. Solve the initial value problem= 2yı + y2 + y +y2y = -5yı + 2y2 + 5y – y2yı(0) = y2(0) = y (0) = 4,y,(0) = -4

Answered: Solve the initial value problem = 2yı +…

Elementary Linear Algebra (MindTap Course List)

8th Edition

ISBN:9781305658004

Author:Ron Larson

Publisher:Ron Larson

Chapter2: Matrices

Section2.2: Properties Of Matrrix Operations

Problem 13E: Solve for X in the Equation, given A=-401-532 and B=12-2144 a 3X+2A=B b 2A5B=3X c X-3A+2B=0 d...

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Question

Solve the initial value problem
= 2yı + y2 + y +y2
y = -5yı + 2y2 + 5y – y2
yı(0) = y2(0) = y (0) = 4,
y,(0) = -4

Expert Solution

Step 1

Given :

$y_{1}^{''} = 2 y_{1} + y_{2} + y_{1}^{'} + y_{2}^{'} y_{2}^{''} = - 5 y_{1} + 2 y_{2} + 5 y_{1}^{'} - y_{2}^{'}$ , $y_{1} (0) = y_{2} (0) = y_{1}^{'} (0) = 4, y_{2}^{'} (0) = - 4$

To Find : Solution of given differential equation

Step 2

Given that the initial value $y_{1}^{'}$ and $y_{2}^{'}$ .

Hence , Let

$y_{1}^{'} = y_{3}$

$y_{2}^{'} = y_{4}$

$y_{3}^{'} = 2 y_{1} + y_{2} + y_{3} + y_{4} y_{4}^{'} = - 5 y_{1} + 2 y_{2} + 5 y_{3} - y_{4}$

So equivalent system of first order ODE

$[\begin{matrix} y_{1}^{'} \\ y_{2}^{'} \\ y_{3}^{'} \\ y_{4}^{'} \end{matrix}] = [\begin{matrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 2 & 1 & 1 & 1 \\ - 5 & 2 & 5 & - 1 \end{matrix}] [\begin{matrix} y_{1} \\ y_{2} \\ y_{3} \\ y_{4} \end{matrix}]$

$y_{1} (0) = y_{2} (0) = y_{3} (0) = 4, y_{4} (0) = - 4$

Here $A = [\begin{matrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 2 & 1 & 1 & 1 \\ - 5 & 2 & 5 & - 1 \end{matrix}]$

We will find eigenvalues and eigenvectors of A

Consider characteristics equation of A is $d e t (A - λ I) = 0$

$λ^{4} - 10 λ^{2} + 9 = 0 λ^{4} - 9 λ^{2} - λ^{2} + 9 = 0 λ^{2} (λ^{2} - 9) - 1 (λ^{2} - 9) = 0 (λ^{2} - 9) (λ^{2} - 1) = 0 λ = 1, - 1, 3, - 3$

Eigenvalues of A are 1 , -1 , -3 , 3

For Eigenvector corresponding $λ = 1$

Consider $v \in N u l l (A - I)$

$(A - I) v = 0$

$[\begin{matrix} - 1 & 0 & 1 & 0 \\ 0 & - 1 & 0 & 1 \\ 2 & 1 & 0 & 1 \\ - 5 & 2 & 5 & - 2 \end{matrix}] [\begin{matrix} x \\ y \\ z \\ w \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \end{matrix}]$

After reducing we get

$[\begin{matrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & - 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{matrix}] [\begin{matrix} x \\ y \\ z \\ w \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \end{matrix}]$

$x + w = 0 y - w = 0 z + w = 0$

$N u l l (A - I) = \{(x, y, z, w) | x = - w, y = w, z = - w\} = \{(- w, w, - w, w) | w \in R\} = s p a n (- 1, 1, - 1, 1)$

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