Solve the given initial value problem. y'' + 2y' +10y=0; y(0)=3, y'(0) = -2 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 an Initial Value Problem
We are given the following differential equation and initial conditions:
\[ y'' + 2y' + 10y = 0; \]
\[ y(0) = 3, \; y'(0) = -2 \]
Our goal is to determine the function \( y(t) \) that satisfies both the differential equation and the initial conditions.
First, we solve the homogeneous differential equation \( y'' + 2y' + 10y = 0 \).
1. **Find the characteristic equation:**
\[ r^2 + 2r + 10 = 0 \]
2. **Solve the characteristic equation:**
Use the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ r = \frac{-2 \pm \sqrt{4 - 40}}{2} \]
\[ r = \frac{-2 \pm \sqrt{-36}}{2} \]
\[ r = \frac{-2 \pm 6i}{2} \]
\[ r = -1 \pm 3i\ ]
The roots are complex, \( r = -1 + 3i \) and \( r = -1 - 3i \).
3. **Write the general solution:**
For complex roots \( \alpha \pm \beta i \), the general solution is:
\[ y(t) = e^{\alpha t} (C_1 \cos(\beta t) + C_2 \sin(\beta t)) \]
In this case, \( \alpha = -1 \) and \( \beta = 3 \), so the solution is:
\[ y(t) = e^{-t} (C_1 \cos(3t) + C_2 \sin(3t)) \]
4. **Apply the initial conditions to determine \( C_1 \) and \( C_2 \):**
- When \( t = 0 \):
\[ y(0) = e^{0} (C_1 \cos(0) + C_2 \sin(0)) \]
\[ y(0) = C_1 = 3 \]
Therefore, \( C_1 = 3 \).
- When](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3a1a8c69-ac20-486b-9f67-8f66504c5494%2F75d2e448-4391-4d0e-975c-5f6bad9a7e99%2F7ya9qx_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Solving an Initial Value Problem
We are given the following differential equation and initial conditions:
\[ y'' + 2y' + 10y = 0; \]
\[ y(0) = 3, \; y'(0) = -2 \]
Our goal is to determine the function \( y(t) \) that satisfies both the differential equation and the initial conditions.
First, we solve the homogeneous differential equation \( y'' + 2y' + 10y = 0 \).
1. **Find the characteristic equation:**
\[ r^2 + 2r + 10 = 0 \]
2. **Solve the characteristic equation:**
Use the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ r = \frac{-2 \pm \sqrt{4 - 40}}{2} \]
\[ r = \frac{-2 \pm \sqrt{-36}}{2} \]
\[ r = \frac{-2 \pm 6i}{2} \]
\[ r = -1 \pm 3i\ ]
The roots are complex, \( r = -1 + 3i \) and \( r = -1 - 3i \).
3. **Write the general solution:**
For complex roots \( \alpha \pm \beta i \), the general solution is:
\[ y(t) = e^{\alpha t} (C_1 \cos(\beta t) + C_2 \sin(\beta t)) \]
In this case, \( \alpha = -1 \) and \( \beta = 3 \), so the solution is:
\[ y(t) = e^{-t} (C_1 \cos(3t) + C_2 \sin(3t)) \]
4. **Apply the initial conditions to determine \( C_1 \) and \( C_2 \):**
- When \( t = 0 \):
\[ y(0) = e^{0} (C_1 \cos(0) + C_2 \sin(0)) \]
\[ y(0) = C_1 = 3 \]
Therefore, \( C_1 = 3 \).
- When
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