Answered: In the following system Problem,…

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

In the following system Problem, categorize the eigenvalues and eigenvectors of the coefficient matrix A and sketch the phase portrait of the system by hand. Then use a computer system or graphing calculator to check your answer.

x'₁ =2x₁ + 3x₂, x'₂ = 2x₁ + x₂

Expert Solution

Step 1

Given that,

The system of differential equation is-

$x_{1}' = 2 x_{1} + 3 x_{2} x_{2}' = 2 x_{1} + x_{2}$

The system can be written in the matrix form as-

$[\begin{matrix} x_{1}' \\ x_{2}' \end{matrix}] = [\begin{matrix} 2 & 3 \\ 2 & 1 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \end{matrix}]$

Here,

The coefficient matrix is-

$A = [\begin{matrix} 2 & 3 \\ 2 & 1 \end{matrix}]$

Step 2

Let,

The eigenvalue of $A$ is $λ$ , then the characteristics equation is-

$|A - λ I| = 0 \Rightarrow |\begin{matrix} 2 - λ & 3 \\ 2 & 1 - λ \end{matrix}| = 0 \Rightarrow (2 - λ) (1 - λ) - 6 = 0 \Rightarrow λ^{2} - 3 λ + 2 - 6 = 0 \Rightarrow λ^{2} - 3 λ - 4 = 0 \Rightarrow (λ - 4) (λ + 1) = 0 \Rightarrow λ = 4, - 1$

For $λ = 4$ :

Let the eigenvector is $U = [\begin{matrix} u_{1} \\ u_{2} \end{matrix}]$ .

Then,

$[\begin{matrix} 2 - 4 & 3 \\ 2 & 1 - 4 \end{matrix}] [\begin{matrix} u_{1} \\ u_{2} \end{matrix}] = [\begin{matrix} 0 \\ 0 \end{matrix}] [\begin{matrix} - 2 & 3 \\ 2 & - 3 \end{matrix}] [\begin{matrix} u_{1} \\ u_{2} \end{matrix}] = [\begin{matrix} 0 \\ 0 \end{matrix}] ~ [\begin{matrix} - 2 & 3 \\ 0 & 0 \end{matrix}] [\begin{matrix} u_{1} \\ u_{2} \end{matrix}] = [\begin{matrix} 0 \\ 0 \end{matrix}]$

So, by comparing the coefficients, it gives-

$- 2 u_{1} + 3 u_{2} = 0 \Rightarrow u_{1} = 3, u_{2} = 2$

So,

$[\begin{matrix} u_{1} \\ u_{2} \end{matrix}] = [\begin{matrix} 3 \\ 2 \end{matrix}]$

Now,

Let, the eigenvector for $λ = - 1$ is $V = [\begin{matrix} v_{1} \\ v_{2} \end{matrix}]$ .

Then,

$[\begin{matrix} 2 + 1 & 3 \\ 2 & 1 + 1 \end{matrix}] [\begin{matrix} v_{1} \\ v_{2} \end{matrix}] = [\begin{matrix} 0 \\ 0 \end{matrix}] [\begin{matrix} 3 & 3 \\ 2 & 2 \end{matrix}] [\begin{matrix} v_{1} \\ v_{2} \end{matrix}] = [\begin{matrix} 0 \\ 0 \end{matrix}] ~ [\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix}] [\begin{matrix} v_{1} \\ v_{2} \end{matrix}] = [\begin{matrix} 0 \\ 0 \end{matrix}]$

Then,

$v_{1} + v_{2} = 0 \Rightarrow v_{1} = 1, v_{2} = - 1$

So,

$V = [\begin{matrix} v_{1} \\ v_{2} \end{matrix}] = [\begin{matrix} 1 \\ - 1 \end{matrix}]$

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