### Solving Homogeneous Linear Differential Equations with Complex Roots When dealing with homogeneous linear differential equations, we frequently encounter scenarios that require us to find the general solution of the differential equation. Consider the following second-order differential equation: \[ ay'' + by' + cy = 0 \] ### Characteristic Equation To solve this differential equation, we first form the characteristic (auxiliary) equation: \[ ar^2 + br + c = 0 \] ### Using the Quadratic Formula The roots of this characteristic equation can be found using the quadratic formula: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] ### Nature of the Roots The nature of the roots of the characteristic equation depends on the discriminant, \( b^2 - 4ac \). 1. **Real and Distinct Roots** \( (b^2 - 4ac > 0) \) 2. **Real and Repeated Roots** \( (b^2 - 4ac = 0) \) 3. **Complex Roots** \( (b^2 - 4ac < 0) \) ### Case with Complex Roots For the case where \( b^2 - 4ac < 0 \), the roots of the characteristic equation are complex and can be expressed as: \[ r = \alpha \pm \beta i \] ### General Solution for Complex Roots The general solution for the differential equation, when the roots are complex, is: \[ y(t) = C_1 e^{\alpha t} \cos(\beta t) + C_2 e^{\alpha t} \sin(\beta t) \] where: - \( \alpha \) and \( \beta \) are real numbers derived from the roots of the characteristic equation. - \( C_1 \) and \( C_2 \) are constants determined by initial conditions. ### Example Selection Given the specific question: **What form will the solution to the differential equation be if \( b^2 - 4ac < 0 \)?** The correct answer is: \[ \boxed{y(t) = C_1 e^{\alpha t} \cos(\beta t) + C_2 e^{\alpha t} \sin(\beta t)} \] ### Explanation of the Provided Options - **First Option (Correct)**: \( y(t)
### Solving Homogeneous Linear Differential Equations with Complex Roots When dealing with homogeneous linear differential equations, we frequently encounter scenarios that require us to find the general solution of the differential equation. Consider the following second-order differential equation: \[ ay'' + by' + cy = 0 \] ### Characteristic Equation To solve this differential equation, we first form the characteristic (auxiliary) equation: \[ ar^2 + br + c = 0 \] ### Using the Quadratic Formula The roots of this characteristic equation can be found using the quadratic formula: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] ### Nature of the Roots The nature of the roots of the characteristic equation depends on the discriminant, \( b^2 - 4ac \). 1. **Real and Distinct Roots** \( (b^2 - 4ac > 0) \) 2. **Real and Repeated Roots** \( (b^2 - 4ac = 0) \) 3. **Complex Roots** \( (b^2 - 4ac < 0) \) ### Case with Complex Roots For the case where \( b^2 - 4ac < 0 \), the roots of the characteristic equation are complex and can be expressed as: \[ r = \alpha \pm \beta i \] ### General Solution for Complex Roots The general solution for the differential equation, when the roots are complex, is: \[ y(t) = C_1 e^{\alpha t} \cos(\beta t) + C_2 e^{\alpha t} \sin(\beta t) \] where: - \( \alpha \) and \( \beta \) are real numbers derived from the roots of the characteristic equation. - \( C_1 \) and \( C_2 \) are constants determined by initial conditions. ### Example Selection Given the specific question: **What form will the solution to the differential equation be if \( b^2 - 4ac < 0 \)?** The correct answer is: \[ \boxed{y(t) = C_1 e^{\alpha t} \cos(\beta t) + C_2 e^{\alpha t} \sin(\beta t)} \] ### Explanation of the Provided Options - **First Option (Correct)**: \( y(t)
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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![### Solving Homogeneous Linear Differential Equations with Complex Roots
When dealing with homogeneous linear differential equations, we frequently encounter scenarios that require us to find the general solution of the differential equation.
Consider the following second-order differential equation:
\[ ay'' + by' + cy = 0 \]
### Characteristic Equation
To solve this differential equation, we first form the characteristic (auxiliary) equation:
\[ ar^2 + br + c = 0 \]
### Using the Quadratic Formula
The roots of this characteristic equation can be found using the quadratic formula:
\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
### Nature of the Roots
The nature of the roots of the characteristic equation depends on the discriminant, \( b^2 - 4ac \).
1. **Real and Distinct Roots** \( (b^2 - 4ac > 0) \)
2. **Real and Repeated Roots** \( (b^2 - 4ac = 0) \)
3. **Complex Roots** \( (b^2 - 4ac < 0) \)
### Case with Complex Roots
For the case where \( b^2 - 4ac < 0 \), the roots of the characteristic equation are complex and can be expressed as:
\[ r = \alpha \pm \beta i \]
### General Solution for Complex Roots
The general solution for the differential equation, when the roots are complex, is:
\[ y(t) = C_1 e^{\alpha t} \cos(\beta t) + C_2 e^{\alpha t} \sin(\beta t) \]
where:
- \( \alpha \) and \( \beta \) are real numbers derived from the roots of the characteristic equation.
- \( C_1 \) and \( C_2 \) are constants determined by initial conditions.
### Example Selection
Given the specific question:
**What form will the solution to the differential equation be if \( b^2 - 4ac < 0 \)?**
The correct answer is:
\[ \boxed{y(t) = C_1 e^{\alpha t} \cos(\beta t) + C_2 e^{\alpha t} \sin(\beta t)} \]
### Explanation of the Provided Options
- **First Option (Correct)**: \( y(t)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3738e16a-10fd-4378-b4a3-b8733d2d5bb5%2Fd9991a68-19c3-4fa1-a5fe-f4853d4bbbe5%2Fwejdw8d_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Solving Homogeneous Linear Differential Equations with Complex Roots
When dealing with homogeneous linear differential equations, we frequently encounter scenarios that require us to find the general solution of the differential equation.
Consider the following second-order differential equation:
\[ ay'' + by' + cy = 0 \]
### Characteristic Equation
To solve this differential equation, we first form the characteristic (auxiliary) equation:
\[ ar^2 + br + c = 0 \]
### Using the Quadratic Formula
The roots of this characteristic equation can be found using the quadratic formula:
\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
### Nature of the Roots
The nature of the roots of the characteristic equation depends on the discriminant, \( b^2 - 4ac \).
1. **Real and Distinct Roots** \( (b^2 - 4ac > 0) \)
2. **Real and Repeated Roots** \( (b^2 - 4ac = 0) \)
3. **Complex Roots** \( (b^2 - 4ac < 0) \)
### Case with Complex Roots
For the case where \( b^2 - 4ac < 0 \), the roots of the characteristic equation are complex and can be expressed as:
\[ r = \alpha \pm \beta i \]
### General Solution for Complex Roots
The general solution for the differential equation, when the roots are complex, is:
\[ y(t) = C_1 e^{\alpha t} \cos(\beta t) + C_2 e^{\alpha t} \sin(\beta t) \]
where:
- \( \alpha \) and \( \beta \) are real numbers derived from the roots of the characteristic equation.
- \( C_1 \) and \( C_2 \) are constants determined by initial conditions.
### Example Selection
Given the specific question:
**What form will the solution to the differential equation be if \( b^2 - 4ac < 0 \)?**
The correct answer is:
\[ \boxed{y(t) = C_1 e^{\alpha t} \cos(\beta t) + C_2 e^{\alpha t} \sin(\beta t)} \]
### Explanation of the Provided Options
- **First Option (Correct)**: \( y(t)
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