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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

Find general solutions of the systems in Problem Attached. Use a computer system or graphing calculator to construct a direction field and typical solution curves for the given system.

Expert Solution

Step 1

The given system is $x' = [\begin{matrix} - 2 & 1 \\ - 1 & - 4 \end{matrix}] x$

We have to find the general solution of this system and sketch its direction field.

Step 2

The given system is $x' = [\begin{matrix} - 2 & 1 \\ - 1 & - 4 \end{matrix}] x$

Let $A = [\begin{matrix} - 2 & 1 \\ - 1 & - 4 \end{matrix}]$

To find eigenvalues of $A$ :

$λ$ be an eigenvalue of $A$ , then $d e t (A - λ I) = 0$

$\begin{array}{rcl} |\begin{matrix} - 2 - λ & 1 \\ - 1 & - 4 - λ \end{matrix}| & = & 0 \\ (- 2 - λ) (- 4 - λ) + 1 & = & 0 \\ 8 + 2 λ + 4 λ + λ^{2} + 1 & = & 0 \\ λ^{2} + 6 λ + 9 & = & 0 \\ {(λ + 3)}^{2} & = & 0 \\ λ + 3 & = & 0 \\ λ & = & - 3 \end{array}$

The eigenvalue is $λ = - 3$ of multiplicity $2$ .

Step 3

To find the eigenvector of $A$ corresponding to the eigenvalue $λ = - 3$ :

If $X$ is an eigenvector of $A$ corresponding to the eigenvalue $λ$ ,then $(A - λ I) X = 0$

Now,

$\begin{array}{rcl} (A - λ I) X & = & 0 \\ (A + 3 I) X & = & 0 \\ [\begin{matrix} - 2 + 3 & 1 \\ - 1 & - 4 + 3 \end{matrix}] [\begin{matrix} x \\ y \end{matrix}] & = & [\begin{matrix} 0 \\ 0 \end{matrix}] \\ [\begin{matrix} 1 & 1 \\ - 1 & - 1 \end{matrix}] [\begin{matrix} x \\ y \end{matrix}] & = & [\begin{matrix} 0 \\ 0 \end{matrix}] \end{array}$

$\begin{array}{rcl} x + y & = & 0 \\ x & = & - y \end{array}$

$X = [\begin{matrix} x \\ y \end{matrix}] = [\begin{matrix} - y \\ y \end{matrix}] = y [\begin{matrix} - 1 \\ 1 \end{matrix}]$

Therefore, $X = [\begin{matrix} - 1 \\ 1 \end{matrix}]$ is the eigenvector of $A$ corresponding to the eigenvalue $λ = - 3$ .

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