1. Determine the general solution for the y" - 4y + 4y = 0, and the solution of the initial value problem y" - 4y + 4y = 0, y(0) = 0 e y'(0) = 1
1. Determine the general solution for the y" - 4y + 4y = 0, and the solution of the initial value problem y" - 4y + 4y = 0, y(0) = 0 e y'(0) = 1
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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![### Differential Equations - Example Problem
**Problem:**
1. Determine the general solution for the differential equation \( y'' - 4y' + 4y = 0 \), and the solution of the initial value problem \( y'' - 4y' + 4y = 0 \), \( y(0) = 0 \), and \( y'(0) = 1 \).
**Steps to Solve:**
1. **Finding the General Solution:**
- Rewrite the differential equation \( y'' - 4y' + 4y = 0 \).
- This is a second-order linear homogeneous differential equation with constant coefficients.
- The characteristic equation is obtained by replacing \( y \) with \( e^{rx} \):
\[ r^2 - 4r + 4 = 0 \]
- Solve the characteristic quadratic equation:
\[ r = \frac{4 \pm \sqrt{16 - 16}}{2} = 2 \]
- Since we have a repeated root \( r = 2 \), the general solution is:
\[ y(t) = (C_1 + C_2 t)e^{2t} \]
2. **Solving the Initial Value Problem:**
- Given the initial conditions \( y(0) = 0 \) and \( y'(0) = 1 \).
- Plug \( t = 0 \) into the general solution:
\[ y(0) = (C_1 + C_2 \cdot 0)e^{2 \cdot 0} = C_1 = 0 \]
- Thus, \( C_1 = 0 \).
- Differentiate the general solution:
\[ y(t) = C_2 t e^{2t} \]
\[ y'(t) = C_2 e^{2t} + C_2 t \cdot 2e^{2t} = C_2 e^{2t} (1 + 2t) \]
- Use initial condition \( y'(0) = 1 \):
\[ y'(0) = C_2 e^{0} \cdot (1 + 2 \cdot 0) = C_2 = 1 \]
- Thus, \( C_2 = 1 \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc47a605b-1167-4b41-bc40-883b330bf596%2F641e9481-0301-4f41-bdb2-e29991ef5426%2Fkpk6p8p_processed.png&w=3840&q=75)
Transcribed Image Text:### Differential Equations - Example Problem
**Problem:**
1. Determine the general solution for the differential equation \( y'' - 4y' + 4y = 0 \), and the solution of the initial value problem \( y'' - 4y' + 4y = 0 \), \( y(0) = 0 \), and \( y'(0) = 1 \).
**Steps to Solve:**
1. **Finding the General Solution:**
- Rewrite the differential equation \( y'' - 4y' + 4y = 0 \).
- This is a second-order linear homogeneous differential equation with constant coefficients.
- The characteristic equation is obtained by replacing \( y \) with \( e^{rx} \):
\[ r^2 - 4r + 4 = 0 \]
- Solve the characteristic quadratic equation:
\[ r = \frac{4 \pm \sqrt{16 - 16}}{2} = 2 \]
- Since we have a repeated root \( r = 2 \), the general solution is:
\[ y(t) = (C_1 + C_2 t)e^{2t} \]
2. **Solving the Initial Value Problem:**
- Given the initial conditions \( y(0) = 0 \) and \( y'(0) = 1 \).
- Plug \( t = 0 \) into the general solution:
\[ y(0) = (C_1 + C_2 \cdot 0)e^{2 \cdot 0} = C_1 = 0 \]
- Thus, \( C_1 = 0 \).
- Differentiate the general solution:
\[ y(t) = C_2 t e^{2t} \]
\[ y'(t) = C_2 e^{2t} + C_2 t \cdot 2e^{2t} = C_2 e^{2t} (1 + 2t) \]
- Use initial condition \( y'(0) = 1 \):
\[ y'(0) = C_2 e^{0} \cdot (1 + 2 \cdot 0) = C_2 = 1 \]
- Thus, \( C_2 = 1 \
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