1.3.3 - Example
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Calculating the Integrating Factor
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Today, we will learn how to solve the equation dy/dx + 2y = e^(-x) using the integrating factor method. First, who can tell me what P(x) is from this equation?
P(x) is 2, right?
Correct! Now, we calculate the integrating factor, $$ \mu(x) = e^{\int P(x) \, dx} = e^{\int 2 \, dx} $$.
So, that means $$ \mu(x) = e^{2x} $$?
Exactly! It’s crucial because it transforms our differential equation into a simpler form.
Transforming the Equation
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Next, we multiply both sides of the equation by the integrating factor. What does that give us?
We get e^{2x} (dy/dx + 2y) = e^{2x} e^{-x}.
Exactly! This allows us to rewrite the left side as a product derivative. Can anyone express it as such?
It becomes $$ \frac{d}{dx}(e^{2x}y) = e^{x} $$.
Great job! Now we can integrate both sides.
Integrating the Equation
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When we integrate, what do we end up with?
We have $$ e^{2x}y = \int e^{x} \, dx $$, which gives us $$ e^{2x}y = e^{x} + C $$?
Excellent! Now we can express y in terms of e^{x}.
So, y = e^{-x} + Ce^{-2x}?
Correct! This is our final solution. Who can summarize why we used the integrating factor?
Final Summary
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Let’s summarize. What are the steps we took to find the solution?
We first identified P(x), then calculated the integrating factor, multiplied by the equation, re-expressed it as a product derivative, integrated both sides, and finally solved for y!
Exactly! This is a powerful method for solving first-order linear differential equations and has many engineering applications.
I appreciate how we applied each step logically; it helps me understand the process better.
Introduction & Overview
Read summaries of the section's main ideas at different levels of detail.
Quick Overview
Standard
In this section, we illustrate the solution of the first-order linear differential equation dy/dx + 2y = e^(-x) by applying the integrating factor method. The example details each step, from calculating the integrating factor to integrating both sides and deriving the final solution.
Detailed
Detailed Summary
In the section 1.3.3 Example, we explore the practical application of solving a first-order linear differential equation using the integrating factor method. The equation presented is:
$$ \frac{dy}{dx} + 2y = e^{-x}. $$
Step 1: Identify P(x) and calculate the integrating factor. Here, we can see that P(x) = 2. We calculate the integrating factor, $$ \mu(x) $$, as follows:
$$ \mu(x) = e^{\int 2 \, dx} = e^{2x}. $$
Step 2: Multiply the entire equation by the integrating factor. This results in:
$$ e^{2x} \frac{dy}{dx} + 2e^{2x}y = e^{x}. $$
Step 3: Recognize that the left side is the derivative of the product:
$$ \frac{d}{dx}(e^{2x}y) = e^{x}. $$
Step 4: Integrate both sides:
$$ \int \frac{d}{dx}(e^{2x}y) \, dx = \int e^{x} \, dx $$
which gives:
$$ e^{2x}y = e^{x} + C $$
Step 5: Solve for y:
$$ y = e^{-x} + Ce^{-2x}. $$
This example effectively demonstrates the use of the integrating factor method in solving first-order linear differential equations, highlighting its importance in engineering applications.
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Problem Statement
Chapter 1 of 4
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Chapter Content
Solve:
dy
+2y =e−x
dx
Detailed Explanation
In this problem, we are tasked with solving a first-order linear differential equation. The equation can be identified by the format of 'dy/dx + P(x)y = Q(x)', where P(x) and Q(x) are functions of x. Here, P(x) is 2 and Q(x) is e^(-x). Our goal is to find the function y that satisfies this equation for all x.
Examples & Analogies
Think of this equation like balancing weights on a seesaw (the left side) against a force that pulls it down (the right side, e^(-x)). The function y is what we want to adjust in order to balance the seesaw at various points.
Identifying the Integrating Factor
Chapter 2 of 4
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Chapter Content
• P(x)=2, so µ(x)=e^(2dx) =e^(2x)
Detailed Explanation
To solve the differential equation, we first identify the integrating factor, µ(x). The integrating factor is calculated using the formula µ(x) = e^(∫P(x)dx), where P(x) is the coefficient of y in the standard format of the differential equation. For P(x) = 2, we integrate: ∫2dx = 2x, which gives us µ(x) = e^(2x). This factor is crucial as it simplifies the equation, making it easier to solve.
Examples & Analogies
Imagine you are trying to prepare a special drink. The integrating factor is like finding the perfect mix of ingredients (here, e^(2x)) that brings all the flavors together, helping you achieve just the right taste.
Multiplying by the Integrating Factor
Chapter 3 of 4
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Chapter Content
• Multiply both sides:
dy d
e^(2x) +2e^(2x)y =e^x ⇒ (e^(2x)y)=e^x
dx dx
Detailed Explanation
Next, we multiply every term of our original equation by the integrating factor we found, e^(2x). This helps to rewrite the left side of the equation in a specific form, allowing it to be expressed as the derivative of a product: d/dx [µ(x)y]. So, we have d/dx [e^(2x)y] = e^x. This is a key step that transforms our equation to a form that can be easily integrated.
Examples & Analogies
Imagine using a special blender that combines all your ingredients (the left-hand side of the equation) into a smooth mixture, simplifying your next steps in the cooking process, just like how this multiplication simplifies the equation.
Integration
Chapter 4 of 4
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Chapter Content
• Integrate:
∫(e^(2x)y)dx = ∫e^xdx = e^x + C ⇒ y = e^(-x) + Ce^(-2x)
Detailed Explanation
Now we take the integral of both sides. The left side gives us e^(2x)y, while integrating e^x on the right side yields e^x, plus an arbitrary constant C (since we are looking for a general solution). After integration, we isolate y by dividing by e^(2x) to derive our final solution: y = e^(-x) + Ce^(-2x). This is our general solution to the differential equation.
Examples & Analogies
Think of this final integration as completing a recipe by carefully mixing all the ingredients until you have the final dish (the solution to the equation). You add a pinch of creativity (the constant C) to make it your own.
Key Concepts
-
Integrating Factor: A technique to simplify first-order linear differential equations into exact equations.
-
Product Derivative: Recognizing the left side of the transformed equation is a derivative of a product.
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Solution Steps: The sequence of finding the integrating factor, transforming, integrating, and solving.
Examples & Applications
Example of solving dy/dx + 2y = e^(-x) with the integrating factor e^(2x).
Demonstrating how to integrate both sides to find y = e^(-x) + Ce^(-2x).
Memory Aids
Interactive tools to help you remember key concepts
Rhymes
To solve with ease, just find A and B, an integrating factor makes it a breeze.
Stories
Imagine you have a treasure map! The integrating factor helps you find the 'X' that marks the spot where the equation solves itself.
Memory Tools
I.G.I.N - Identify P(x), Calculate G, Integrate, Solve for y, it's neat!
Acronyms
FIND - Factor, Integrate, New Form, Derive the answer.
Flash Cards
Glossary
- Differential Equation
An equation involving an unknown function and its derivatives.
- Integrating Factor
A function that is multiplied to a differential equation to transform it into an exact equation.
- Complementary Function
The solution to the homogeneous part of a differential equation.
Reference links
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