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:

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