The Central Idea

In ordinary algebra, an equation such as

\[ 2x+3=11 \]

asks us to find a number \(x\).

A differential equation such as

\[ \frac{dy}{dx}=2x \]

asks us to find a function \(y(x)\) whose derivative is \(2x\).

Because

\[ \frac{d}{dx}(x^2+C)=2x, \]

every function of the form

\[ \boxed{y=x^2+C} \]

satisfies the differential equation.

Differential equations therefore move us from finding unknown values to finding unknown functions.

What Is a Differential Equation?

A differential equation is an equation containing an unknown function together with one or more of its derivatives.

Examples include

\[ y'=3x^2, \] \[ \frac{dy}{dt}=5y, \]

and

\[ y''+4y=0. \]

Each equation specifies a rule governing how a function changes.

Ordinary Differential Equations

An ordinary differential equation, or ODE, involves derivatives with respect to a single independent variable.

For example,

\[ \frac{dy}{dx}=x+y \]

describes how \(y\) changes with respect to \(x\).

Likewise,

\[ \frac{dP}{dt}=0.04P \]

describes how a quantity \(P\) changes with respect to time \(t\).

This lesson focuses on ordinary differential equations.

Dependent and Independent Variables

In

\[ \frac{dy}{dx}=x+y, \]

\(x\) is the independent variable, while \(y\) is understood to be a function of \(x\).

We may write this explicitly as

\[ y=y(x). \]

Then

\[ \frac{dy}{dx}=y'(x) \]

measures the instantaneous rate at which the dependent variable changes as the independent variable changes.

What Does First Order Mean?

The order of a differential equation is the order of the highest derivative appearing in the equation.

For example,

\[ y'+3y=x \]

is first order because the highest derivative is \(y'\).

The equation

\[ y''+5y'-6y=0 \]

is second order because the highest derivative is \(y''\).

A first-order differential equation contains no second derivative, third derivative, or higher derivative.

Common First-Order Notation

The first derivative can be written in several equivalent ways.

\[ y', \] \[ \frac{dy}{dx}, \]

or, when time is the independent variable,

\[ \frac{dy}{dt}. \]

In applied work, notation often reflects the meaning of the variable. Population might be written \(P(t)\), temperature \(T(t)\), position \(x(t)\), concentration \(C(t)\), or capital \(K(t)\).

The General First-Order Form

A first-order differential equation may be written abstractly as

\[ F(x,y,y')=0. \]

When the equation can be solved algebraically for \(y'\), it takes the form

\[ \boxed{\frac{dy}{dx}=f(x,y)}. \]

This form is especially useful because \(f(x,y)\) tells us the slope that a solution curve must have at each point \((x,y)\).

A Differential Equation Specifies Slopes

Consider

\[ \frac{dy}{dx}=x+y. \]

At the point \((1,2)\), the required slope is

\[ y'=1+2=3. \]

At \((0,2)\), the slope is

\[ y'=0+2=2. \]

At \((2,-2)\), the slope is

\[ y'=2-2=0. \]

The differential equation therefore assigns a local slope to every point where its right-hand side is defined.

Solutions Are Functions

A function is a solution of a differential equation if substituting the function and its required derivatives into the equation produces a true statement throughout the relevant interval.

Consider

\[ y'=2x. \]

The function

\[ y=x^2+5 \]

is a solution because

\[ y'=2x. \]

The function

\[ y=x^2-11 \]

is also a solution.

General Solutions

A first-order differential equation often has an entire family of solutions.

For

\[ y'=2x, \]

integration gives

\[ y=x^2+C. \]

The arbitrary constant \(C\) represents infinitely many possible solution curves.

This family is called a general solution.

\[ \boxed{y=x^2+C} \]

Why an Arbitrary Constant Appears

Differentiation removes constant terms:

\[ \frac{d}{dx}(x^2+7)=2x, \] \[ \frac{d}{dx}(x^2-4)=2x, \]

and

\[ \frac{d}{dx}(x^2+C)=2x. \]

Consequently, the differential equation \(y'=2x\) cannot determine the vertical position of the curve by itself.

Additional information is needed to select one member of the family.

Particular Solutions

A particular solution is one specific member of a general solution family.

Suppose

\[ y'=2x \]

and we know

\[ y(1)=5. \]

The general solution is

\[ y=x^2+C. \]

Apply the condition:

\[ 5=1^2+C. \]

Therefore,

\[ C=4. \]

The particular solution is

\[ \boxed{y=x^2+4}. \]

Initial Conditions

An initial condition specifies the value of the unknown function at a particular value of the independent variable.

For example,

\[ y(0)=3 \]

says that the solution curve passes through the point

\[ (0,3). \]

Initial conditions frequently arise from the starting state of a physical, biological, financial, or engineering system.

Initial-Value Problems

A differential equation together with one or more initial conditions is called an initial-value problem, often abbreviated IVP.

A first-order initial-value problem typically has the form

\[ \boxed{ \frac{dy}{dx}=f(x,y), \qquad y(x_0)=y_0 }. \]

The differential equation specifies the allowed slopes, while the initial condition specifies which solution curve we want.

Example: Solving a Simple Initial-Value Problem

Solve

\[ \frac{dy}{dx}=6x^2, \qquad y(1)=4. \]

Integrate:

\[ y=2x^3+C. \]

Apply the initial condition:

\[ 4=2(1)^3+C. \]

Thus,

\[ C=2. \]

Therefore,

\[ \boxed{y=2x^3+2}. \]

Verification Is Essential

A proposed solution should be checked against the original differential equation.

Suppose we claim that

\[ y=2x^3+2 \]

solves

\[ y'=6x^2, \qquad y(1)=4. \]

Differentiate:

\[ y'=6x^2. \]

The differential equation is satisfied.

Next check the initial condition:

\[ y(1)=2(1)^3+2=4. \]

Both requirements hold.

Checking a Proposed Solution

Determine whether

\[ y=e^{2x} \]

solves

\[ y'=2y. \]

Differentiate:

\[ y'=2e^{2x}. \]

The right side of the differential equation is

\[ 2y=2e^{2x}. \]

Because both expressions are equal,

\[ \boxed{y=e^{2x}} \]

is a solution.

A Function Can Fail to Be a Solution

Consider the same differential equation

\[ y'=2y. \]

Test

\[ y=e^x. \]

Then

\[ y'=e^x, \]

while

\[ 2y=2e^x. \]

These are not equal.

Therefore,

\[ \boxed{y=e^x\text{ is not a solution of }y'=2y}. \]

Explicit Solutions

An explicit solution expresses the dependent variable directly as a function of the independent variable.

For example,

\[ y=x^2+C \]

is explicit because \(y\) has been isolated.

Explicit form is often convenient for graphing, evaluating, and interpreting the solution.

Implicit Solutions

A solution does not always need to be written with \(y\) isolated.

An equation such as

\[ x^2+y^2=C \]

may describe a family of solution curves implicitly.

Differentiate implicitly:

\[ 2x+2y\frac{dy}{dx}=0. \]

Therefore,

\[ \frac{dy}{dx}=-\frac{x}{y}. \]

Thus

\[ \boxed{x^2+y^2=C} \]

is an implicit family of solutions to

\[ y'=-\frac{x}{y} \]

wherever the derivative is defined.

Solutions Live on Intervals

A differential-equation solution is normally considered on an interval where the function and the expressions appearing in the differential equation are defined.

For example,

\[ y'=\frac{1}{x} \]

leads to

\[ y=\ln|x|+C. \]

Because the differential equation is undefined at \(x=0\), a solution interval cannot simply pass through zero.

Typical intervals include

\[ (-\infty,0) \]

or

\[ (0,\infty). \]

First-Order Equations as Rate Laws

Many first-order differential equations have a natural interpretation as statements about rates.

If

\[ \frac{dP}{dt}=20, \]

then \(P\) increases at a constant rate of \(20\) units per unit time.

If

\[ \frac{dP}{dt}=0.03P, \]

then the rate of change is proportional to the current value of \(P\).

If

\[ \frac{dP}{dt}=0.03P\left(1-\frac{P}{1000}\right), \]

then the rate depends both on the current population and on its relationship to a limiting scale.

These different rate laws produce very different long-term behavior.

Autonomous Differential Equations

A first-order equation is called autonomous when the independent variable does not appear explicitly.

It has the form

\[ \boxed{\frac{dy}{dt}=f(y)}. \]

Examples include

\[ y'=3y, \] \[ y'=y(1-y), \]

and

\[ y'=y^2-4. \]

In an autonomous equation, the rate of change depends only on the current state of the dependent variable.

Equilibrium Solutions

An equilibrium solution is a constant solution of an autonomous differential equation.

For

\[ y'=f(y), \]

equilibrium values occur where

\[ f(y)=0. \]

At such values,

\[ y'=0, \]

so the solution can remain constant.

Example: Finding Equilibria

Consider

\[ y'=y(4-y). \]

Set the right side equal to zero:

\[ y(4-y)=0. \]

Therefore,

\[ y=0 \]

or

\[ y=4. \]

The equilibrium solutions are

\[ \boxed{y(t)=0} \]

and

\[ \boxed{y(t)=4}. \]

Positive and Negative Derivatives Tell Us Direction

If

\[ y'>0, \]

then a solution is locally increasing.

If

\[ y'<0, \]

then a solution is locally decreasing.

If

\[ y'=0, \]

then the solution has a horizontal tangent at that point.

A differential equation can therefore provide qualitative information even before we obtain an explicit solution formula.

Example: Qualitative Analysis

Consider

\[ y'=y(4-y). \]

If

\[ 0 then both \(y\) and \(4-y\) are positive, so

\[ y'>0. \]

Solutions increase.

If

\[ y>4, \]

then \(y>0\) but \(4-y<0\), so

\[ y'<0. \]

Solutions decrease.

This already suggests that the equilibrium \(y=4\) plays an important role in the system.

Stable and Unstable Behavior

An equilibrium is informally called stable when nearby solutions tend to move toward it.

It is called unstable when nearby solutions tend to move away from it.

For

\[ y'=y(4-y), \]

solutions immediately below \(4\) increase toward \(4\), while solutions immediately above \(4\) decrease toward \(4\).

Thus \(y=4\) behaves as a stable equilibrium.

More systematic stability analysis will become important when we study systems of differential equations and phase-plane representations.

Slope Fields

A slope field, also called a direction field, provides a visual representation of a first-order differential equation

\[ y'=f(x,y). \]

At many points \((x,y)\), we draw a short line segment whose slope equals

\[ f(x,y). \]

The resulting field shows the directions that solution curves must follow.

Reading a Slope Field

Suppose the equation is

\[ y'=x-y. \]

At \((0,0)\),

\[ y'=0. \]

At \((1,0)\),

\[ y'=1. \]

At \((0,1)\),

\[ y'=-1. \]

Each point contributes a small directional segment. A solution curve must pass through the field while remaining tangent to these local directions.

Isoclines

An isocline is a curve along which the differential equation assigns the same slope.

For

\[ y'=x-y, \]

points having slope \(0\) satisfy

\[ x-y=0. \]

Therefore,

\[ y=x \]

is the zero-slope isocline.

Points having slope \(2\) satisfy

\[ x-y=2, \]

or

\[ y=x-2. \]

Isoclines help organize the geometry of a slope field.

Differential Equations Need Not Be Solved to Be Useful

An explicit formula is valuable, but it is not the only useful outcome.

From a differential equation we may be able to determine:

Differential equations are therefore both computational and qualitative tools.

Classification Helps Choose a Method

First-order equations appear in several important forms.

A separable equation can be rearranged so that the variables can be separated.

A linear first-order equation can be written as

\[ y'+P(x)y=Q(x). \]

An exact equation has a structure connected to the total differential of a function of two variables.

Recognizing the structure of an equation is often the first step toward selecting an efficient solution method.

Separable Structure

A first-order equation is separable when it can be expressed in a form such as

\[ \frac{dy}{dx}=g(x)h(y). \]

For example,

\[ \frac{dy}{dx}=xy \]

is separable because its dependence on \(x\) and \(y\) occurs as a product.

The next lesson, Separable and Linear Equations, develops this method systematically.

Linear First-Order Structure

A first-order differential equation is linear in \(y\) when it can be written

\[ \boxed{y'+P(x)y=Q(x)}. \]

For example,

\[ y'+3y=x \]

is linear.

By contrast,

\[ y'+y^2=x \]

is not linear in \(y\) because \(y\) appears squared.

Linear Does Not Mean the Graph Is a Line

The word linear in differential equations describes the algebraic way the unknown function and its derivatives appear.

It does not mean that the solution graph must be a straight line.

For example,

\[ y'-2y=0 \]

is a linear first-order differential equation, but its nonzero solutions are exponential:

\[ y=Ce^{2x}. \]

Nonlinear First-Order Equations

A first-order equation is nonlinear when the dependent variable or its first derivative appears in a nonlinear way.

Examples include

\[ y'=y^2-x, \] \[ y'=\sin y+x, \]

and

\[ (y')^2+y=x. \]

Nonlinear equations can display behaviors not present in simple linear models, including multiple equilibria, thresholds, and more complicated long-term dynamics.

Growth as a Differential Equation

Suppose a quantity grows at a rate proportional to its current size.

If \(P(t)\) represents the quantity, the verbal statement becomes

\[ \frac{dP}{dt}=kP, \]

where \(k\) is a constant of proportionality.

When

\[ k>0, \]

the model describes exponential growth.

When

\[ k<0, \]

it describes exponential decay.

Growth and decay models receive their own focused treatment later in this section.

Logistic Change

Unlimited exponential growth is not appropriate for every system.

A common model with a limiting level is

\[ \frac{dP}{dt} = rP\left(1-\frac{P}{K}\right), \]

where \(r\) is a growth-rate parameter and \(K\) is a limiting scale often called the carrying capacity.

This is still a first-order differential equation, but its behavior is nonlinear.

Logistic models will be developed in a later lesson.

From Verbal Rules to Differential Equations

Many applications begin with a verbal statement about change.

The statement

The rate of change of \(Q\) is proportional to \(Q\)

becomes

\[ \frac{dQ}{dt}=kQ. \]

The statement

The rate of change of \(T\) is proportional to the difference between \(T\) and an environmental temperature \(T_e\)

can be represented as

\[ \frac{dT}{dt}=-k(T-T_e). \]

Translating relationships into rate equations is one of the central modeling skills of differential equations.

Units Provide a Powerful Check

Differential equations must be dimensionally consistent.

If \(P\) is measured in people and \(t\) in years, then

\[ \frac{dP}{dt} \]

has units of people per year.

In

\[ \frac{dP}{dt}=kP, \]

the product \(kP\) must have the same units.

Therefore, \(k\) has units of

\[ \frac{1}{\text{year}}. \]

Unit analysis often detects modeling errors before any calculus is performed.

Example: Constant Accumulation

Suppose water enters a reservoir at a net rate of \(12\) cubic meters per hour.

If \(V(t)\) is the volume, then

\[ \frac{dV}{dt}=12. \]

Integrating gives

\[ V(t)=12t+C. \]

If

\[ V(0)=300, \]

then

\[ C=300. \]

Therefore,

\[ \boxed{V(t)=300+12t}. \]

Example: Motion from Velocity

Suppose an object's velocity is

\[ v(t)=3t^2-4. \]

Since velocity is the derivative of position,

\[ \frac{ds}{dt}=3t^2-4. \]

Integrating gives

\[ s=t^3-4t+C. \]

If

\[ s(0)=10, \]

then

\[ C=10. \]

Thus,

\[ \boxed{s(t)=t^3-4t+10}. \]

Existence of a Solution

Before trying to calculate a solution, it is reasonable to ask whether a solution actually exists.

Consider an initial-value problem

\[ y'=f(x,y), \qquad y(x_0)=y_0. \]

Under appropriate continuity conditions on \(f\) near \((x_0,y_0)\), a solution exists at least on some interval around \(x_0\).

The precise hypotheses matter, but the basic message is important: differential-equation formulas are not automatically guaranteed to generate well-behaved solutions everywhere.

Uniqueness of a Solution

A second question is whether the initial condition determines exactly one solution.

For many equations of the form

\[ y'=f(x,y), \]

suitable regularity conditions on \(f\), often including sufficient control of its dependence on \(y\), guarantee local uniqueness.

Informally, when uniqueness holds, two different solution curves cannot pass through the same initial point and then immediately follow different paths.

A Useful Introductory Existence-Uniqueness Criterion

For an initial-value problem

\[ y'=f(x,y), \qquad y(x_0)=y_0, \]

a commonly used introductory criterion says that if \(f\) and

\[ \frac{\partial f}{\partial y} \]

are continuous in a rectangle containing \((x_0,y_0)\), then the problem has a unique solution on some interval containing \(x_0\).

This is a local statement. It does not automatically guarantee that the solution can be extended for all possible values of \(x\).

Solutions Can Develop Singular Behavior

Consider

\[ y'=y^2. \]

One family of nonzero solutions is

\[ y=-\frac{1}{x+C}. \]

Such a solution becomes unbounded when

\[ x=-C. \]

Thus even a simple first-order equation can produce a solution that cannot be continued through every value of the independent variable.

The interval of validity is part of the mathematical solution.

Exact and Approximate Solutions

Some differential equations can be solved symbolically in closed form.

Others cannot be expressed conveniently using elementary functions.

In those cases, we may use:

Differential-equation analysis therefore combines symbolic, graphical, and numerical methods.

Euler's Method

One of the simplest numerical methods for

\[ y'=f(x,y) \]

is Euler's method.

Beginning from a known point

\[ (x_n,y_n), \]

we use the current slope

\[ f(x_n,y_n) \]

to estimate the next value.

With step size \(h\),

\[ \boxed{ y_{n+1} = y_n+h\,f(x_n,y_n) }. \]

Where Euler's Formula Comes From

The derivative gives a local linear approximation:

\[ \Delta y\approx y'\Delta x. \]

If the horizontal step is

\[ \Delta x=h, \]

and

\[ y'=f(x_n,y_n), \]

then

\[ \Delta y\approx h f(x_n,y_n). \]

Therefore,

\[ y_{n+1} \approx y_n+h f(x_n,y_n). \]

Euler's method repeatedly applies this local tangent-line approximation.

Example: One Euler Step

Suppose

\[ y'=x+y, \qquad y(0)=1. \]

Use step size

\[ h=0.1. \]

At the initial point,

\[ f(0,1)=0+1=1. \]

Euler's method gives

\[ y_1 = 1+0.1(1). \]

Therefore,

\[ \boxed{y(0.1)\approx1.1}. \]

Example: A Second Euler Step

Continue from

\[ (0.1,1.1). \]

The new slope is

\[ f(0.1,1.1)=0.1+1.1=1.2. \]

Therefore,

\[ y_2 = 1.1+0.1(1.2). \]

Thus,

\[ \boxed{y(0.2)\approx1.22}. \]

Repeating the process creates a numerical approximation to the solution curve.

Step Size Matters

Euler's method approximates a curved solution using short tangent-line movements.

A smaller step size generally follows the changing slope more closely, but requires more computational steps.

A larger step size requires less computation but can accumulate more error.

This creates a recurring numerical-analysis tradeoff between computational cost and accuracy.

Local and Global Numerical Error

Numerical methods introduce approximation error.

Local error concerns the error introduced during an individual step.

Global error concerns the accumulated difference between the numerical approximation and the true solution after many steps.

More advanced numerical methods are designed to achieve greater accuracy than basic Euler stepping.

These methods will be developed in Numerical Solution Methods.

Analytical, Qualitative, and Numerical Views

A single differential equation can be studied from several perspectives.

Analytical methods seek formulas.

Qualitative methods examine signs, equilibria, stability, and geometry.

Numerical methods calculate approximate solution values.

These methods complement one another rather than compete with one another.

Parameters and Solution Behavior

Differential equations frequently contain parameters.

Consider

\[ y'=ky. \]

The parameter \(k\) changes the behavior of the entire solution family.

If \(k>0\), nonzero positive solutions grow.

If \(k<0\), they decay toward zero.

If \(k=0\), then

\[ y'=0, \]

so every solution is constant.

Parameter analysis is therefore central to mathematical modeling.

State, Rate, and Parameter

It is useful to distinguish three roles in an equation such as

\[ \frac{dP}{dt}=kP. \]

The quantity

\[ P \]

is the current state.

The quantity

\[ \frac{dP}{dt} \]

is the rate of change of that state.

The quantity

\[ k \]

is a parameter controlling the relationship between state and rate.

This state-rate-parameter structure appears throughout applied mathematics.

Common Error: Treating \(y'\) as an Independent Variable

In

\[ y'=x+y, \]

\(y'\) is not an unrelated algebraic symbol.

It represents the derivative of the same unknown function \(y(x)\).

A valid solution must therefore satisfy the relationship

\[ y'(x)=x+y(x) \]

throughout its solution interval.

Common Error: Forgetting the Constant of Integration

If

\[ y'=4x^3, \]

then integration gives

\[ \boxed{y=x^4+C}, \]

not merely

\[ y=x^4. \]

The constant represents the family of possible solution curves.

Common Error: Ignoring the Initial Condition

Suppose

\[ y'=4x^3, \qquad y(1)=7. \]

The expression

\[ y=x^4+C \]

is only the general solution.

Use the initial condition:

\[ 7=1+C. \]

Thus,

\[ C=6, \]

and the requested particular solution is

\[ \boxed{y=x^4+6}. \]

Common Error: Checking Only the Initial Condition

A function can pass through the required initial point without satisfying the differential equation.

For example, both

\[ y=x^2+1 \]

and

\[ y=x^3+1 \]

satisfy

\[ y(0)=1. \]

But only the function whose derivative matches the differential equation can be the correct solution.

Always verify both the differential equation and the initial condition.

Common Error: Confusing Order with Exponents

The order of a differential equation is determined by the highest derivative order, not by ordinary exponents.

The equation

\[ y'+y^5=x \]

is still first order because the highest derivative is \(y'\).

The exponent \(5\) makes the equation nonlinear, but it does not make the equation fifth order.

Common Error: Assuming Every Equation Has an Elementary Formula

Many important differential equations cannot be expressed using a simple combination of familiar elementary functions.

Failure to find a closed-form solution does not mean the model is useless.

We may still analyze its direction field, equilibria, stability, asymptotic behavior, and numerical approximations.

Common Error: Ignoring the Domain

Suppose a solution contains

\[ \ln|x|. \]

The solution cannot be treated as though it were defined at

\[ x=0. \]

Similarly, rational expressions may introduce denominator restrictions, and square roots or logarithms may impose additional domain conditions.

A complete differential-equation solution includes an appropriate interval of validity.

A Reliable First-Order Workflow

  1. Identify the independent and dependent variables.
  2. Determine the order of the differential equation.
  3. Rewrite the equation in a useful form when possible.
  4. Inspect the equation for recognizable structure.
  5. Check for equilibrium solutions when appropriate.
  6. Use an analytical method if one applies.
  7. Apply the initial condition or other supplied data.
  8. Determine an appropriate interval of validity.
  9. Differentiate and substitute to verify the solution.
  10. Interpret the result in the context of the model.
  11. Use qualitative or numerical analysis when an explicit formula is unavailable.

Example: Full Workflow

Solve

\[ y'=3x^2-2x, \qquad y(2)=5. \]

This is a first-order ordinary differential equation.

Integrate:

\[ y=x^3-x^2+C. \]

Apply the initial condition:

\[ 5=2^3-2^2+C. \] \[ 5=8-4+C. \] \[ C=1. \]

Therefore,

\[ \boxed{y=x^3-x^2+1}. \]

Verify:

\[ y'=3x^2-2x. \]

Also,

\[ y(2)=8-4+1=5. \]

Both conditions are satisfied.

Example: Recognizing an Equilibrium Without Solving

Consider

\[ y'=(y-2)(y+5). \]

Equilibria occur where

\[ (y-2)(y+5)=0. \]

Therefore,

\[ y=2 \]

or

\[ y=-5. \]

Thus

\[ \boxed{y(t)=2} \]

and

\[ \boxed{y(t)=-5} \]

are constant solutions.

We obtained useful information without solving the complete differential equation.

Example: Sign Analysis Without Solving

Continue with

\[ y'=(y-2)(y+5). \]

If

\[ y>2, \]

both factors are positive, so

\[ y'>0. \]

If

\[ -5 one factor is negative and the other positive, so

\[ y'<0. \]

If

\[ y<-5, \]

both factors are negative, so

\[ y'>0. \]

These signs describe the direction of solution motion in each region.

Differential Equations as Dynamic Models

Algebraic equations typically describe relationships among quantities at a particular state.

Differential equations add a dynamic layer by describing how those quantities evolve.

This makes them foundational in:

The Modeling Cycle

In applications, solving the equation is only one stage of a larger modeling process.

  1. Identify the changing quantities.
  2. Choose state variables and parameters.
  3. Translate assumptions into rate relationships.
  4. Construct the differential equation.
  5. Specify initial or boundary information.
  6. Analyze or approximate the solution.
  7. Interpret the mathematical behavior.
  8. Compare predictions with the real system.
  9. Revise assumptions or parameters when necessary.

Differential equations therefore connect calculus directly to mathematical modeling.

Practice 1: Identify the Order

Determine the order of

\[ y'+4y=x^2. \]

Practice 2: Identify the Order

Determine the order of

\[ y''+3y'-7y=0. \]

Practice 3: Verify a Solution

Determine whether

\[ y=5e^{3x} \]

solves

\[ y'=3y. \]

Practice 4: Reject an Incorrect Solution

Determine whether

\[ y=x^3 \]

solves

\[ y'=2x. \]

Practice 5: Find a General Solution

Solve

\[ y'=8x^3. \]

Practice 6: Initial-Value Problem

Solve

\[ y'=4x, \qquad y(1)=7. \]

Practice 7: Find the Equilibria

Find all equilibrium solutions of

\[ y'=y(6-y). \]

Practice 8: Determine Increasing and Decreasing Regions

For

\[ y'=y(6-y), \]

determine the sign of \(y'\) when:

  1. \(y<0\);
  2. \(0
  3. \(y>6\).

Practice 9: Translate a Rate Statement

A population \(P(t)\) changes at a rate proportional to the current population.

Write a differential equation representing this statement.

Practice 10: One Euler Step

For

\[ y'=x+2y, \qquad y(0)=1, \]

use Euler's method with

\[ h=0.1 \]

to approximate \(y(0.1)\).

Practice 11: Units

Suppose \(M(t)\) is measured in kilograms and \(t\) is measured in days.

In the model

\[ \frac{dM}{dt}=kM, \]

what units must \(k\) have?

Practice 12: Initial Condition

The general solution of a differential equation is

\[ y=3x^2+C. \]

Find the particular solution satisfying

\[ y(2)=17. \]

Practice Answers

Practice 1

The highest derivative is \(y'\), so the equation is

\[ \boxed{\text{first order}}. \]

Practice 2

The highest derivative is \(y''\), so the equation is

\[ \boxed{\text{second order}}. \]

Practice 3

Differentiate:

\[ y'=15e^{3x}. \]

Also,

\[ 3y=3(5e^{3x})=15e^{3x}. \]

Therefore,

\[ \boxed{y=5e^{3x}\text{ is a solution}}. \]

Practice 4

If

\[ y=x^3, \]

then

\[ y'=3x^2. \]

This does not equal \(2x\) in general.

Therefore,

\[ \boxed{y=x^3\text{ is not a solution}}. \]

Practice 5

Integrate:

\[ y=\int 8x^3\,dx. \]

Therefore,

\[ \boxed{y=2x^4+C}. \]

Practice 6

Integrate:

\[ y=2x^2+C. \]

Apply

\[ y(1)=7. \]

Then

\[ 7=2+C, \]

so

\[ C=5. \]

Therefore,

\[ \boxed{y=2x^2+5}. \]

Practice 7

Set

\[ y(6-y)=0. \]

Thus,

\[ y=0 \]

or

\[ y=6. \]

The equilibrium solutions are

\[ \boxed{y(t)=0} \]

and

\[ \boxed{y(t)=6}. \]

Practice 8

For

\[ y'=y(6-y): \]

if \(y<0\), then

\[ \boxed{y'<0}; \]

if \(0 \[ \boxed{y'>0}; \]

if \(y>6\), then

\[ \boxed{y'<0}. \]

Thus solutions move upward between \(0\) and \(6\) and downward above \(6\).

Practice 9

If the rate of change is proportional to the current population, then

\[ \boxed{\frac{dP}{dt}=kP}, \]

where \(k\) is a proportionality constant.

Practice 10

Euler's method uses

\[ y_{n+1}=y_n+h f(x_n,y_n). \]

Here,

\[ f(x,y)=x+2y. \]

At \((0,1)\),

\[ f(0,1)=0+2(1)=2. \]

Therefore,

\[ y_1=1+0.1(2). \]

Thus,

\[ \boxed{y(0.1)\approx1.2}. \]

Practice 11

The derivative

\[ \frac{dM}{dt} \]

has units of kilograms per day.

Since

\[ kM \]

must have the same units and \(M\) is measured in kilograms, \(k\) must have units of

\[ \boxed{\frac{1}{\text{day}}}. \]

Practice 12

Start with

\[ y=3x^2+C. \]

Apply

\[ y(2)=17: \] \[ 17=3(2)^2+C. \] \[ 17=12+C. \]

Therefore,

\[ C=5. \]

The particular solution is

\[ \boxed{y=3x^2+5}. \]

What to Remember

A first-order differential equation involves an unknown function and its first derivative, but no higher derivative.

A common form is

\[ \boxed{y'=f(x,y)}. \]

The equation tells us the slope that a solution must have at each point.

The central viewpoint is:

\[ \boxed{ \text{state} \longrightarrow \text{rate law} \longrightarrow \text{evolution} }. \]

A differential equation specifies how the current state of a system governs its change, and the resulting solution describes how that system evolves.

Where We Go Next

First-order differential equations come in several recognizable forms, and their structure determines which analytical technique is most efficient.

The next lesson, Separable and Linear Equations, develops two of the most important first-order solution methods: separation of variables and the integrating-factor method for linear equations.

Those techniques will provide the computational foundation for later models involving growth, decay, logistic behavior, forcing, coupled systems, and numerical approximation.