Answered: Matrix System 3 -2 x' = -1 3 -2 X; X1 =…

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

First verify that the given vectors are solutions of the given system. Then use the Wronskian to show that they are linearly independent. Finally, write the general solution of the system. find a particular solution of the indicatedlinear system that satisfies the below initial conditions

x₁(0) = 0 , x₂ (0) = 0, x₃(0)=4

Matrix System
3 -2
x' =
-1
3
-2
X; X1 = et
-1
3
X2 = e3t
X3 = e5t
-2

Expert Solution

Step 1

Given that, the system is-

$x' = [\begin{matrix} 3 & - 2 & 0 \\ - 1 & 3 & - 2 \\ 0 & - 1 & 3 \end{matrix}] x$ with

$x_{1} = e^{t} [\begin{matrix} 2 \\ 2 \\ 1 \end{matrix}]$ , $x_{2} = e^{3 t} [\begin{matrix} - 2 \\ 0 \\ 1 \end{matrix}]$ , $x_{3} = e^{5 t} [\begin{matrix} 2 \\ - 2 \\ 1 \end{matrix}]$ .

Step 2

Since,

First given vector is $x_{1} = e^{t} [\begin{matrix} 2 \\ 2 \\ 1 \end{matrix}]$ .

Now,

$x_{1}' = e^{t} [\begin{matrix} 2 \\ 2 \\ 1 \end{matrix}]$

and,

$\begin{array}{rcl} A x_{1} & = & e^{t} [\begin{matrix} 3 & - 2 & 0 \\ - 1 & 3 & - 2 \\ 0 & - 1 & 3 \end{matrix}] [\begin{matrix} 2 \\ 2 \\ 1 \end{matrix}] \\ = & e^{t} [\begin{matrix} 2 \\ 2 \\ 1 \end{matrix}] \\ = & x_{1}^{'} \end{array}$

So,

$x_{1}$ is a solution of the system.

Step 3

Since,

The second given vector is $x_{2} = e^{3 t} [\begin{matrix} - 2 \\ 0 \\ 1 \end{matrix}]$ .

Now,

$x_{2}' = 3 e^{3 t} [\begin{matrix} - 2 \\ 0 \\ 1 \end{matrix}]$

and,

$\begin{array}{rcl} A x_{2} & = & e^{3 t} [\begin{matrix} 3 & - 2 & 0 \\ - 1 & 3 & - 2 \\ 0 & - 1 & 3 \end{matrix}] [\begin{matrix} - 2 \\ 0 \\ 1 \end{matrix}] \\ = & e^{3 t} [\begin{matrix} - 6 \\ 0 \\ 3 \end{matrix}] \\ = & 3 e^{3 t} [\begin{matrix} - 2 \\ 0 \\ 1 \end{matrix}] \\ = & x_{2}^{'} \end{array}$

So,

$x_{2}$ is a solution of the system.

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