Find the solution the initial value problem x' -10 %3D -5 x(0) = (). cos(t) – 5 sin(t) X = -2t e cos(t) – sin(t)
Find the solution the initial value problem x' -10 %3D -5 x(0) = (). cos(t) – 5 sin(t) X = -2t e cos(t) – sin(t)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![**Solution to the Initial Value Problem**
Given the initial value problem:
\[
x' = \begin{pmatrix} 1 & -10 \\ 1 & -5 \end{pmatrix} x,
\]
with initial condition:
\[
x(0) = \begin{pmatrix} 1 \\ 1 \end{pmatrix}.
\]
### Incorrect Solution Attempt
An incorrect attempt is shown as:
\[
x = e^{-2t} \begin{pmatrix} \cos(t) - 5 \sin(t) \\ \cos(t) - \sin(t) \end{pmatrix}.
\]
### Correct Asymptotic Behavior
Describe the behavior of the solution as \( t \to \infty \):
\[
x = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \text{ as } t \to \infty.
\]
This solution indicates that as time progresses, the solution vector \( x \) approaches zero.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbb658db9-2bac-4a84-9686-e5fcb3cfbacc%2F43c68433-238a-4773-9c5e-a1189a665d95%2Fi8yil6_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Solution to the Initial Value Problem**
Given the initial value problem:
\[
x' = \begin{pmatrix} 1 & -10 \\ 1 & -5 \end{pmatrix} x,
\]
with initial condition:
\[
x(0) = \begin{pmatrix} 1 \\ 1 \end{pmatrix}.
\]
### Incorrect Solution Attempt
An incorrect attempt is shown as:
\[
x = e^{-2t} \begin{pmatrix} \cos(t) - 5 \sin(t) \\ \cos(t) - \sin(t) \end{pmatrix}.
\]
### Correct Asymptotic Behavior
Describe the behavior of the solution as \( t \to \infty \):
\[
x = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \text{ as } t \to \infty.
\]
This solution indicates that as time progresses, the solution vector \( x \) approaches zero.
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