$$\frac{dT}{dt} = -k(T - T_a)$$

$T(t)$ = object temperature, $T_a$ = ambient temperature (constant), $k>0$ = cooling constant

The substitution $u = T - T_a$ transforms this into the exponential decay equation $\dfrac{du}{dt} = -ku$, with solution $u = u_0 e^{-kt}$. Substituting back yields the full solution:

Solution to Newton's Law of Cooling
$$T(t) = T_a + (T_0 - T_a)\,e^{-kt}$$

As $t\to\infty$: $T(t)\to T_a$. The object asymptotically approaches ambient temperature.

Example 3.9.1
Coffee Cooling
  • ProblemCoffee starts at $90°C$. Room temperature is $20°C$. After $10$ min the coffee is $70°C$. Find $k$ and $T(30)$.
  • Setup$T(t) = 20 + 70\,e^{-kt}$
  • Find k Apply $T(10)=70$: $\quad 70 = 20 + 70e^{-10k}$

    $50 = 70e^{-10k} \;\Rightarrow\; e^{-10k} = \dfrac{5}{7}$

    $k = \dfrac{1}{10}\ln\!\left(\dfrac{7}{5}\right) \approx 0.03365$ per minute
  • Evaluate$T(30) = 20 + 70\,e^{-1.0095} \approx 20 + 70(0.3645) \approx 45.5°C$
  • Check$T(0)=90°C$ ✓, $T(10)=70°C$ ✓, $T\to 20°C$ as $t\to\infty$ ✓
  • InterpretAfter $30$ minutes the coffee is approximately $45.5°C$ — still warm but cooling steadily. It asymptotically approaches room temperature and will never cool below $20°C$ under ideal conditions.
$k \approx 0.03365$ per min. $T(t) = 20 + 70e^{-0.03365\,t}$. $T(30) \approx 45.5°C$.
Section 3.10

Simple Motion with Separable Equations

Newton's second law $F=ma$ becomes a differential equation when force depends on velocity. A common model for a falling object subject to linear air drag is:

Motion with Linear Drag
$$m\frac{dv}{dt} = mg - bv$$

$m$ = mass, $g$ = gravitational acceleration, $b>0$ = drag coefficient, $v$ = velocity

Letting $r = b/m$, this separates to $\dfrac{dv}{g-rv} = dt$. Integrating and applying $v(0)=v_0$ yields:

Velocity with Drag
$$v(t) = \frac{g}{r} + \!\left(v_0 - \frac{g}{r}\right)e^{-rt}$$

Terminal velocity: $v_{\text{term}} = \dfrac{g}{r} = \dfrac{mg}{b}$

Example 3.10.1
Skydiver Terminal Velocity
  • ProblemSkydiver mass $70$ kg, drag coefficient $b=14$ kg/s, $v_0=0$, $g=9.8$ m/s². Find $v(t)$ and terminal velocity.
  • Compute$r = 14/70 = 0.2$ per second. $v_{\text{term}} = 9.8/0.2 = 49$ m/s.
  • Solve$v(t) = 49\!\left(1 - e^{-0.2t}\right)$
  • Check$v(0)=0$ ✓. As $t\to\infty$: $v\to 49$ m/s ✓.
  • InterpretThe skydiver accelerates from rest and asymptotically approaches $49$ m/s ($\approx 176$ km/h). Acceleration is greatest at the start, when drag is small, and diminishes as speed increases.
$v(t) = 49(1-e^{-0.2t})$ m/s. Terminal velocity: $49$ m/s.
Sections 3.11 – 3.14
Implicit Solutions, Domain, Mistakes, and What Comes Next
Implicit Solutions · Domain Restrictions · Equilibria · Common Errors · Bridge to Chapter 4
Section 3.11

Implicit Solutions

Not every separable equation can be solved explicitly for $y$ as a function of $x$. When the implicit equation relating $y$ and $x$ cannot be solved in closed form, the implicit form is accepted as the general solution.

An implicit solution is an equation $F(x,y) = C$ such that any function $y(x)$ satisfying this equation also satisfies the original differential equation.

Example 3.11.1
An equation with no explicit solution
  • Equation$\dfrac{dy}{dx} = \dfrac{x^2}{1-y^2}$
  • Separate$(1-y^2)\,dy = x^2\,dx$
  • Integrate$y - \dfrac{y^3}{3} = \dfrac{x^3}{3} + C$
  • ResultThere is no elementary formula isolating $y$. The implicit equation $y - \dfrac{y^3}{3} = \dfrac{x^3}{3} + C$ is the general solution.
General implicit solution: $\;y - \dfrac{y^3}{3} = \dfrac{x^3}{3} + C$
Tip — Implicit Solutions Are Valid

The inability to write $y$ explicitly is a limitation of algebraic notation, not a failure of the solution method. Implicit solutions define solution curves completely, allow numerical computation of specific values, and can be verified by implicit differentiation.

Section 3.12

Domain and Validity of Solutions

The general solution to a separable equation is valid only where it is defined and where the original equation makes sense. Several issues commonly restrict the domain:

IssueExampleEffect on Domain
Logarithm arguments$\ln|y|$ requires $y\neq 0$Solution invalid where $y=0$
Rational denominators$y = \dfrac{2}{B-x^2}$Undefined at $x=\pm\sqrt{B}$
Division by $h(y)$$h(y)=0$ at some $y_0$Requires separate check for equilibrium solution $y=y_0$

Equilibrium solutions: If $h(y_0)=0$, the constant function $y=y_0$ satisfies $y'=g(x)\cdot 0 = 0$. Verify this independently — the general formula will not include it.

Example 3.12.1
Equilibrium solutions for $\dfrac{dy}{dx} = xy^2$
  • Check $h(y)=0$$h(y)=y^2=0 \;\Rightarrow\; y=0$. Test: if $y=0$, then $y'=0$ and $g(x)\cdot 0=0$. So $y=0$ is an equilibrium solution for all $x$.
  • General solutionFrom Section 3.5, $y = \dfrac{2}{B-x^2}$. This is never zero — confirming that the equilibrium $y=0$ does not appear in this family.
Equilibrium solution: $y = 0$. General solution: $y = \dfrac{2}{B-x^2}$, valid on any interval excluding $x=\pm\sqrt{B}$.
Section 3.13

Common Separable Equation Mistakes

Mistake 1 — Separating When Addition Is Present
$\dfrac{dy}{dx}=x+y$ separated as $\dfrac{dy}{y}=x\,dx$
Correct: $x+y$ cannot be factored as $g(x)\cdot h(y)$. This equation is not separable. Use the integrating factor method from Chapter 4.

Addition does not allow variable separation. Only products can be separated.

Mistake 2 — Forgetting the Constant of Integration
$\ln|y|=kx$ (no $+C$)
Correct: $\ln|y| = kx + C$, giving $y = Ae^{kx}$ where $A$ is arbitrary.

Omitting $C$ gives one particular solution without justification. The result is incomplete until an initial condition is applied.

Mistake 3 — Adding Instead of Multiplying After Exponentiation
$e^{kx+C} = e^{kx} + e^C$
Correct: $e^{kx+C} = e^C \cdot e^{kx}$. Exponentiation of a sum becomes a product. Rename $e^C$ as $A$ to write $y = Ae^{kx}$.
Mistake 4 — Dropping the Absolute Value Too Early
Writing $\ln y$ instead of $\ln|y|$, restricting to $y>0$ with no justification
Correct: Maintain $\ln|y|$ until an initial condition or the context establishes the sign of $y$. Allow $A$ in $y=Ae^{kx}$ to be negative when the initial value is negative.
Mistake 5 — Forgetting Equilibrium Solutions
Presenting only the general formula without checking whether $h(y)=0$ gives additional solutions
Correct: Always check if $h(y_0)=0$ for any constant $y_0$. If so, verify that $y=y_0$ satisfies the original equation and include it as a separate solution.
Mistake 6 — Applying the Initial Condition to an Intermediate Step
Substituting $y(0)=y_0$ into $\ln|y|=kx+C$ instead of into $y=Ae^{kx}$
Correct: Apply the initial condition after converting to the clearest form of the general solution. In this case, substitute into $y=Ae^{kx}$ to find $A=y_0$ directly.
Mistake 7 — Sign Error in Decay Problems
Writing $y=y_0 e^{0.5t}$ for a decay problem with rate $r=0.5$ (growth instead of decay)
Correct: For decay, $\dfrac{dy}{dt}=-ry$ with $r>0$, giving $y=y_0 e^{-rt}$. The negative exponent ensures the quantity decreases over time.
Section 3.14

Preparing for Linear First-Order Equations

Separable equations are powerful, but they require the right side of the equation to factor as a product $g(x)h(y)$. Many important equations, including all equations of the form $\dfrac{dy}{dx} + p(x)y = q(x)$, are not separable. The equation $\dfrac{dy}{dx} = x - y$, for instance, cannot be separated because $x - y$ is a difference, not a product.

These are linear first-order equations. They have special algebraic structure: the dependent variable and its derivative appear only to the first power and are not multiplied together. Chapter 4 introduces the integrating factor method, designed precisely for this class of equations.

The progression is natural: separable equations are solvable when variables can be algebraically separated; linear equations are solvable through an integrating factor that converts the left side into a recognizable product derivative. Every skill from this chapter — separating, integrating, managing constants, applying initial conditions, interpreting solutions — carries forward directly.

MGU Library Connection
  • Exponential growth and decay models (Section 3.7) connect to MGU Biology: Population Dynamics and MGU Finance: Continuous Compounding
  • Newton's Law of Cooling (Section 3.9) connects to MGU Physics: Thermodynamics and MGU Engineering: Heat Transfer
  • Terminal velocity (Section 3.10) connects to MGU Physics: Classical Mechanics and MGU Engineering: Fluid Drag
  • Radioactive decay (Section 3.8) connects to MGU Chemistry: Nuclear Reactions and MGU Geology: Radiometric Dating
  • Separable equations underlie the earliest compartment models for disease spread, extended to systems in Chapter 9: Systems of Differential Equations
Chapter 3 — Back Matter

Chapter Summary

Key Terms Review

separable equation
A first-order equation of the form $\frac{dy}{dx}=g(x)h(y)$ that can be reorganized so variables appear on opposite sides.
separation of variables
The method of dividing by $h(y)$ and multiplying by $dx$ to isolate $y$ terms on one side and $x$ terms on the other, enabling separate integration.
constant of integration
The arbitrary constant $C$ that appears when integrating a differential equation, representing the family of all possible solutions.
general solution
The family of all solutions to a differential equation, expressed with an undetermined constant $C$.
particular solution
A specific member of the solution family, determined by an initial condition that fixes the value of $C$.
initial value problem
A differential equation paired with an initial condition $y(x_0)=y_0$, together determining a unique particular solution.
explicit solution
A solution in which $y$ is expressed directly and alone as a function of $x$.
implicit solution
A solution in which $y$ and $x$ appear together in an equation that cannot be solved cleanly for $y$ alone.
exponential growth
Behavior described by $y=y_0 e^{kt}$ with $k>0$, where the quantity increases without bound over time.
exponential decay
Behavior described by $y=y_0 e^{-rt}$ with $r>0$, where the quantity decreases toward zero over time.
growth constant
The proportionality constant $k$ in $\frac{dy}{dt}=ky$ that determines the rate of exponential growth.
decay constant
The proportionality constant $r$ in $\frac{dy}{dt}=-ry$ that determines the rate of exponential decay.
half-life $T_{1/2}$
The time for a decaying quantity to fall to half its current value: $T_{1/2}=\frac{\ln 2}{r}$. Depends only on the rate constant.
doubling time $T_2$
The time for a growing quantity to double: $T_2=\frac{\ln 2}{k}$. Depends only on the growth constant.
Newton's Law of Cooling
The model $\frac{dT}{dt}=-k(T-T_a)$, describing temperature change proportional to the difference between object and ambient temperature.
terminal velocity
The limiting velocity of a falling object when gravity and drag balance: $v_{\text{term}}=mg/b$. Approached asymptotically.
equilibrium solution
A constant function $y=y_0$ satisfying the differential equation, arising when $h(y_0)=0$. Not captured by the general formula.

Concept Review Questions

  1. What property of a first-order equation makes it separable? Give one example of a separable equation and one that is not, and explain the difference.
  2. Why is it valid to write $\frac{1}{h(y)}\,dy = g(x)\,dx$ when separating variables? What assumption must hold for this step?
  3. Explain why the constant of integration appears on only one side after integrating a separated equation.
  4. What is the difference between the general solution and a particular solution? What is needed to move from one to the other?
  5. In the equation $\frac{dy}{dt}=ky$, what does $k$ represent? Under what condition does this produce growth? Under what condition decay?
  6. State the doubling time and half-life formulas. Explain why neither depends on the initial value.
  7. What is an equilibrium solution, and why might it not appear in the general solution formula found by separation of variables?
  8. What is an implicit solution, and when should it be accepted rather than solved explicitly?
  9. Explain the physical meaning of terminal velocity and how it arises from the differential equation for motion with drag.
  10. Newton's Law of Cooling predicts that an object approaches the ambient temperature asymptotically. What feature of the solution formula $T(t)=T_a+(T_0-T_a)e^{-kt}$ confirms this?
Practice AClassification Practice

Determine whether each equation is separable. If separable, identify $g$ and $h$.

  1. $\dfrac{dy}{dx} = 5y$
  2. $\dfrac{dy}{dx} = x^2 - y$
  3. $\dfrac{dy}{dx} = x^2 y^2$
  4. $\dfrac{dy}{dx} = \dfrac{\cos x}{y^2}$
  5. $\dfrac{dy}{dx} = xy + x$
  6. $\dfrac{dy}{dt} = e^t\sin y$
  7. $\dfrac{dP}{dt} = rP(1-P/K)$
  8. $\dfrac{dy}{dx} = \dfrac{x^2+1}{y^3+1}$
  9. $\dfrac{dy}{dx} = \ln(x+y)$
  10. $\dfrac{dv}{dt} = g - kv$
Practice BSkill Practice — General Solutions

Solve each separable equation. Write the general solution.

  1. $\dfrac{dy}{dx} = 4y$
  2. $\dfrac{dy}{dx} = \dfrac{x}{y}$
  3. $\dfrac{dy}{dx} = x^2 y$
  4. $\dfrac{dy}{dx} = e^x e^y$
  5. $\dfrac{dy}{dx} = \dfrac{1}{y^2}$
  6. $\dfrac{dy}{dt} = -3y$
  7. $y\,dy = (x^2+1)\,dx$
  8. $\dfrac{dv}{dx} = v^2 x$
  9. $\dfrac{dy}{dx} = \dfrac{y}{x}$
  10. $\dfrac{dP}{dt} = 0.05P$
Practice CInitial Value Problems
  1. $y'=2y$, $y(0)=3$
  2. $y'=-0.1y$, $y(0)=50$
  3. $y'=xy$, $y(0)=2$
  4. $\dfrac{dy}{dx}=\dfrac{x}{y}$, $y(1)=3$
  5. $\dfrac{dy}{dt}=4-2y$, $y(0)=0$
  6. $\dfrac{dT}{dt}=-0.05(T-25)$, $T(0)=100$
  7. $\dfrac{dy}{dx}=y^2$, $y(0)=1$
  8. $\dfrac{dP}{dt}=0.03P$, $P(0)=1000$
  9. $(1+x^2)\,dy=xy\,dx$, $y(0)=4$
  10. $\dfrac{dy}{dx}=e^{x-y}$, $y(0)=0$
Practice DSolution Verification

Verify each claimed solution by differentiating and substituting into the equation.

  1. Claim: $y = 5e^{3x}$ satisfies $\dfrac{dy}{dx}=3y$.
  2. Claim: $y = \sqrt{x^2+4}$ satisfies $\dfrac{dy}{dx}=\dfrac{x}{y}$.
  3. Claim: $y = \dfrac{1}{1-x}$ satisfies $\dfrac{dy}{dx}=y^2$.
  4. Claim: $T(t)=20+60e^{-0.05t}$ satisfies $\dfrac{dT}{dt}=-0.05(T-20)$ with $T(0)=80$.
  5. Claim: $P(t)=200e^{0.1t}$ satisfies $\dfrac{dP}{dt}=0.1P$ with $P(0)=200$.
Practice EGrowth, Decay, and Half-Life Applications
  1. A bank account earns $6\%$ annual interest, compounded continuously. If $\$2000$ is deposited today, find the balance after $10$ years.
  2. A radioactive substance has a half-life of $20$ days. A sample begins with $100$ mg. Find the decay model and determine when only $10$ mg remains.
  3. A colony of bacteria doubles every $5$ hours. The initial population is $1000$. Find the population after $24$ hours.
  4. A drug is eliminated from the bloodstream at a rate proportional to the amount present, with decay constant $0.12$ per hour. An initial dose is $200$ mg. Find when the amount drops below $50$ mg.
  5. Carbon-14 decays with a half-life of $5730$ years. An artifact contains $30\%$ of its original carbon-14. Estimate the age of the artifact.
Practice FNewton's Law of Cooling Problems
  1. A cup of tea is initially at $85°C$. Ambient temperature is $22°C$. After $5$ minutes the tea is $75°C$. Find the cooling constant and the temperature after $20$ minutes.
  2. A metal rod at $200°C$ is placed in a room at $25°C$. It cools to $100°C$ in $30$ minutes. How long does it take to cool to $50°C$?
  3. A body is discovered in a room at $18°C$. At discovery, the body's temperature is $29°C$; one hour later it is $26°C$. Assuming normal body temperature $37°C$, estimate the time of death.
  4. A frozen dinner at $-10°C$ is placed in a $200°C$ oven. After $20$ minutes its temperature is $80°C$. Find the model and predict the temperature after $40$ minutes.
Practice GMotion and Terminal Velocity Problems
  1. A ball of mass $0.5$ kg is dropped from rest. Drag force is $0.1v$ N. Find velocity as a function of time and the terminal velocity. Use $g=9.8$ m/s².
  2. A parachutist of mass $80$ kg descends with drag coefficient $b=40$ kg/s. Find the terminal velocity and the time to reach $90\%$ of terminal velocity.
  3. A boat traveling at $10$ m/s has its engine cut. Drag is proportional to velocity with $b=200$ kg/s; boat mass $500$ kg. Find velocity as a function of time and how long until speed drops to $1$ m/s.
Practice HImplicit Solution Practice

Find the general implicit solution to each equation.

  1. $\dfrac{dy}{dx} = \dfrac{1-y^2}{x}$
  2. $\dfrac{dy}{dx} = \dfrac{x^2}{1+y^2}$
  3. $y^2\,dy = \sin x\,dx$
  4. $\dfrac{dy}{dx} = \dfrac{e^x}{e^y+1}$
Practice IDomain and Equilibrium Practice
  1. For $\dfrac{dy}{dx}=y(y-2)$, identify all equilibrium solutions. Separate variables and find the general non-equilibrium solution. Describe the domain of each solution.
  2. For $\dfrac{dy}{dx}=y^2-4$, find all equilibrium solutions. Separate and integrate. What happens to solutions starting near $y=2$? Near $y=-2$?
  3. The solution to $\dfrac{dy}{dx}=y^2$, $y(0)=1$ is $y=\dfrac{1}{1-x}$. What is the domain? What happens at $x=1$?
  4. Explain why the general solution $y=\dfrac{2}{B-x^2}$ from Section 3.5 is not defined for all real $x$ for any finite value of $B$.
Practice JModeling Practice
  1. A lake contains $5000$ kg of pollutant that degrades at a rate proportional to the amount present, with constant $k=0.02$ per year. Write and solve the initial value problem. When will the pollutant fall to $500$ kg?
  2. A savings account grows at a continuous rate of $4.5\%$ per year. How long until the account doubles?
  3. A yeast culture starts with $200$ cells and grows at $30\%$ per hour continuously. Write the model and find the population after $8$ hours. When does it exceed $10{,}000$ cells?
  4. An animal population decays exponentially due to habitat loss. If $20\%$ is lost every $10$ years, find the decay constant and estimate the time until only $50\%$ of the original population remains.
  5. Hot chocolate at $95°C$ cools in a room at $18°C$ with cooling constant $k=0.04$ per minute. At what time does the temperature reach $55°C$? Is the drink still enjoyably warm after $30$ minutes?
Practice KInterpretation Problems
  1. A solution is $y = 8e^{-0.3t}$. What is the initial value? What is the decay constant? In the context of drug concentration (mg), describe what happens over time.
  2. Two bacteria cultures have the same growth constant $k=0.5$ per hour. Culture A starts with $200$ organisms; culture B starts with $500$. Which culture is larger after $3$ hours? After $10$ hours? Will one ever catch up to the other?
  3. A cooling model gives $T(t) = 22 + 63e^{-0.04t}$. What is the ambient temperature? What is the initial object temperature? How does the object approach equilibrium?
  4. The general solution to a separable equation is $x^2 + y^2 = C^2$. What curve does each particular solution trace? Is $y$ an explicit function of $x$?

Error Analysis

Each problem contains a mathematical error. Identify and correct it.

Error Analysis 1

Solving $\dfrac{dy}{dx}=xy$, a student wrote $\dfrac{dy}{y}=dx$ (dropped the $x$ from the right side), then obtained $\ln|y|=x+C$, giving $y=Ae^x$.

$\dfrac{dy}{y} = dx \;\Rightarrow\; y = Ae^x$
Correct: $\dfrac{dy}{y} = x\,dx \;\Rightarrow\; \ln|y| = \dfrac{x^2}{2}+C \;\Rightarrow\; y = Ae^{x^2/2}$

The $x$ is part of $g(x)$ and must remain on the right side when separating.

Error Analysis 2

Solving $\dfrac{dy}{dx}=2y$, $y(0)=5$, a student wrote $\dfrac{dy}{y}=2\,dx$, then $\ln y=2x$, then $y=e^{2x}$. No constant was included; the answer grew from $y(0)=1$ instead of $5$.

$\ln y = 2x \;\Rightarrow\; y = e^{2x}$ (omitting $C$, gives $y(0)=1$)
Correct: $\ln|y|=2x+C \;\Rightarrow\; y=Ae^{2x}$. Apply $y(0)=5$: $A=5$. Particular solution: $y=5e^{2x}$.
Error Analysis 3

A student solved $\dfrac{dy}{dx}=x+y$ by separating to get $\dfrac{dy}{y}=x\,dx$.

$\dfrac{dy}{dx}=x+y$ treated as separable; $\dfrac{dy}{y}=x\,dx$
Correct: $x+y$ cannot be factored as $g(x)\cdot h(y)$. This equation is not separable. Use the integrating factor method from Chapter 4: write in standard form $y'-y=-x$ with $p(x)=-1$.
Error Analysis 4

A student solved $\dfrac{dy}{dx}=-0.5y$, $y(0)=100$ and obtained $y=100e^{0.5t}$ (a growing solution instead of a decaying one).

$y = 100\,e^{0.5t}$ (grows, but the equation has a negative sign)
Correct: The equation $y'=-0.5y$ has $k=-0.5$, so $y=100e^{-0.5t}$. The negative sign must be preserved throughout: $\ln|y|=-0.5t+C \;\Rightarrow\; y=Ae^{-0.5t}$.
Error Analysis 5

After finding $y=Ae^{kx}$ and applying $y(0)=3$, a student wrote $3=Ae^0=A+0=A$ and got $A=3$. Is the reasoning correct?

$3 = A e^0 = A + 0$ (wrong: $e^0 = 1$, not $0$)
Correct reasoning: $e^0 = 1$, so $3 = A \cdot 1 = A$. The answer $A=3$ is correct, but the intermediate step "$A+0$" is wrong. Write $3 = A \cdot e^0 = A \cdot 1 = A$.

Chapter 3 Checkpoint

Chapter 3 Checkpoint
Complete without assistance — assesses all key skills from Chapter 3 before proceeding to Chapter 4
Section A — Classification (4 problems)
  1. Is $\dfrac{dy}{dx} = 5xy^2$ separable? Explain and identify $g(x)$ and $h(y)$ if so.
  2. Is $\dfrac{dy}{dx} = x + 3y$ separable? Explain why or why not.
  3. Is $\dfrac{dy}{dt} = t\cos y$ separable? If so, identify $g$ and $h$.
  4. Is $\dfrac{dP}{dt} = rP$ separable? What is the general solution?
Section B — Solving Separable Equations (4 problems)
  1. Solve $\dfrac{dy}{dx} = -6y$. Write the general solution.
  2. Solve $\dfrac{dy}{dx} = \dfrac{x^3}{y}$. Write the general solution.
  3. Solve $\dfrac{dy}{dx} = y(1-y)$. Write the general solution. Are there equilibrium solutions?
  4. Solve $\dfrac{dy}{dt} = 2y - 6$. Write the general solution.
Section C — Initial Value Problems (3 problems)
  1. Solve $\dfrac{dy}{dx} = 3y$, $y(0)=7$.
  2. Solve $\dfrac{dT}{dt} = -0.04(T-20)$, $T(0)=90$.
  3. Solve $\dfrac{dy}{dx} = \dfrac{x}{y}$, $y(2)=1$.
Section D — Application Problems (3 problems)
  1. A substance decays at a rate proportional to its mass, with decay constant $r=0.035$ per year. An initial sample has $80$ grams. Write the decay model and find the half-life.
  2. A population of $400$ animals grows exponentially. After $5$ years, the population is $520$. Find the growth constant and estimate the population after $15$ years.
  3. A cup of soup at $80°C$ cools in a kitchen at $20°C$. After $10$ minutes the soup is $65°C$. Find $T(t)$ and determine when the soup reaches $40°C$.
Section E — Interpretation (2 problems)
  1. A solution to a differential equation is $y = 200e^{-0.02t}$. Explain what this says about the quantity $y$ in physical terms. What is the long-term behavior? At what time does $y$ first drop below $50$?
  2. The general solution to a separable equation is $y^2 = x^2 + C$. Apply the initial condition $y(1)=2$ to find a particular solution. Is the solution explicit or implicit?
Looking Ahead
Chapter 4: Linear First-Order Differential Equations

Separable equations are a powerful tool, but they require the right side to factor as a product $g(x)h(y)$. Many important equations cannot be separated this way. The equation $\dfrac{dy}{dx} + p(x)y = q(x)$ describes a first-order linear structure: $y$ and $y'$ appear only to the first power, and they are not multiplied together. Chapter 4 shows that this linear structure supports a precise and elegant solution technique — the integrating factor method — which converts the left side of the equation into a recognizable derivative of a product. Applications will include mixing problems, RC circuits, motion with linear resistance, and financial and population models. Every skill from Chapter 3 carries forward directly.

MGU Library Connection
  • Calculus: Integration Techniques — Appendix B of this textbook (antiderivatives required throughout Chapter 3)
  • Calculus: The Chain Rule — foundational to understanding why separation of variables is valid
  • MGU Biology: Population Dynamics — exponential and logistic growth models (Chapter 3 and 5)
  • MGU Finance: Continuous Compounding — derivation of $A=Pe^{rt}$ via separable equations
  • MGU Physics: Classical Mechanics — terminal velocity and drag force models
  • MGU Chemistry: Nuclear Reactions — radioactive decay chains modeled by separable equations
  • Chapter 5: Modeling with First-Order Differential Equations — extends separable methods to logistic growth, harvesting, mixing, and chemical concentration