**Part 3 of a Three-Part Problem****Consider the System of Differential Equations**We are given the following system of differential equations:\[y_1' = 5y_1 + 3y_2,\]\[y_2' = 3y_1 + 5y_2,\]**Solutions to the System**The solutions to this system are expressed as:\[y_1(t) = c_1 e^{2t} + c_2 e^{8t},\]\[y_2(t) = -c_1 e^{2t} + c_2 e^{8t},\]where $ c_1 $ and $ c_2 $ are arbitrary constants.**Rewrite the Solution in Vector Form**The solution can be rewritten in vector form as:\[\mathbf{y}(t) = c_1 \mathbf{y}_1(t) + c_2 \mathbf{y}_2(t).\]This is equivalent to expressing it as a linear combination of the two solution vectors:\[\mathbf{y}(t) = c_1 \begin{bmatrix} [\ \ ] \\ [\ \ ] \end{bmatrix} + c_2 \begin{bmatrix} [\ \ ] \\ [\ \ ] \end{bmatrix}.\]

Answered: This is the third part of a three-part…

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

**Part 3 of a Three-Part Problem**

**Consider the System of Differential Equations**

We are given the following system of differential equations:

\[
y_1' = 5y_1 + 3y_2,
\]

\[
y_2' = 3y_1 + 5y_2,
\]

**Solutions to the System**

The solutions to this system are expressed as:

\[
y_1(t) = c_1 e^{2t} + c_2 e^{8t},
\]

\[
y_2(t) = -c_1 e^{2t} + c_2 e^{8t},
\]

where $ c_1 $ and $ c_2 $ are arbitrary constants.

**Rewrite the Solution in Vector Form**

The solution can be rewritten in vector form as:

\[
\mathbf{y}(t) = c_1 \mathbf{y}_1(t) + c_2 \mathbf{y}_2(t).
\]

This is equivalent to expressing it as a linear combination of the two solution vectors:

\[
\mathbf{y}(t) = c_1 \begin{bmatrix} [\ \ ] \\ [\ \ ] \end{bmatrix} + c_2 \begin{bmatrix} [\ \ ] \\ [\ \ ] \end{bmatrix}.
\]

Expert Solution

Step 1: The system of differential equation is:

$y subscript 1 superscript apostrophe equals 5 y subscript 1 plus 3 y subscript 2 y subscript 2 superscript apostrophe equals 3 y subscript 1 plus 5 y subscript 2$

With solution is

$y subscript 1 open parentheses t close parentheses equals c subscript 1 e to the power of 2 t end exponent plus c subscript 2 e to the power of 8 t end exponent y subscript 2 open parentheses t close parentheses equals negative c subscript 1 e to the power of 2 t end exponent plus c subscript 2 e to the power of 8 t end exponent$