Mathematical modeling / English

The Method of Integrating Factors for Linear 1st Order ODEs full example

Dr. Trefor Bazett · YouTube · 5:07

YouTube · Dr. Trefor BazettOriginal video

IN THIS VIDEO

01Standard Form of a First-Order Linear ODE
02Integrating Factor Formula
03Applying the Integrating Factor

FROM THE LEARNING NOTES

y' + p(x)y = f(x)

Standard Form of a First-Order Linear ODE

SELECTED MOMENT

Applying the Initial Condition

SUGGESTED PREPARATION

First-order derivatives
Indefinite integration
Product rule for differentiation
Exponent rules
Product Rule for Differentiation
Definition of Integrating Factor
Basic Indefinite Integration Rules

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Reviewed learning material · Video analysis · English

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This video segment introduces the method of integrating factors for solving first-order linear differential equations, using the initial value problem y' + $4y = e^{-x}$ , $y(0) = 4/3$ as an example. The instructor identifies the standard form, calculates the integrating factor $r(x) = e^{4x}$ , multiplies the entire equation by it, and begins to simplify both sides, demonstrating how the left side becomes the exact derivative of the product $e^{4x}y$ . This video segment demonstrates the final stages of solving a first-order linear ordinary differential equation using the integrating factor method. Starting from the exact derivative form $d/dx[e^{4x}y] = e^{3x}$ , the instructor integrates both sides to obtain the general solution $y = 1/3 e^{-x}$ + Ce^{-4x}. A key pedagogical point is made regarding the inclusion of the constant of integration (+C), distinguishing its optional nature when finding an integrating factor from its necessity when solving the differential equation. Finally, the initial condition $y(0) = 4/3$ is applied to determine $C=1$ , yielding the particular solution $y = 1/3 e^{-x} + e^{-4x}$ .

Use the learning inspector for key ideas and moments, or open the reading tabs for the complete notes.

Chapters

0:00Introduction to Integrating Factors 0:20Standard Form of Linear ODEs 1:08Calculating the Integrating Factor 1:40Multiplying the Equation 2:23Simplifying via the Product Rule 2:45Integrating Both Sides 3:08The Role of the Constant of Integration 3:38Finding the General Solution 4:05Applying the Initial Condition 4:41Final Particular Solution

Learning script

Generated from the video's visuals and explanation; not verbatim speech.

The video begins by introducing the method of integrating factors as a technique for solving first-order linear differential equations. The specific example presented on the board is the initial value problem $y' + 4y = e^{-x}$ with the condition $y(0) = \frac{4}{3}$ .

The instructor explains the criteria for a linear first-order equation. By adding an implicit $1$ to the right side, the equation is viewed as $1 \cdot y' + 4 \cdot y = e^{-x} \cdot 1$ . Because the coefficients of $y'$ , $y$ , and $1$ are all functions of $x$ and the variables are to the first power, it fits the standard form. The coefficient of $y$ is labeled $p(x)$ , which is $4$ , and the right-hand side is labeled $f(x)$ , which is $e^{-x}$ .

Next, the formula for the integrating factor, $r(x) = e^{\int p(x) dx}$ , is stated. For this problem, substituting $p(x) = 4$ yields $r(x) = e^{\int 4 dx} = e^{4x}$ . The instructor clarifies that the constant of integration is omitted because any valid integrating factor will suffice, so $C=0$ is chosen for simplicity.

The core step of the method is then applied: multiplying every term in the original differential equation by the integrating factor $e^{4x}$ . This results in the equation $e^{4x}y' + 4e^{4x}y = e^{4x}e^{-x}$ . The right-hand side is simplified using exponent rules to $e^{3x}$ .

Finally, the instructor focuses on the left-hand side, $e^{4x}y' + 4e^{4x}y$ . He points out that this expression is the exact result of applying the product rule to the derivative of $e^{4x}y$ . Thus, the left side is rewritten as $\frac{d}{dx}[e^{4x}y]$ , setting up the equation for direct integration in subsequent steps.

We begin with the equation in the form of a total derivative on the left side: $d/dx[e^{4x}y] = e^{3x}$ . To solve for y, we integrate both sides with respect to x.

On the left side, the integral of the derivative cancels out, leaving simply $e^{4x}y$ . On the right side, we compute the indefinite integral of $e^{3x}$ , which results in ( $1/3$ ) $e^{3x}$ .

Crucially, we must add the constant of integration, +C, to the right side. While finding the integrating factor earlier allowed us to ignore constants, solving the differential equation requires the general family of solutions represented by C.

Now we have $e^{4x}y = (1/3)e^{3x} + C$ . To isolate y, we multiply the entire equation by $e^{-4x}$ . Distributing this term gives us the general solution: $y = (1/3)e^{-x}$ + Ce^{-4x}.

Next, we apply the initial condition given in the problem: $y(0) = 4/3$ . Substituting $x=0$ and $y=4/3$ into our general solution allows us to solve for the specific value of C.

The equation becomes $4/3 = (1/3)e^0$ + Ce^0. Since $e^0 = 1$ , this simplifies to $4/3 = 1/3 + C$ . Subtracting $1/3$ from both sides yields $C = 1$ .

Finally, substituting $C=1$ back into the general solution, we arrive at the particular solution: $y = (1/3)e^{-x} + e^{-4x}$ . This completes the method of integrating factors for this initial value problem.

Knowledge cards

Standard Form of a First-Order Linear ODE

A first-order linear differential equation can be written as $y' + p(x)y = f(x)$ . In this form, the dependent variable $y$ and its derivative $y'$ appear only to the first power, and their coefficients, $p(x)$ and $f(x)$ , are functions of the independent variable $x$ only.

y' + p(x)y = f(x)

Integrating Factor Formula

The integrating factor $r(x)$ is defined as the exponential of the indefinite integral of $p(x)$ . When calculating it, the constant of integration is typically set to zero, as any non-zero multiple of an integrating factor also works.

r(x) = e^{\int p(x) dx}

Applying the Integrating Factor

To solve the differential equation, multiply every term on both sides by the integrating factor $r(x)$ . This transforms the left-hand side into the exact derivative of the product $r(x)y$ , allowing both sides to be integrated directly with respect to $x$ .

\frac{d}{dx}[r(x)y] = r(x)f(x)

Integrating the Derivative Side

When solving the equation $d/dx[μ(x)y] = g(x)$ , integrating both sides with respect to x removes the derivative on the left, leaving $μ(x)y$ . This relies on the Fundamental Theorem of Calculus.

\int \frac{d}{dx}[\mu(x)y] dx = \mu(x)y

Integral of Exponential Functions

The standard rule for integrating an exponential function with a linear exponent is $\int e$ ^{kx} $dx = (1/k)e$ ^{kx} + C. In this example, $k=3$ .

\int e^{3x} dx = \frac{1}{3}e^{3x} + C

When to Include +C

Constants of integration can be omitted when finding an integrating factor because any non-zero multiple works. However, they are mandatory when integrating the differential equation itself to define the general solution space.

Isolating y for General Solution

After integration, the equation is $μ(x)y = \int$ ... + C. To get $y(x)$ , multiply both sides by $1/μ(x)$ (or $e^{-\int P(x)dx}$ ) and distribute.

y = \frac{1}{\mu(x)} \left[ \int \mu(x)f(x) dx + C \right]

Solving for C using Initial Conditions

Substitute the given $x_0$ and $y_0$ values into the general solution. Solve the resulting algebraic equation for C to find the specific constant that satisfies the initial state.

y(x_0) = y_0 \rightarrow C = \dots

Particular Solution

The final answer to an Initial Value Problem (IVP) is the specific function obtained by plugging the calculated C back into the general solution.

y = \frac{1}{3}e^{-x} + e^{-4x}

Detailed learning notes

Explore conditions, steps and evidence. Supplementary explanations are labeled separately from content shown in the video.

Symbols · 15

y'

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board shows y' in the equation y' + $4y = e^{-x}$ .

Symbol

Meaning

Derivative of the unknown function y with respect to x.

Domain

Function of x

y

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board shows y in the equation y' + $4y = e^{-x}$ and in the initial condition $y(0) = 4/3$ .

Symbol

Meaning

Unknown dependent variable or function to be solved for.

Domain

Function of x

x

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board shows x as the independent variable in $e^{-x}$ , $p(x)$ , $f(x)$ , and $r(x)$ .

Symbol

Meaning

Independent variable.

Domain

Real numbers

$p(x)$

Clear evidence

Shown in the video

Evidence

Formula
Observation
The instructor writes $p(x)$ under the coefficient 4 in the equation y' + $4y = e^{-x}$ .
Audio
Observation
The instructor states that the coefficient function of y is typically called $p(x)$ .

Symbol

$p(x)$

Meaning

Coefficient function of y in the standard form of a first-order linear differential equation.

Domain

Function of x

$f(x)$

Clear evidence

Shown in the video

Evidence

Formula
Observation
The instructor writes $f(x)$ under the term $e^{-x}$ on the right-hand side of the equation.
Audio
Observation
The instructor states that the coefficient function of the one is referred to as $f(x)$ .

Symbol

$f(x)$

Meaning

Non-homogeneous term or forcing function in the standard form of a first-order linear differential equation.

Domain

Function of x

$r(x)$

Clear evidence

Shown in the video

Evidence

Formula
Observation
The instructor writes $r(x) = e^{\int p(x) dx}$ .
Audio
Observation
The instructor refers to $r(x)$ as the integrating factor.

Symbol

$r(x)$

Meaning

Integrating factor used to solve the first-order linear differential equation.

Domain

Function of x

e

Clear evidence

Shown in the video

Evidence

Formula
Observation
The base of the natural exponential function is written as e in $r(x) = e^{\int p(x) dx}$ .

Symbol

Meaning

Base of the natural logarithm.

Domain

Mathematical constant

$\int$

Clear evidence

Shown in the video

Evidence

Formula
Observation
The integral sign $\int$ is used in the expression for the integrating factor.

Symbol

$\int$

Meaning

Indefinite integral operator.

Domain

Calculus operator

dx

Clear evidence

Shown in the video

Evidence

Formula
Observation
The differential dx is written at the end of the integral in the integrating factor formula.

Symbol

Meaning

Differential indicating integration with respect to x.

Domain

Calculus notation

$\frac{4}{3}$

Clear evidence

Shown in the video

Evidence

Formula
Observation
The initial condition is written as $y(0) = 4/3$ .

Symbol

$\frac{4}{3}$

Meaning

Initial value of the function y at $x = 0$ .

Domain

Rational number

$\frac{d}{dx}$

Clear evidence

Shown in the video

Evidence

Formula
Observation
The operator $\frac{d}{dx}$ is written on the board to denote differentiation with respect to x.

Symbol

$\frac{d}{dx}$

Meaning

Derivative operator with respect to x.

Domain

Calculus operator

y

Clear evidence

Shown in the video

Evidence

Formula
Observation
The symbol y appears throughout the worked solution as the dependent variable in expressions such as $e^{4x}y$ and $y(0)=4/3$ .

Symbol

Meaning

Dependent variable (unknown function) of the first-order linear ODE.

Domain

Function of x

Knowledge points · 9

Standard Form of a First-Order Linear Differential Equation

Clear evidence

Shown in the video

Evidence

Audio
Observation
The instructor explains that the equation is a linear first-order equation because the coefficients of y', y, and 1 are functions of x.
Formula
Observation
The instructor labels the coefficient of y as $p(x)$ and the right-hand side as $f(x)$ .

Definition

Explanation

A first-order linear differential equation has the form y' + $p(x)y = f(x)$ , where the dependent variable y and its derivative y' appear to the first power, and their coefficients are functions of the independent variable x.

Formula

y' + p(x)y = f(x)

Conditions

The equation must be first-order (highest derivative is y').
The equation must be linear in y and y' (they are raised to the first power and not multiplied together).

Formula for the Integrating Factor

Clear evidence

Shown in the video

Evidence

Audio
Observation
The instructor states the formula for the integrating factor.
Formula
Observation
The instructor writes $r(x) = e^{\int p(x) dx}$ .

Formula

Explanation

The integrating factor $r(x)$ is calculated by taking the exponential of the indefinite integral of the coefficient function $p(x)$ . An arbitrary constant of integration is typically omitted because any valid integrating factor will work.

Formula

r(x) = e^{\int p(x) dx}

Conditions

The differential equation must be in the standard form y' + $p(x)y = f(x)$ .

Prerequisites

Standard Form of a First-Order Linear Differential Equation

Multiplying the Differential Equation by the Integrating Factor

Clear evidence

Shown in the video

Evidence

Audio
Observation
The instructor explains that the method involves multiplying every single term by the integrating factor.
Formula
Observation
The instructor writes $e^{4x}y$ ' + $e^{4x}4y = e^{4x}e^{-x}$ .

Method

Explanation

To apply the method of integrating factors, multiply both sides of the standard form differential equation by the calculated integrating factor $r(x)$ .

Formula

r(x)y' + r(x)p(x)y = r(x)f(x)

Conditions

The integrating factor $r(x)$ must be correctly calculated from $p(x)$ .

Prerequisites

Formula for the Integrating Factor

Integrating a Derivative

Clear evidence

Shown in the video

Evidence

Audio
Observation
Speaker states: 'now I'm going to integrate and that's going to give me on the left e to the 4x times y all I'm doing here is just integrating a derivative so it cancels'.
Formula
Observation
Board shows transition from $d/dx[e^{4x}y] = e^{3x}$ to $e^{4x}y = \int e^{3x}dx$ .

Method

Explanation

When both sides of an equation are integrated with respect to x, the integral of a derivative cancels out, leaving the original function. Here, $\int (d/dx)[e^{4x}y] dx = e^{4x}y$ .

Formula

\int \frac{d}{dx}[f(x)] dx = f(x)

Conditions

The integrand must be a total derivative with respect to the variable of integration.

Integral of Exponential Function

Clear evidence

Shown in the video

Evidence

Audio
Observation
Speaker states: 'this is the same thing as the integral of e to the 3x dx which is one third times e to the 3x'.
Formula
Observation
Board shows $\int e^{3x}dx = 1/3 e^{3x} + C$ .

Formula

Explanation

The indefinite integral of an exponential function e^{kx} is ( $1/k$ )e^{kx} plus a constant of integration.

Formula

\int e^{kx} dx = \frac{1}{k}e^{kx} + C

Conditions

k is a non-zero constant.

Necessity of Constant of Integration

Clear evidence

Shown in the video

Evidence

Audio
Observation
Speaker explains: 'you didn't need to write down the plus c earlier because it was all about finding an integrating factor... But now I'm going to my solution and what I've gotten is this expression that has an indefinite integral I do want my plus c and my plus c will be useful for my initial condition'.

Method

Explanation

When finding an integrating factor, any valid factor works, so the constant of integration can be omitted. However, when solving the differential equation via indefinite integration, the constant C must be included to form the general solution and apply initial conditions.

Formula

Conditions

Applies when transitioning from finding an integrating factor to solving the full ODE.

Prerequisites

Integral of Exponential Function

Isolating the Dependent Variable

Clear evidence

Shown in the video

Evidence

Audio
Observation
Speaker states: 'I notice that I have this e to the 4x stuck still on the left hand side I don't want that so I'll say that y is going to be e to the minus 4x multiplied by...'.
Formula
Observation
Board shows algebraic manipulation: $e^{4x}y = 1/3 e^{3x} + C$ becomes $y = e^{-4x}[1/3 e^{3x} + C] = 1/3 e^{-x}$ + Ce^{-4x}.

Method

Explanation

To find the explicit solution $y(x)$ , multiply both sides of the equation by the reciprocal of the integrating factor ( $e^{-4x}$ ) and distribute it across the terms on the right-hand side.

Formula

y = e^{-\mu(x)} \left[ \int e^{\mu(x)} f(x) dx + C \right]

Conditions

The equation is in the form $μ(x)y = \int$ ... + C.

Prerequisites

Integrating a Derivative

Applying Initial Conditions

Clear evidence

Shown in the video

Evidence

Audio
Observation
Speaker states: 'we actually also had this initial condition that y of zero was equal to four thirds... let's see how we can deal with that... plug in is that y at x equal to zero is equal to four thirds'.
Formula
Observation
Board shows substitution: $y(0) = 4/3 = 1/3 e^0 + C e^0$ => $4/3 = 1/3 + C$ => $C = 1$ .

Method

Explanation

Substitute the given initial values ( $x=0$ , $y=4/3$ ) into the general solution to solve for the specific value of the constant of integration C.

Formula

y(x_0) = y_0 \implies \text{solve for } C

Conditions

An initial condition $y(x_0)=y_0$ is provided.

Prerequisites

Isolating the Dependent Variable

Particular Solution

Clear evidence

Shown in the video

Evidence

Audio
Observation
Speaker states: 'if I want to put this together with the value of y that I had just computed I can get my final answer here it is y is one third e to the minus x plus c is equal to one now e to the minus four x final answer'.
Formula
Observation
Board shows final boxed answer: $y = 1/3 e^{-x} + e^{-4x}$ .

Definition

Explanation

The specific solution to a differential equation obtained by substituting the calculated constant C back into the general solution.

Formula

y = \frac{1}{3}e^{-x} + e^{-4x}

Conditions

C has been determined from an initial condition.

Prerequisites

Applying Initial Conditions

Claims and conditions · 1

Reverse Product Rule for the Left-Hand Side

Clear evidence

Shown in the video

Evidence

Audio
Observation
The instructor states that the left-hand side is the same as the derivative with respect to x of $e^{4x}$ multiplied by y, and can be verified by the product rule.
Formula
Observation
The instructor writes $\frac{d}{dx}[e^{4x}y]$ .

Proposition

Statement

After multiplying the standard form differential equation by its integrating factor $r(x)$ , the left-hand side simplifies to the exact derivative $\frac{d}{dx}[r(x)y]$ .

Hypotheses

The differential equation is in standard form y' + $p(x)y = f(x)$ .
$r(x)$ is the integrating factor defined as $e^{\int p(x) dx}$ .

Quantifiers

For all valid first-order linear differential equations.

Derivations and proofs · 4

Calculating the Integrating Factor for the Example

Clear evidence

Shown in the video

Evidence

Audio
Observation
The instructor walks through substituting $p(x) = 4$ into the formula and evaluating the integral.
Formula
Observation
The instructor writes $r(x) = e^{\int 4 dx} = e^{4x}$ .

Proof

Steps

Expression
$p(x) = 4$
Explanation
Identify the coefficient of y in the given equation y' + $4y = e^{-x}$ .
Justification
Definition of standard form.
Shown in the video
Expression
$r(x) = e^{\int 4 dx}$
Explanation
Substitute $p(x) = 4$ into the integrating factor formula.
Justification
Formula for the integrating factor.
Shown in the video
Expression
$r(x) = e^{4x}$
Explanation
Evaluate the indefinite integral of 4 with respect to x, omitting the constant of integration.
Justification
Power rule for integration and choice of arbitrary constant $C=0$ .
Shown in the video

Conclusion

The integrating factor for the example equation is $r(x) = e^{4x}$ .

Simplifying the Right-Hand Side

Clear evidence

Shown in the video

Evidence

Audio
Observation
The instructor notes a little simplification on the far right-hand side.
Formula
Observation
The instructor writes $e^{4x}e^{-x} = e^{3x}$ .

Proof

Steps

Expression
$e^{4x}e^{-x}$
Explanation
Multiply the integrating factor by the original right-hand side function.
Justification
Method of multiplying by the integrating factor.
Shown in the video
Expression
$e^{3x}$
Explanation
Combine the exponents using the laws of exponents.
Justification
Product rule for exponents: $a^m a^n = a^{m+n}$ .
Shown in the video

Conclusion

The right-hand side of the modified equation simplifies to $e^{3x}$ .

Derivation of General Solution

Clear evidence

Shown in the video

Evidence

Audio
Observation
Continuous narration explaining the integration and isolation steps.
Formula
Observation
Sequential writing on the digital blackboard showing the algebraic transformations.

Proof

Steps

Expression
$\frac{d}{dx}[e^{4x}y] = e^{3x}$
Explanation
Start with the exact derivative form achieved after multiplying by the integrating factor.
Justification
Previous step in the video (not fully shown in this clip but implied by board state).
Shown in the video
Expression
$e^{4x}y = \int e^{3x} dx$
Explanation
Integrate both sides with respect to x.
Justification
Fundamental Theorem of Calculus / Definition of Antiderivative.
Shown in the video
Expression
$e^{4x}y = \frac{1}{3}e^{3x} + C$
Explanation
Evaluate the integral on the right side and add the constant of integration.
Justification
Integration rule for exponentials.
Shown in the video
Expression
$y = e^{-4x}\left[\frac{1}{3}e^{3x} + C\right]$
Explanation
Multiply both sides by $e^{-4x}$ to isolate y.
Justification
Algebraic manipulation.
Shown in the video
Expression
$y = \frac{1}{3}e^{-x} + Ce^{-4x}$
Explanation
Distribute $e^{-4x}$ and simplify exponents ( $e^{-4x} * e^{3x} = e^{-x}$ ).
Justification
Exponent laws.
Shown in the video

Conclusion

The general solution to the differential equation is $y = \frac{1}{3}e^{-x}$ + Ce^{-4x}.

Derivation of Particular Solution

Clear evidence

Shown in the video

Evidence

Audio
Observation
Narration describing the substitution of $x=0$ and $y=4/3$ .
Formula
Observation
Writing of the equation $4/3 = 1/3 + C$ and subsequent boxing of $C=1$ and the final answer.

Proof

Steps

Expression
$y(0) = \frac{4}{3}$
Explanation
State the initial condition.
Justification
Given in the problem statement (visible on board).
Shown in the video
Expression
$\frac{4}{3} = \frac{1}{3}e^{0} + Ce^{0}$
Explanation
Substitute $x=0$ and $y=4/3$ into the general solution.
Justification
Definition of satisfying an initial condition.
Shown in the video
Expression
$\frac{4}{3} = \frac{1}{3} + C$
Explanation
Simplify using $e^0 = 1$ .
Justification
Arithmetic simplification.
Shown in the video
Expression
$C = 1$
Explanation
Solve for C by subtracting $1/3$ from both sides.
Justification
Algebraic manipulation.
Shown in the video
Expression
$y = \frac{1}{3}e^{-x} + e^{-4x}$
Explanation
Substitute $C=1$ back into the general solution.
Justification
Definition of particular solution.
Shown in the video

Conclusion

The particular solution satisfying the initial condition is $y = \frac{1}{3}e^{-x} + e^{-4x}$ .

Worked examples · 2

Solving a First-Order Linear Initial Value Problem

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board displays Ex/ y' + $4y = e^{-x}$ , $y(0) = 4/3$ .

Uncertainties

The final solution to the initial value problem is not reached within this video segment.

Problem

Solve the initial value problem y' + $4y = e^{-x}$ with $y(0) = 4/3$ using the method of integrating factors.

Given

Differential equation: y' + $4y = e^{-x}$
Initial condition: $y(0) = 4/3$

Goal

Find the explicit function $y(x)$ that satisfies both the differential equation and the initial condition.

Steps

Expression
$y' + 4y = e^{-x}$
Explanation
Write down the differential equation in standard form.
Justification
Given in the problem statement.
Shown in the video
Expression
$p(x) = 4$
Explanation
Identify the coefficient function $p(x)$ .
Justification
Definition of standard form.
Shown in the video
Expression
$r(x) = e^{\int 4 dx} = e^{4x}$
Explanation
Calculate the integrating factor.
Justification
Formula for the integrating factor.
Shown in the video
Expression
$e^{4x}y' + 4e^{4x}y = e^{4x}e^{-x}$
Explanation
Multiply every term in the differential equation by the integrating factor.
Justification
Method of integrating factors.
Shown in the video
Expression
$\frac{d}{dx}[e^{4x}y] = e^{3x}$
Explanation
Rewrite the left-hand side as the derivative of a product and simplify the right-hand side.
Justification
Reverse product rule and laws of exponents.
Shown in the video

Answer

The segment ends before the final answer for $y(x)$ is derived.

Verification

Not performed in this segment.

Solving Linear First-Order ODE with Initial Value

Clear evidence

Shown in the video

Evidence

Formula
Observation
Problem statement visible at top of board: y' + $4y = e^{-x}$ , $y(0) = 4/3$ .
Audio
Observation
Speaker refers to 'my solution' and 'initial condition' throughout.

Problem

Solve the initial value problem: y' + $4y = e^{-x}$ , with $y(0) = 4/3$ .

Given

Differential equation: y' + $4y = e^{-x}$
Initial condition: $y(0) = 4/3$
Integrating factor (pre-calculated): $r(x) = e^{4x}$

Goal

Find the explicit function $y(x)$ that satisfies both the differential equation and the initial condition.

Steps

Expression
$\frac{d}{dx}[e^{4x}y] = e^{3x}$
Explanation
Recognize the left side as the derivative of the product of the integrating factor and y.
Justification
Product rule reverse application.
Shown in the video
Expression
$e^{4x}y = \frac{1}{3}e^{3x} + C$
Explanation
Integrate both sides to remove the derivative.
Justification
Integration.
Shown in the video
Expression
$y = \frac{1}{3}e^{-x} + Ce^{-4x}$
Explanation
Isolate y to get the general solution.
Justification
Algebra.
Shown in the video
Expression
$C = 1$
Explanation
Use the initial condition $y(0)=4/3$ to find the constant.
Justification
Substitution.
Shown in the video
Expression
$y = \frac{1}{3}e^{-x} + e^{-4x}$
Explanation
Write the final particular solution.
Justification
Substitution of C.
Shown in the video

Answer

$y = \frac{1}{3}e^{-x} + e^{-4x}$

Verification

Check $y(0) = 1/3 + 1 = 4/3$ . Check ODE: y' = -1/3 $e^{-x} - 4e^{-4x}$ ; y'+ $4y = (-1/3 e^{-x} - 4e^{-4x}) + 4(1/3 e^{-x} + e^{-4x}) = e^{-x}$ . Correct.

Visual events · 3

Writing the Solution Steps on the Whiteboard

Clear evidence

Shown in the video

Evidence

Animation
Observation
The instructor progressively writes the mathematical steps on a black digital whiteboard using colored pens.

Objects

Digital whiteboard
Colored pens
Mathematical text

Changes

New equations and labels are added sequentially from top to bottom.
Previous steps remain visible as the solution progresses.

Invariants

The background remains black.
The instructor remains in the bottom right corner.

Interpretation

The visual progression mirrors the logical steps of the mathematical derivation, allowing the viewer to follow the method in real-time.

Sequential Equation Writing

Clear evidence

Shown in the video

Evidence

Animation
Observation
Digital ink appears sequentially on the black background as the speaker writes equations.

Objects

Digital blackboard
Handwritten mathematical expressions

Changes

Equations appear line by line corresponding to the speaker's derivation steps.
Colors change between white, pink/red, and blue to distinguish parts of the equation or emphasis.

Invariants

The speaker remains in the bottom right corner.
The background remains black.

Interpretation

Visual representation of the step-by-step algebraic and calculus manipulations required to solve the ODE.

Highlighting Final Answer

Clear evidence

Shown in the video

Evidence

Animation
Observation
A red box is drawn around the final equation $y = 1/3 e^{-x} + e^{-4x}$ .

Objects

Red rectangular box
Final solution equation

Changes

Box is drawn around the text.

Invariants

Text inside box remains unchanged.

Interpretation

Emphasizes the completion of the problem and the final result.

Misconceptions · 2

Omitting the Constant of Integration in the Integrating Factor

Clear evidence

Shown in the video

Evidence

Audio
Observation
The instructor addresses the potential confusion about the missing +C in the indefinite integral for the integrating factor.

Misconception

Students often wonder why the constant of integration (+C) is not included when calculating the indefinite integral for the integrating factor.

Clarification

Any valid integrating factor will work for the method. Choosing the arbitrary constant $C=0$ simplifies the calculation without loss of generality.

Omitting Constant in Integrating Factor vs Solution

Clear evidence

Shown in the video

Evidence

Audio
Observation
Speaker addresses potential confusion: 'You might think this is a little bit odd I told you earlier you didn't need to write down the plus c but you didn't need to write down the plus c earlier because it was all about finding an integrating factor'.

Misconception

Students may think the constant of integration (+C) should always be omitted or always included, confusing the step of finding an integrating factor with the step of solving the differential equation.

Clarification

When finding an integrating factor, any single valid function works, so +C can be dropped (it would cancel out anyway). When solving the differential equation via integration, +C is mandatory to represent the family of solutions before applying initial conditions.

Concept relations · 4

Reverse Product Rule for the Left-Hand Side → Multiplying the Differential Equation by the Integrating Factor

Clear evidence

Shown in the video

Evidence

Audio
Observation
The instructor explicitly mentions verifying the simplification using the product rule.

Proof dependency

Explanation

The claim that the left-hand side simplifies to a single derivative relies directly on the reverse application of the product rule from calculus.

Integrating a Derivative → Isolating the Dependent Variable

Clear evidence

Shown in the video

Evidence

Audio
Observation
Speaker mentions 'integrating a derivative so it cancels'.

Prerequisite

Explanation

Understanding that integration reverses differentiation is necessary before isolating y.

Isolating the Dependent Variable → Applying Initial Conditions

Clear evidence

Shown in the video

Evidence

Audio
Observation
Speaker moves from general solution to applying initial condition.

Prerequisite

Explanation

The general solution containing C must be found before the initial condition can be used to solve for C.

Applying Initial Conditions → Particular Solution

Clear evidence

Shown in the video

Evidence

Formula
Observation
Substitution of $C=1$ into the general equation.

Application

Explanation

The value of C derived from the initial condition is applied to the general solution to create the particular solution.

Find an answer · 5

Why is there no +C when calculating the integrating factor?

Clear evidence

Shown in the video

Evidence

Audio
Observation
The instructor explains the reasoning behind omitting the constant.

Knowledge points

Formula for the Integrating Factor
Omitting the Constant of Integration in the Integrating Factor

How does multiplying by the integrating factor allow the left side to become a single derivative?

Clear evidence

Shown in the video

Evidence

Audio
Observation
The instructor explains the simplification of the left-hand side.

Knowledge points

Multiplying the Differential Equation by the Integrating Factor
Reverse Product Rule for the Left-Hand Side

Why did the instructor not include +C when calculating the integrating factor earlier?

Clear evidence

Shown in the video

Evidence

Audio
Observation
Explanation of why +C was skipped for the integrating factor.

Knowledge points

Necessity of Constant of Integration
Omitting Constant in Integrating Factor vs Solution

How do you isolate y after integrating both sides of the equation?

Clear evidence

Shown in the video

Evidence

Formula
Observation
Algebraic steps showing multiplication by $e^{-4x}$ .

Knowledge points

Isolating the Dependent Variable

How is the constant C determined from the initial condition $y(0)=4/3$ ?

Clear evidence

Shown in the video

Evidence

Formula
Observation
Substitution of $x=0$ and $y=4/3$ .

Knowledge points

Applying Initial Conditions

Coverage and review notes

Covered · Introduction to the topic of integrating factors for first-order linear differential equations.

Covered · Explanation of why the given equation is linear and identification of $p(x)$ and $f(x)$ .

Covered · Presentation of the integrating factor formula and its calculation for the specific example.

Covered · Multiplying the differential equation by the integrating factor and simplifying the right-hand side.

Covered · Rewriting the left-hand side as the derivative of a product using the reverse product rule.

Covered · Setup of the integration step.

Covered · Performing the integration on both sides.

Covered · Discussion on the necessity of the constant of integration.

Covered · Algebraic manipulation to find the general solution $y(x)$ .

Covered · Using the initial condition to solve for C.

Covered · Writing the final particular solution and conclusion.

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Initial value problems ApplicationAt 0:00
Why this connection?
Reviewed English material v1: example "Solving a First-Order Linear Initial Value Problem" at 0:00 — Solve the initial value problem y' + $4y = e^{-x}$ with $y(0) = 4/3$ using the method of integrating factors; the givens are recorded as the differential equation y' + $4y = e^{-x}$ and the initial condition $y(0) = 4/3$ , and the stated goal is to find the explicit function $y(x)$ that satisfies both the differential equation and the initial condition.
Integrating factors ExplanationAt 1:08
Why this connection?
Reviewed English material v1: card "Integrating Factor Formula" at 1:08 — The integrating factor $r(x)$ is defined as the exponential of the indefinite integral of $p(x)$ . When calculating it, the constant of integration is typically set to zero, as any non-zero multiple of an integrating factor also works.

The Method of Integrating Factors for Linear 1st Order ODEs **full example**

The explanation, unpacked.

Chapters

Learning script

Knowledge cards

Standard Form of a First-Order Linear ODE

Integrating Factor Formula

Applying the Integrating Factor

Integrating the Derivative Side

Integral of Exponential Functions

When to Include +C

Isolating y for General Solution

Solving for C using Initial Conditions

Particular Solution

Detailed learning notes

y'

y

x

p(x)p(x)p(x)

f(x)f(x)f(x)

r(x)r(x)r(x)

e

∫\int∫

dx

43\frac{4}{3}34​

ddx\frac{d}{dx}dxd​

y

Standard Form of a First-Order Linear Differential Equation

Formula for the Integrating Factor

Multiplying the Differential Equation by the Integrating Factor

Integrating a Derivative

Integral of Exponential Function

Necessity of Constant of Integration

Isolating the Dependent Variable

Applying Initial Conditions

Particular Solution

Reverse Product Rule for the Left-Hand Side

Calculating the Integrating Factor for the Example

Simplifying the Right-Hand Side

Derivation of General Solution

Derivation of Particular Solution

Solving a First-Order Linear Initial Value Problem

Solving Linear First-Order ODE with Initial Value

Writing the Solution Steps on the Whiteboard

Sequential Equation Writing

Highlighting Final Answer

Omitting the Constant of Integration in the Integrating Factor

Omitting Constant in Integrating Factor vs Solution

Reverse Product Rule for the Left-Hand Side → Multiplying the Differential Equation by the Integrating Factor

Integrating a Derivative → Isolating the Dependent Variable

Isolating the Dependent Variable → Applying Initial Conditions

Applying Initial Conditions → Particular Solution

Why is there no +C when calculating the integrating factor?

How does multiplying by the integrating factor allow the left side to become a single derivative?

Why did the instructor not include +C when calculating the integrating factor earlier?

How do you isolate y after integrating both sides of the equation?

How is the constant C determined from the initial condition y(0)=4/3y(0)=4/3y(0)=4/3?

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The Method of Integrating Factors for Linear 1st Order ODEs full example

$p(x)$

$f(x)$

$r(x)$

$\int$

$\frac{4}{3}$

$\frac{d}{dx}$

How is the constant C determined from the initial condition $y(0)=4/3$ ?