Solve the given initial value problem. y'"' + 2y' + y = 0; y(0) = -5, y'(0) = 6 What is the auxiliary equation associated with the given differential equation? (Type an equation using r as the variable.)
Solve the given initial value problem. y'"' + 2y' + y = 0; y(0) = -5, y'(0) = 6 What is the auxiliary equation associated with the given differential equation? (Type an equation using r as the variable.)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
![**Solving Initial Value Problems in Differential Equations**
### Problem Statement
Solve the given initial value problem.
\[
y'' + 2y' + y = 0; \quad y(0) = -5, \quad y'(0) = 6
\]
### Step-by-Step Breakdown
1. **Identify the Differential Equation:**
The given differential equation is a second-order linear homogeneous differential equation.
2. **Initial Conditions:**
The initial conditions provided are:
- \( y(0) = -5 \)
- \( y'(0) = 6 \)
3. **Auxiliary Equation:**
To solve the differential equation, we need to find the auxiliary equation associated with it. This can be obtained by assuming a solution of the form \( y = e^{rt} \), leading to the characteristic equation:
\[
r^2 + 2r + 1 = 0
\]
4. **Solving the Auxiliary Equation:**
Solving the characteristic equation will give us the roots that are essential in forming the general solution to the differential equation. (Here, you type an equation using \( r \) as the variable.)
### Question
What is the auxiliary equation associated with the given differential equation?
**Answer Box** (Type an equation using \( r \) as the variable.)
---
**Explanation:**
The auxiliary equation, sometimes referred to as the characteristic equation, is derived from the coefficients of the differential equation. It transforms the differential equation into an algebraic equation that can be solved to find the roots, which are then used to form the general solution to the differential equation.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe57d9078-3a26-4ab9-a5c6-f67f173c178d%2Fa5e2a6ee-df9d-4e8a-bc8a-9eb5802537f2%2Fn69oodb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Solving Initial Value Problems in Differential Equations**
### Problem Statement
Solve the given initial value problem.
\[
y'' + 2y' + y = 0; \quad y(0) = -5, \quad y'(0) = 6
\]
### Step-by-Step Breakdown
1. **Identify the Differential Equation:**
The given differential equation is a second-order linear homogeneous differential equation.
2. **Initial Conditions:**
The initial conditions provided are:
- \( y(0) = -5 \)
- \( y'(0) = 6 \)
3. **Auxiliary Equation:**
To solve the differential equation, we need to find the auxiliary equation associated with it. This can be obtained by assuming a solution of the form \( y = e^{rt} \), leading to the characteristic equation:
\[
r^2 + 2r + 1 = 0
\]
4. **Solving the Auxiliary Equation:**
Solving the characteristic equation will give us the roots that are essential in forming the general solution to the differential equation. (Here, you type an equation using \( r \) as the variable.)
### Question
What is the auxiliary equation associated with the given differential equation?
**Answer Box** (Type an equation using \( r \) as the variable.)
---
**Explanation:**
The auxiliary equation, sometimes referred to as the characteristic equation, is derived from the coefficients of the differential equation. It transforms the differential equation into an algebraic equation that can be solved to find the roots, which are then used to form the general solution to the differential equation.
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