Solve the given initial value problem. y' + 4y' + 20y=0; y(0) = 3, y'(0) = -5 y(t) = 0
Solve the given initial value problem. y' + 4y' + 20y=0; y(0) = 3, y'(0) = -5 y(t) = 0
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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![### Initial Value Problem Solution
#### Problem Statement:
Solve the given initial value problem:
\[ y'' + 4y' + 20y = 0 \; ; \; y(0) = 3, \; y'(0) = -5 \]
#### Solution:
1. **Establish the characteristic equation:**
The differential equation \( y'' + 4y' + 20y = 0 \) can be converted into its characteristic form by substituting \( y = e^{rt} \). This gives:
\[ r^2 + 4r + 20 = 0 \]
2. **Solve the characteristic equation:**
Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 4 \), and \( c = 20 \):
\[ r = \frac{-4 \pm \sqrt{16 - 80}}{2} \]
\[ r = \frac{-4 \pm \sqrt{-64}}{2} \]
\[ r = \frac{-4 \pm 8i}{2} \]
\[ r = -2 \pm 4i \]
3. **Write the general solution:**
Since the roots are complex, the general solution is:
\[ y(t) = e^{\alpha t} (C_1 \cos(\beta t) + C_2 \sin(\beta t)) \]
Here, \( \alpha = -2 \) and \( \beta = 4 \), so:
\[ y(t) = e^{-2t} (C_1 \cos(4t) + C_2 \sin(4t)) \]
4. **Apply initial conditions:**
Use \( y(0) = 3 \) and \( y'(0) = -5 \) to find \( C_1 \) and \( C_2 \).
- For \( y(0) = 3 \):
\[ y(0) = e^{-2(0)} (C_1 \cos(4(0)) + C_2 \sin(4(0))) \]
\[ 3 = C_1 \cdot 1 + C_2 \cd](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe57d9078-3a26-4ab9-a5c6-f67f173c178d%2Fda3de5c8-8281-41e2-93c8-09970f770abc%2Fx63fc4d_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Initial Value Problem Solution
#### Problem Statement:
Solve the given initial value problem:
\[ y'' + 4y' + 20y = 0 \; ; \; y(0) = 3, \; y'(0) = -5 \]
#### Solution:
1. **Establish the characteristic equation:**
The differential equation \( y'' + 4y' + 20y = 0 \) can be converted into its characteristic form by substituting \( y = e^{rt} \). This gives:
\[ r^2 + 4r + 20 = 0 \]
2. **Solve the characteristic equation:**
Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 4 \), and \( c = 20 \):
\[ r = \frac{-4 \pm \sqrt{16 - 80}}{2} \]
\[ r = \frac{-4 \pm \sqrt{-64}}{2} \]
\[ r = \frac{-4 \pm 8i}{2} \]
\[ r = -2 \pm 4i \]
3. **Write the general solution:**
Since the roots are complex, the general solution is:
\[ y(t) = e^{\alpha t} (C_1 \cos(\beta t) + C_2 \sin(\beta t)) \]
Here, \( \alpha = -2 \) and \( \beta = 4 \), so:
\[ y(t) = e^{-2t} (C_1 \cos(4t) + C_2 \sin(4t)) \]
4. **Apply initial conditions:**
Use \( y(0) = 3 \) and \( y'(0) = -5 \) to find \( C_1 \) and \( C_2 \).
- For \( y(0) = 3 \):
\[ y(0) = e^{-2(0)} (C_1 \cos(4(0)) + C_2 \sin(4(0))) \]
\[ 3 = C_1 \cdot 1 + C_2 \cd
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