Solve the Attachment? 36A- -[% 1)-[* ]-36in the system x' = Ax; thus A is the negative of the matrix in Example 4. Once again we cansolve the system using the principle of time reversal: Replacing t with -t in the right-handside of the solution in Eq. (19) of Example 4 leads tox(t) = c135t + c2-6

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Description: Solve the Attachment? 36A- -[% 1)-[* ]-36in the system x' = Ax; thus A is the negative of the matrix in Example 4. Once again we cansolve the system using the principle of time reversal: Replacing t with -t in the right-handside of the solution in Eq

Answered: 36 A- -[% 1)-[* ] -36 in the system x'…

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter9: Systems Of Equations And Inequalities

Section9.7: The Inverse Of A Matrix

Problem 30E

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Solve the Attachment?

36
A- -[% 1)-[* ]
-36
in the system x' = Ax; thus A is the negative of the matrix in Example 4. Once again we can
solve the system using the principle of time reversal: Replacing t with -t in the right-hand
side of the solution in Eq. (19) of Example 4 leads to
x(t) = c1
35t + c2
-6

Expert Solution

Step 1

What is System of Differential Equation:

A set of finite differential equations is known as a system of differential equations in mathematics. Such a system can be either linear or non-linear. A system of partial differential equations or a system of ordinary differential equations can both be used to describe this one. Eigenvalues and eigenvectors are used to determine the solution.

Given:

Given system is $x' = A x$ where $\begin{array}{rcl} A & = & - [\begin{matrix} - 36 & - 6 \\ 6 & 1 \end{matrix}] = [\begin{matrix} 36 & 6 \\ - 6 & - 1 \end{matrix}] \end{array}$ .

To Determine:

We solve the system.

Step 2

Given matrix is

$\begin{array}{rcl} A & = & - [\begin{matrix} - 36 & - 6 \\ 6 & 1 \end{matrix}] \\ = & [\begin{matrix} 36 & 6 \\ - 6 & - 1 \end{matrix}] \end{array}$

In order to determine the solution to the homogeneous system of differential equations $x' = A x$ , we need to determine the eigenvalues and corresponding eigenvectors of the matrix $A$ .

The characteristic polynomial is

$\begin{array}{rcl} P (λ) & = & |\begin{matrix} 36 - λ & 6 \\ - 6 & - 1 - λ \end{matrix}| \\ = & - (36 - λ) (1 + λ) + 36 \\ = & - (36 + 35 λ - λ^{2}) + 36 \\ = & λ^{2} - 35 λ \end{array}$

Now, solve the characteristic equation $P (λ) = 0$ . It follows,

$\begin{array}{rcl} λ^{2} - 35 λ & = & 0 \\ λ & = & 0, 35 \end{array}$

Therefore, two eigenvalues are $λ_{1} = 0, λ_{2} = 35$ .

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