Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Solving Linear Ordinary Differential Equations (ODEs) Using an Integrating Factor
#### Task:
**Problem Statement:**
Solve the linear Ordinary Differential Equation (ODE) using an Integrating Factor.
\[ x^2 \cdot \frac{dy}{dx} + 2xy = 3 \sin(8x) \]
### Step-by-Step Solution:
1. **Rewrite the ODE:**
The given ODE is:
\[ x^2 \frac{dy}{dx} + 2xy = 3 \sin(8x) \]
First, we need to get the equation into a standard linear form. By dividing through by \(x^2\), we have:
\[ \frac{dy}{dx} + \frac{2}{x} y = \frac{3 \sin(8x)}{x^2} \]
2. **Identify the Integrating Factor:**
The standard form of a linear ODE is:
\[ \frac{dy}{dx} + P(x)y = Q(x) \]
By comparison, \(P(x) = \frac{2}{x}\) and \(Q(x) = \frac{3 \sin(8x)}{x^2}\).
The integrating factor, \( \mu(x) \), is given by:
\[ \mu(x) = e^{\int P(x) \, dx} = e^{\int \frac{2}{x} \, dx} \]
\[ \mu(x) = e^{2 \ln |x|} = |x|^2 = x^2 \]
(Assuming \(x\) is positive here for simplicity, so \(|x| = x\)).
3. **Multiply through by the Integrating Factor:**
Multiply both sides of the ODE by \(x^2\):
\[ x^2 \cdot \frac{dy}{dx} + x^2 \cdot \frac{2}{x} y = x^2 \cdot \frac{3 \sin(8x)}{x^2} \]
Simplifying, we get:
\[ x^2 \frac{dy}{dx} + 2xy = 3 \sin(8x) \]
Notice this is the original](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5dfcde0a-cc8e-4f7e-b5e2-9ec6ad36a3c9%2Fbb8e78a6-3b42-4267-9941-f100c91877e0%2Fc2ysc5p_processed.png&w=3840&q=75)
Transcribed Image Text:### Solving Linear Ordinary Differential Equations (ODEs) Using an Integrating Factor
#### Task:
**Problem Statement:**
Solve the linear Ordinary Differential Equation (ODE) using an Integrating Factor.
\[ x^2 \cdot \frac{dy}{dx} + 2xy = 3 \sin(8x) \]
### Step-by-Step Solution:
1. **Rewrite the ODE:**
The given ODE is:
\[ x^2 \frac{dy}{dx} + 2xy = 3 \sin(8x) \]
First, we need to get the equation into a standard linear form. By dividing through by \(x^2\), we have:
\[ \frac{dy}{dx} + \frac{2}{x} y = \frac{3 \sin(8x)}{x^2} \]
2. **Identify the Integrating Factor:**
The standard form of a linear ODE is:
\[ \frac{dy}{dx} + P(x)y = Q(x) \]
By comparison, \(P(x) = \frac{2}{x}\) and \(Q(x) = \frac{3 \sin(8x)}{x^2}\).
The integrating factor, \( \mu(x) \), is given by:
\[ \mu(x) = e^{\int P(x) \, dx} = e^{\int \frac{2}{x} \, dx} \]
\[ \mu(x) = e^{2 \ln |x|} = |x|^2 = x^2 \]
(Assuming \(x\) is positive here for simplicity, so \(|x| = x\)).
3. **Multiply through by the Integrating Factor:**
Multiply both sides of the ODE by \(x^2\):
\[ x^2 \cdot \frac{dy}{dx} + x^2 \cdot \frac{2}{x} y = x^2 \cdot \frac{3 \sin(8x)}{x^2} \]
Simplifying, we get:
\[ x^2 \frac{dy}{dx} + 2xy = 3 \sin(8x) \]
Notice this is the original
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