Consider the following differential equation. Find the coefficient function P(x) when the given differential equation is written in the standard form dy + P(x) = f(x). dx P(x)= Find the integrating factor for the differential equation. P(x) dx_ y(x) = Proceed as in Example 6 in Section 2.3 to solve the given initial-value problem. +y = f(x), y(0) 1, where f(x)= -{-4 dx y 3 2 Use a graphing utility to graph the continuous function y(x). 1 , 0≤x≤1 2 X> 1 1, 0≤x≤ 1 x>1 3 4 5 1 2 3 4 4 2 3 4 5

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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### Differential Equations Problem and Solution

Consider the following differential equation:

\[ \frac{dy}{dx} + y = f(x), \quad y(0) = 1, \quad \text{where} \ f(x) = 
\begin{cases} 
1, & 0 \leq x \leq 1 \\
-1, & x > 1 
\end{cases}
\]

#### Step 1: Identifying the Coefficient Function \(P(x)\)

Find the coefficient function \(P(x)\) when the given differential equation is written in the standard form:

\[ \frac{dy}{dx} + P(x)y = f(x) \]

\[ P(x) = \underline{\hspace{2cm}} \]

#### Step 2: Finding the Integrating Factor

Find the integrating factor for the differential equation:

\[ e^{\int P(x) \, dx} = \underline{\hspace{6cm}} \]

#### Step 3: Solving the Initial-Value Problem

Proceed as in Example 6 in Section 2.3 to solve the given initial-value problem.

\[ y(x) = 
\begin{cases} 
\underline{\hspace{2cm}}, & 0 \leq x \leq 1 \\
\underline{\hspace{2cm}}, & x > 1 
\end{cases}
\]

#### Step 4: Graphing the Continuous Function

Use a graphing utility to graph the continuous function \( y(x) \).

### Graph Explanation

There are four different graphs provided, each showing a blue curve representing different possible solutions or behaviors for the function \( y(x) \) given different \( P(x) \) and \( f(x) \) values.

1. **First Graph (Selected with a Check)**
   - **X-Axis Range:** 0 to 5
   - **Y-Axis Range:** -2 to 3
   - Description: The graph shows the function beginning with a value of 1 at \( x=0 \). It increases slightly and maintains a flat line around \( y=2 \). After \( x=1 \), the function decreases slightly. This graph likely represents a scenario consistent with the initial value problem where \( f(x) \) switches behavior at \( x=1 \).

2. **Second Graph**
   - **X-Axis Range
Transcribed Image Text:### Differential Equations Problem and Solution Consider the following differential equation: \[ \frac{dy}{dx} + y = f(x), \quad y(0) = 1, \quad \text{where} \ f(x) = \begin{cases} 1, & 0 \leq x \leq 1 \\ -1, & x > 1 \end{cases} \] #### Step 1: Identifying the Coefficient Function \(P(x)\) Find the coefficient function \(P(x)\) when the given differential equation is written in the standard form: \[ \frac{dy}{dx} + P(x)y = f(x) \] \[ P(x) = \underline{\hspace{2cm}} \] #### Step 2: Finding the Integrating Factor Find the integrating factor for the differential equation: \[ e^{\int P(x) \, dx} = \underline{\hspace{6cm}} \] #### Step 3: Solving the Initial-Value Problem Proceed as in Example 6 in Section 2.3 to solve the given initial-value problem. \[ y(x) = \begin{cases} \underline{\hspace{2cm}}, & 0 \leq x \leq 1 \\ \underline{\hspace{2cm}}, & x > 1 \end{cases} \] #### Step 4: Graphing the Continuous Function Use a graphing utility to graph the continuous function \( y(x) \). ### Graph Explanation There are four different graphs provided, each showing a blue curve representing different possible solutions or behaviors for the function \( y(x) \) given different \( P(x) \) and \( f(x) \) values. 1. **First Graph (Selected with a Check)** - **X-Axis Range:** 0 to 5 - **Y-Axis Range:** -2 to 3 - Description: The graph shows the function beginning with a value of 1 at \( x=0 \). It increases slightly and maintains a flat line around \( y=2 \). After \( x=1 \), the function decreases slightly. This graph likely represents a scenario consistent with the initial value problem where \( f(x) \) switches behavior at \( x=1 \). 2. **Second Graph** - **X-Axis Range
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