Find a solution to the IVP y" + 4y' + 4y = 0 if y(0) = 1 anc y'(0) = 3. 1400
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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![**Problem Statement:**
Find a solution to the Initial Value Problem (IVP) for the differential equation:
\( y'' + 4y' + 4y = 0 \),
subject to the initial conditions:
\( y(0) = 1 \) and \( y'(0) = 3 \).
**Explanation:**
To solve this differential equation with the given initial conditions, we follow these steps:
1. **Solve the Characteristic Equation**:
- For the differential equation \( y'' + 4y' + 4y = 0 \), we first find the characteristic equation, which is \( r^2 + 4r + 4 = 0 \).
- Factor the characteristic equation: \( (r + 2)^2 = 0 \).
- This implies \( r = -2 \) is a repeated root.
2. **General Solution**:
- Because there is a repeated root, the general solution takes the form:
\[
y(t) = C_1 e^{-2t} + C_2 t e^{-2t}
\]
3. **Apply Initial Conditions**:
- Use \( y(0) = 1 \):
\[
1 = C_1 e^{0} + C_2 \cdot 0 \cdot e^{0} = C_1
\]
Thus, \( C_1 = 1 \).
- Use \( y'(0) = 3 \):
- First, find the derivative:
\[
y'(t) = -2C_1 e^{-2t} + C_2 e^{-2t} - 2C_2 t e^{-2t}
\]
- Substitute \( t = 0 \) and \( y'(0) = 3 \):
\[
3 = -2C_1 + C_2
\]
- Substitute \( C_1 = 1 \):
\[
3 = -2(1) + C_2 \implies 3 = -2 + C_2 \implies C_2 = 5
\]
4. **Particular Solution**:
- With the determined constants, the particular solution is:
\[
y(t) = e^{-2t} + 5t e](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb26ca370-3b85-410b-82ca-0e03b921dc6b%2Fb557db59-10c9-4cf4-82d6-8282033c2c4d%2Fyr04nxg_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find a solution to the Initial Value Problem (IVP) for the differential equation:
\( y'' + 4y' + 4y = 0 \),
subject to the initial conditions:
\( y(0) = 1 \) and \( y'(0) = 3 \).
**Explanation:**
To solve this differential equation with the given initial conditions, we follow these steps:
1. **Solve the Characteristic Equation**:
- For the differential equation \( y'' + 4y' + 4y = 0 \), we first find the characteristic equation, which is \( r^2 + 4r + 4 = 0 \).
- Factor the characteristic equation: \( (r + 2)^2 = 0 \).
- This implies \( r = -2 \) is a repeated root.
2. **General Solution**:
- Because there is a repeated root, the general solution takes the form:
\[
y(t) = C_1 e^{-2t} + C_2 t e^{-2t}
\]
3. **Apply Initial Conditions**:
- Use \( y(0) = 1 \):
\[
1 = C_1 e^{0} + C_2 \cdot 0 \cdot e^{0} = C_1
\]
Thus, \( C_1 = 1 \).
- Use \( y'(0) = 3 \):
- First, find the derivative:
\[
y'(t) = -2C_1 e^{-2t} + C_2 e^{-2t} - 2C_2 t e^{-2t}
\]
- Substitute \( t = 0 \) and \( y'(0) = 3 \):
\[
3 = -2C_1 + C_2
\]
- Substitute \( C_1 = 1 \):
\[
3 = -2(1) + C_2 \implies 3 = -2 + C_2 \implies C_2 = 5
\]
4. **Particular Solution**:
- With the determined constants, the particular solution is:
\[
y(t) = e^{-2t} + 5t e
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