Consider the initial value problem-41y,0 -4F0) = .a. Find the eigenvalue X, an eigenvector v1, and a generalized eigenvector v2 for the coefficient matrix of this linear system.v1b. Find the most general real-valued solution to the linear system of differential equations. Use t as the independent variable in your answers.y(t) = c1+ c2%3Dc. Solve the original initial value problem.Y1 (t) :Y2(t) =

Given y '→=-410-4y → and y →(0) = 1-4

Answered: i' = [, T0) = ý, j(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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Question

Consider the initial value problem
-4
1
y,
0 -4
F0) = .
a. Find the eigenvalue X, an eigenvector v1, and a generalized eigenvector v2 for the coefficient matrix of this linear system.
v1
b. Find the most general real-valued solution to the linear system of differential equations. Use t as the independent variable in your answers.
y(t) = c1
+ c2
%3D
c. Solve the original initial value problem.
Y1 (t) :
Y2(t) =

Expert Solution

Step 1

Given

$\vec{y'} = [\begin{matrix} - 4 & 1 \\ 0 & - 4 \end{matrix}] \vec{y}$ and $\vec{y} (0) = [\begin{matrix} 1 \\ - 4 \end{matrix}]$

Step 2

(a)

Let $λ$ be the eigenvalue of the coefficient matrix $A = [\begin{matrix} - 4 & 1 \\ 0 & - 4 \end{matrix}]$

Then

$\det (A - λI) = |\begin{matrix} - 4 - λ & 1 \\ 0 & - 4 - λ \end{matrix}| = 0 \det (A - λI) = {(- 4 - λ)}^{2} = 0 {(4 + λ)}^{2} = 0 λ = - 4, - 4$

this implies -4 is the repeated eigenvalue of matrix A

Step 3

To find the eigenvector $x_{1}$ corresponding to $λ = - 4$

$(A - λI) x_{1} = [\begin{matrix} - 4 - λ & 1 \\ 0 & - 4 - λ \end{matrix}] [\begin{matrix} a \\ b \end{matrix}] = 0$ where $x = [\begin{matrix} a \\ b \end{matrix}]$ and $λ = - 4$

$(A - (- 4) I) x_{1} = [\begin{matrix} - 4 - (- 4) & 1 \\ 0 & - 4 - (- 4) \end{matrix}] [\begin{matrix} a \\ b \end{matrix}] = 0 (A + 4) I) x_{1} = [\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}] [\begin{matrix} a \\ b \end{matrix}] = 0$

$[\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}] [\begin{matrix} a \\ b \end{matrix}] = 0 \Rightarrow b = 0 x_{1} = [\begin{matrix} a \\ 0 \end{matrix}] x_{1} = [\begin{matrix} 1 \\ 0 \end{matrix}], a = 1$

This implies there is only one Linearly Independent vector corresponding to the eigenvalue $λ = - 4$

Hence the matrix $A = [\begin{matrix} - 4 & 1 \\ 0 & - 4 \end{matrix}]$ is defective

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