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 the Differential Equation
**Problem Statement**:
Solve the differential equation:
\[ \frac{dy}{dx} = 80 - 2y, \]
with the initial condition \( y = 70 \) when \( x = 0 \).
**Analysis**:
This is a first-order linear differential equation. To solve it, we can use the method of finding an integrating factor or recognizing it as a linear differential equation and solving it accordingly.
1. **Step 1: Identify the Integrating Factor**
The standard form of a first-order linear differential equation is:
\[ \frac{dy}{dx} + P(x)y = Q(x), \]
where \( P(x) \) and \( Q(x) \) are functions of \( x \). For our equation, \( P(x) = 2 \) and \( Q(x) = 80 \).
2. **Step 2: Calculate the Integrating Factor**
The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{\int P(x) dx} = e^{\int 2 \, dx} = e^{2x}. \]
3. **Step 3: Multiply Through by the Integrating Factor**
Multiply both sides of the differential equation by \( \mu(x) \):
\[ e^{2x} \frac{dy}{dx} + 2e^{2x}y = 80e^{2x}. \]
4. **Step 4: Simplify and Integrate**
The left side of the equation can be recognized as the derivative of the product \( y e^{2x} \):
\[ \frac{d}{dx} (y e^{2x}) = 80e^{2x}. \]
Now, integrate both sides:
\[ y e^{2x} = \int 80e^{2x} dx = 40e^{2x} + C, \]
where \( C \) is the constant of integration.
5. **Step 5: Solve for \( y \)**
\[ y e^{2x} = 40e^{2x} + C, \]
Divide by \( e^{2x} \):
\[ y = 40 + Ce^{-2x}. \]
6. **Step 6](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fedc85e3b-dc2d-4f8b-a097-9d7c53c76aad%2F834a86f9-dc1d-41bb-80eb-a2671dbf4cde%2Fv79iiye.jpeg&w=3840&q=75)
Transcribed Image Text:### Solving the Differential Equation
**Problem Statement**:
Solve the differential equation:
\[ \frac{dy}{dx} = 80 - 2y, \]
with the initial condition \( y = 70 \) when \( x = 0 \).
**Analysis**:
This is a first-order linear differential equation. To solve it, we can use the method of finding an integrating factor or recognizing it as a linear differential equation and solving it accordingly.
1. **Step 1: Identify the Integrating Factor**
The standard form of a first-order linear differential equation is:
\[ \frac{dy}{dx} + P(x)y = Q(x), \]
where \( P(x) \) and \( Q(x) \) are functions of \( x \). For our equation, \( P(x) = 2 \) and \( Q(x) = 80 \).
2. **Step 2: Calculate the Integrating Factor**
The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{\int P(x) dx} = e^{\int 2 \, dx} = e^{2x}. \]
3. **Step 3: Multiply Through by the Integrating Factor**
Multiply both sides of the differential equation by \( \mu(x) \):
\[ e^{2x} \frac{dy}{dx} + 2e^{2x}y = 80e^{2x}. \]
4. **Step 4: Simplify and Integrate**
The left side of the equation can be recognized as the derivative of the product \( y e^{2x} \):
\[ \frac{d}{dx} (y e^{2x}) = 80e^{2x}. \]
Now, integrate both sides:
\[ y e^{2x} = \int 80e^{2x} dx = 40e^{2x} + C, \]
where \( C \) is the constant of integration.
5. **Step 5: Solve for \( y \)**
\[ y e^{2x} = 40e^{2x} + C, \]
Divide by \( e^{2x} \):
\[ y = 40 + Ce^{-2x}. \]
6. **Step 6
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