matrix - 3

[logistic_guy]

Solve.

\(\displaystyle \bold{X}' = \begin{bmatrix}6 & -1 \\5 & 2 \end{bmatrix} \bold{X}\)

 

[khansaheb]

Solve.

\(\displaystyle \bold{X}' = \begin{bmatrix}6 & -1 \\5 & 2 \end{bmatrix} \bold{X}\)

Please show us what you have tried and exactly where you are stuck.

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Please share your work/thoughts about this problem

 

[logistic_guy]

\(\displaystyle \bold{X}' = \begin{bmatrix}6 & -1 \\5 & 2 \end{bmatrix} \bold{X}\)

Finding the eigenvalues is now a piece of cake!



\(\displaystyle \left| \begin{array}{cc}6 - \lambda & -1 \\5 & 2 - \lambda \\\end{array} \right| = (6 - \lambda)(2 - \lambda) - (-1)(5) = 0 \)

This gives:

\(\displaystyle \lambda_1 = 4 + i\)
\(\displaystyle \lambda_2 = 4 - i\)

FYI. When one eigenvalue is the conjugate of the other eigenvalue, this can be written as:

\(\displaystyle \lambda_1 = \overline{\lambda_2}\)

Welcome to the world of complex analysis!

 

[logistic_guy]

\(\displaystyle (2 - i)k_1 - k_2 = 0\)
\(\displaystyle 5k_1 - (2 + i)k_2 = 0\)

If we choose \(\displaystyle k_1 = 1\) then \(\displaystyle k_2 = 2 - i\).

Then, our first solution is:

\(\displaystyle \bold{X}_1 = \begin{bmatrix}1 \\2 - i \end{bmatrix}e^{(4+i)t}\)

 

[logistic_guy]

Now we use the second eigenvalue \(\displaystyle \lambda_2 = 4 - i\) to get:

\(\displaystyle (2 + i)k_1 - k_2 = 0\)
\(\displaystyle 5k_1 - (2 - i)k_2 = 0\)

If we choose \(\displaystyle k_1 = 1\), then \(\displaystyle k_2 = 2 + i\).

This gives the second solution as:

\(\displaystyle \bold{X}_2 = \begin{bmatrix}1 \\2 + i \end{bmatrix} e^{(4 - i)t}\)

 

[logistic_guy]

The general solution is:

\(\displaystyle \bold{X} = c_1\bold{X}_1 + c_2\bold{X}_2 = c_1\begin{bmatrix}1 \\2 - i \end{bmatrix}e^{(4+i)t}+ c_2\begin{bmatrix}1 \\2 + i \end{bmatrix} e^{(4 - i)t}\)

But this is not the best notation to use. We would love to write the solution in terms of trigonometric functions and without the imaginary \(\displaystyle i\).

Therefore, we will use the following Theorem \(\displaystyle (\textcolor{blue}{\bold{Theorem}} \textcolor{red}{\bold{69}})\):

\(\displaystyle \bold{X}_1 = (\bold{B}_1\cos \beta t - \bold{B}_2\sin \beta t)e^{\alpha t}\)
\(\displaystyle \bold{X}_2 = (\bold{B}_2\cos \beta t + \bold{B}_1\sin \beta t)e^{\alpha t}\)

where \(\displaystyle \bold{B}_1\) and \(\displaystyle \bold{B}_2\) are the real and imaginary parts of \(\displaystyle \bold{K}_1\) respectively. And \(\displaystyle \alpha\) and \(\displaystyle \beta\) are the real and imaginary parts of \(\displaystyle \lambda_1\) respectively, where \(\displaystyle \lambda_1 = \alpha + i\beta\) is the eigenvalue of \(\displaystyle \bold{K}_1\).

Let us write \(\displaystyle \bold{K}_1\) again.

\(\displaystyle \bold{K}_1 = \begin{bmatrix}1 \\2 - i \end{bmatrix} = \begin{bmatrix}1 \\2 \end{bmatrix} + i\begin{bmatrix}0 \\-1 \end{bmatrix}\)

Then,

\(\displaystyle \bold{X}_1 = \left(\begin{bmatrix}1 \\2 \end{bmatrix}\cos t - \begin{bmatrix}0 \\-1 \end{bmatrix}\sin t\right)e^{4 t}\)

\(\displaystyle \bold{X}_2 = \left(\begin{bmatrix}0 \\-1 \end{bmatrix}\cos t + \begin{bmatrix}1 \\2 \end{bmatrix}\sin t\right)e^{4 t}\)

And the general solution is:

\(\displaystyle \bold{X} = c_1\left(\begin{bmatrix}1 \\2 \end{bmatrix}\cos t - \begin{bmatrix}0 \\-1 \end{bmatrix}\sin t\right)e^{4 t} + c_2\left(\begin{bmatrix}0 \\-1 \end{bmatrix}\cos t + \begin{bmatrix}1 \\2 \end{bmatrix}\sin t\right)e^{4 t}\)

With a little simplification, the general solution becomes:

\(\displaystyle \bold{X} = \textcolor{blue}{c_1\begin{bmatrix}\cos t \\2\cos t + \sin t \end{bmatrix}e^{4 t} + c_2\begin{bmatrix}\sin t \\2\sin t - \cos t \end{bmatrix}e^{4 t}}\)