Introduction to Differential Equations

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Textbook: Logan, A First Course in Differential Equations, 3rd ed. • Chapter: 9 • Section: 1

Quick Reference


Textbook	Logan, A First Course in Differential Equations, 3rd ed.
Chapter.Section.Subsection	1.1 First-Order Equations / 1.1.1 Notation and Terminology
Pages	pp. 2 to 6
Course	MATH347

Page range is faculty-verified against the Logan 3rd edition PDF listed in Resources.

Try This First

A small ball is dropped. Its downward velocity $v$ starts at $0$ and grows. Suppose someone tells you only this one fact about the motion:

At every instant, the velocity is increasing at a steady rate of $9.8$ meters per second each second.

Before any formula, try to answer three things on scratch paper.

After $1$ second, what is the velocity? After $2$ seconds?
Sketch velocity against time. Is the graph a line or a curve?
The sentence above describes the slope of the velocity graph, not the velocity itself. Which function $v(t)$ has that slope everywhere and starts at $0$?

Check your thinking

After $1$ second the velocity is about $9.8$ meters per second, and after $2$ seconds about $19.6$. The graph is a straight line through the origin with slope $9.8$, so $v(t) = 9.8\,t$.

The third question is the whole subject in miniature. A sentence about a rate (“velocity increases at $9.8$ each second”) is a statement about a derivative ($v'(t) = 9.8$). Finding the actual velocity function means finding a function whose derivative matches the sentence and whose starting value matches the situation. That search is what it means to solve a differential equation.

The Idea in Plain Words

An ordinary equation such as $x^2 - 5x + 6 = 0$ asks for a number. A differential equation asks for a function. That single shift is the reason this material feels new.

In most applied settings, the law that governs a quantity is not a formula for the quantity itself. It is a formula for how fast the quantity changes. A population grows in proportion to its current size. A hot cup of coffee cools in proportion to how much hotter it is than the room. A falling object speeds up under gravity. Each of these sentences ties a quantity to its own rate of change, and a rate of change is a derivative. Writing the sentence in symbols produces a differential equation. Recovering the original quantity from that equation is the work of this course.

Read a differential equation as a process, not a static fact. The unknown $x(t)$ is a function that takes an input (often time $t$) and gives back an output (a position, a temperature, a population). The equation is a rule the process has to obey at every instant.

Prerequisite Hub

Builds on

Skill	Course	Why it is needed
`math347-w0-antiderivatives-review`	MATH347	A solution is recovered by reversing differentiation.
`math347-w0-exponential-growth-decay`	MATH347	The first models in the course are growth and decay.
`derivative-function`	MATH161	A differential equation relates a function to its own derivative.
`function-definition`	MATH161	The unknown is a function, not a number.
`rates-of-change`	MATH161	A rate sentence becomes a derivative symbol.

Unlocks

Skill	What it adds
`de-modeling`	Turning a real situation into a differential equation.
`direction-fields`	Seeing the family of solutions as a flow before solving.
`separable-de`	The first general solution method.
`antiderivatives-ivp`	Solving the simplest equations by direct integration.

Official Definitions

Differential equation, unknown, solution

A differential equation is an equation involving an unknown function (for example a velocity $v = v(t)$) and its derivative or derivatives. To solve it is to find a function that works in the equation. A solution that also satisfies a stated initial condition, such as $v(0) = V$, determines the unknown function completely.

Source: Logan, 3rd ed., section 1.1.1, p. 3 (worked from $m\,v'(t) = -k\,v(t)$, $v(0) = V$, Eqs. (1.1) to (1.2)).

Order of a differential equation

The order of a differential equation is the order of the highest derivative appearing in the equation. For example $x' + 2x = t$ is a first-order equation, and $x'' = -x' - 7x$ is a second-order equation.

Source: Logan, 3rd ed., section 1.1.1, p. 4.

Normal form of a first-order equation

A generic first-order equation is written

\[ x' = f(t, x), \]

and this form is called the normal form of a first-order equation. A function $x = x(t)$ is a solution if it satisfies the equation identically on an interval.

Source: Logan, 3rd ed., section 1.1.1, p. 4, Eq. (1.3).

Reading the symbols

Symbol	Meaning	Plain reading
$x = x(t)$	the unknown function	the quantity you are solving for
$x'$ or $\frac{dx}{dt}$	its derivative	how fast that quantity changes
$f(t, x)$	a known rule	what the rate is allowed to depend on
$x(0) = x_0$	initial condition	the starting value that pins down one solution

A first-order equation in normal form reads as a single sentence: the rate of change of $x$ at time $t$ is whatever $f$ says it is, given the current time and the current value of $x$.

Worked Examples

Example 1: Naming the parts

Consider the equation

\[ x'' + 3x' - 10x = \sin t . \]

Predict first. Before reading on, predict the order. The order is the order of the highest derivative present.

Work. The highest derivative is $x''$, a second derivative, so the equation is second order. The unknown is the function $x = x(t)$. The terms $3x'$ and $-10x$ depend on the unknown and its first derivative, while $\sin t$ depends only on the input $t$.

Check. The prediction (second order) matches, because $x''$ outranks $x'$ and $x$.

Example 2: Verifying a proposed solution

Claim: the function $x(t) = e^{2t}$ is a solution of the first-order equation $x' = 2x$.

Predict first. A solution must make the left side equal the right side for every $t$, not just at one point. Predict whether $e^{2t}$ has the property that its derivative equals twice itself.

Work. Differentiate the candidate:

\[ x'(t) = \frac{d}{dt}\,e^{2t} = 2 e^{2t}. \]

Now compute twice the candidate:

\[ 2 x(t) = 2 e^{2t}. \]

The two expressions agree for every value of $t$, so $x(t) = e^{2t}$ satisfies $x' = 2x$ identically on the whole real line.

Check. The prediction holds. Substituting back into the equation gives $2 e^{2t} = 2 e^{2t}$, a true statement, which is the meaning of “is a solution.”

Example 3: From a rate sentence to an equation, then to a solution

A quantity $P = P(t)$ grows so that its rate of change is always three times its current value, and at time $0$ the quantity equals $5$.

Predict first. Predict the form of the answer. A quantity whose rate of change is proportional to itself grows at a percentage rate, which points toward an exponential.

Translate. “Rate of change is three times the current value” becomes

\[ P'(t) = 3\,P(t), \]

and the starting value gives the initial condition $P(0) = 5$.

Solve by checking a family. Try $P(t) = C e^{3t}$ for a constant $C$. Then $P'(t) = 3 C e^{3t} = 3 P(t)$, so every member of this family satisfies the equation. The initial condition selects one member:

\[ P(0) = C e^{0} = C = 5, \]

so $C = 5$ and $P(t) = 5 e^{3t}$.

Check. At $t = 0$ the formula gives $5 e^{0} = 5$, matching the initial condition. Differentiating gives $P'(t) = 15 e^{3t} = 3 \cdot 5 e^{3t} = 3 P(t)$, matching the equation. The initial condition picked out exactly one solution from the whole family, which is what an initial condition is for.

Named Misconceptions

Common misconception

the solution of a differential equation is a number. Students fresh from algebra expect an answer like $x = 4$. A differential equation involves a derivative, so its unknown is a function, and its answer is a function. The check “$x = e^{2t}$ is a solution of $x' = 2x$” is not a numerical check at one point. It is the statement that the two sides agree as functions, for every input. Predict-then-check: predict the type of object the answer will be (function, not number) before solving, and the substitution step will make sense.

Common misconception

(input-output confusion): a rate of change is a single fixed number. In $P' = 3P$, the rate $P'$ is not one number. It is a function that changes as $P$ changes. Early on, the population grows slowly because $P$ is small, and later it grows quickly because $P$ is large, even though the proportionality stays fixed at $3$. Watch the tempting small example: at $t = 0$ the rate is $3 \cdot 5 = 15$, but a moment later $P$ is bigger, so the rate is bigger too. The constant in the equation is the proportionality, not the rate.

Common misconception

the order counts how many terms or how high a power appears. The order of a differential equation counts only the highest derivative, never an algebraic power. The equation $(x')^3 + x = t$ has highest derivative $x'$, so it is first order, even though a first derivative is cubed. Test it on the small case $x'' + (x)^{10} = 0$: the tenth power is on $x$, not on a derivative, and the highest derivative is $x''$, so the equation is second order.

Practice Problems

Level 1 State the order

Give the order of each equation.

(a) $x' + 2x = t$    (b) $x'' = -x' - 7x$    (c) $x''' + x = \cos t$

Show Answer

(a) First order (highest derivative is $x'$). (b) Second order (highest derivative is $x''$). (c) Third order (highest derivative is $x'''$).

The order is the order of the highest derivative present, by the definition above.

Level 2 Identify the unknown and the rule

In the normal-form equation $x' = f(t, x)$ applied to $x' = t - x^2$, name the unknown function, and state what the rate of change is allowed to depend on.

Thought Process

The unknown is whatever function the equation solves for. The right-hand side is the rule $f(t, x)$, so read off which letters appear in it.

Show Answer

The unknown function is $x = x(t)$. The rule is $f(t, x) = t - x^2$, so the rate of change depends on both the current time $t$ and the current value $x$. This is exactly the normal form $x' = f(t, x)$ with that particular $f$.

Level 2 Verify a solution

Show that $x(t) = e^{-7t}$ does not solve $x' = 2x$, and find the value of $k$ for which $x(t) = e^{kt}$ does solve it.

Show Answer

For $x(t) = e^{-7t}$, the derivative is $x'(t) = -7 e^{-7t}$, while $2 x(t) = 2 e^{-7t}$. Since $-7 e^{-7t}$ does not equal $2 e^{-7t}$, the function is not a solution.

For the general form $x(t) = e^{kt}$, the derivative is $x'(t) = k e^{kt}$ and $2 x(t) = 2 e^{kt}$. These agree for all $t$ exactly when $k = 2$. So $x(t) = e^{2t}$ is the exponential solution.

Level 3 Translate a sentence into an equation

A culture of bacteria has mass $m = m(t)$. The mass increases at a rate equal to $0.4$ times the current mass. Write the differential equation, then write the initial condition if the starting mass is $20$ grams.

Thought Process

“Rate of change” is $m'$. “Equal to $0.4$ times the current mass” sets $m'$ equal to $0.4\,m$. The starting mass becomes a condition at $t = 0$.

Show Answer

The equation is

\[ m'(t) = 0.4\,m(t), \]

a first-order equation in normal form with $f(t, m) = 0.4\,m$. The initial condition is $m(0) = 20$.

Level 4 Solve a proportional-growth model

Solve $m'(t) = 0.4\,m(t)$ with $m(0) = 20$ by testing the family $m(t) = C e^{0.4t}$, then state the mass at $t = 5$.

Show Answer

Test the family. For $m(t) = C e^{0.4t}$, the derivative is $m'(t) = 0.4 C e^{0.4t} = 0.4\,m(t)$, so every member satisfies the equation.

Apply the initial condition: $m(0) = C e^{0} = C = 20$, so $C = 20$ and

\[ m(t) = 20\, e^{0.4t}. \]

At $t = 5$, the mass is $m(5) = 20\, e^{0.4 \cdot 5} = 20\, e^{2} \approx 20 \cdot 7.389 \approx 147.8$ grams.

Check. At $t = 0$ the formula gives $20$, matching the start. The derivative is $0.4 \cdot 20 e^{0.4t} = 0.4\,m(t)$, matching the equation.

Level 5 A family with two constants

The second-order equation $x'' = -x$ has solutions of the form $x(t) = A \cos t + B \sin t$. Verify that this family works for any constants $A$ and $B$, then find the member satisfying $x(0) = 2$ and $x'(0) = 3$.

Thought Process

A second-order equation typically needs two conditions to pin down one solution, which is why two free constants appear. Differentiate twice, substitute, then use the two given conditions to solve for $A$ and $B$.

Show Answer

Differentiate twice:

\[ x'(t) = -A \sin t + B \cos t, \qquad x''(t) = -A \cos t - B \sin t. \]

Then $x''(t) = -(A \cos t + B \sin t) = -x(t)$, so the family satisfies $x'' = -x$ for any $A$ and $B$.

Apply the conditions. From $x(0) = A \cos 0 + B \sin 0 = A$, the condition $x(0) = 2$ gives $A = 2$. From $x'(0) = -A \sin 0 + B \cos 0 = B$, the condition $x'(0) = 3$ gives $B = 3$. The member is

\[ x(t) = 2 \cos t + 3 \sin t. \]

Check. $x(0) = 2$ and $x'(0) = 3$, and substituting back gives $x'' = -x$. The two initial conditions were both needed because the equation is second order, so the family had two free constants to fix.

More Than One Valid Path

There is more than one way to confirm that a function solves a differential equation. One way substitutes the candidate into both sides and checks that they agree as functions, as in the worked examples above. Another way rewrites the equation so that one side is zero, then checks that the candidate makes the whole left side simplify to zero. For $x' - 2x$ with $x = e^{2t}$, the second path computes $2 e^{2t} - 2 e^{2t} = 0$. Both paths give the same verdict. Which one feels clearer often depends on how the equation is written, so it is worth trying both and noticing which you prefer.

A wrong turn here is common. Many readers first reach for a single number as the answer, then realize the unknown is a function. That realization is the point of the section, not a sign of being behind. There is no need to rush through it.

Mastery Checklist

Novice (Level 1 to 2):

I can state what a differential equation is in one sentence (an equation involving an unknown function and its derivatives).
I can give the order of an equation by finding the highest derivative.
I can name the unknown function and the rule $f(t, x)$ in normal form.

Competent (Level 3 to 4):

I can verify whether a proposed function is a solution by substituting it into the equation.
I can translate a rate sentence into a first-order equation with an initial condition.
I can solve a proportional-growth model by testing an exponential family and fixing the constant.

Proficient (Level 5):

I can explain why a solution is a function, not a number.
I can explain why a second-order equation needs two initial conditions while a first-order equation needs one.
I can read $x' = f(t, x)$ as a sentence about how a quantity changes given the current time and value.

Connections

Looking back:

Reversing differentiation (the antiderivatives review) is the mechanical core of solving the simplest equations.
The derivative as a function (MATH161) is the object every differential equation talks about.

Looking ahead:

de-modeling builds the equation from a real situation, the step before solving.
direction-fields pictures the whole family of solutions as a flow.
separable-de gives the first systematic solution method.
antiderivatives-ivp solves the equations whose right side depends only on $t$ by direct integration.

Real-world connections:

Population dynamics, radioactive decay, Newtonian cooling, and falling-body motion are all first written as differential equations before they are solved.

Resources

Resource	What it covers
Logan 3rd ed. section 1.1 / 1.1.1 (pp. 2 to 6)	First-order equations, notation and terminology: the primary source for the definitions above.
OpenStax Calculus Vol 2, section 4.1: Basics of Differential Equations	Order, solution, general versus particular solution, openly licensed.
Logan 1.1 tutor guide (`~/math347-ingest/math_guides/MATH_347/Logan DEs/New Guides/Chapter 1/logan_1_1.tex`)	Concept structure for this section.

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Last updated: 2026-06-16