matrix - 2

[logistic_guy]

Solve.

$\displaystyle \bold{X}' = \begin{bmatrix}2 & 1 \\-1 & 0 \end{bmatrix} \bold{X}$

 

[logistic_guy]

$\displaystyle \bold{X}' = \begin{bmatrix}2 & 1 \\-1 & 0 \end{bmatrix} \bold{X}$

We are experts now and we know how to find the eigenvalues.

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

This gives:

$\displaystyle \lambda_1 = \lambda_2 = 1$

 

[logistic_guy]

As usual next we construct two equations and solve for $\displaystyle k_1$ and $\displaystyle k_2$.

$\displaystyle k_1 + k_2 = 0$
$\displaystyle -k_1 - k_2 = 0$

We can choose $\displaystyle k_1 = 1$ then $\displaystyle k_2 = -1$, but for F U N we will choose $\displaystyle k_1 = 4$ so that $\displaystyle k_2 = -4$. This is a good demonstration of that it does not matter what we choose as long as the two equations are satisfied! (Except of course the choice $\displaystyle k_1 = k_2 = 0$ which leads to the trivial solution that we don't want!)

This gives us the first solution which is:

$\displaystyle \bold{X}_1 = \bold{K}_1e^{\lambda_1t} =\begin{bmatrix}4 \\-4 \end{bmatrix}e^{t}$

But now we have a problem to find the second solution as $\displaystyle \lambda_1 = \lambda_2$.



With a careful study it was found out that when we have repeated eigenvalues, a second solution can be found with the following form:

$\displaystyle \bold{X}_2 = \bold{P}e^{\lambda_1 t} + \bold{K}te^{\lambda_1 t}$

This means that we only have to find $\displaystyle \bold{P}$ and we are done!

Finding $\displaystyle \bold{P}$ is the same way as finding $\displaystyle \bold{K}$, but its equations are non-homogeneous by $\displaystyle \bold{K}$ values.

That is:

$\displaystyle (\bold{A} - \lambda\bold{I})\bold{P} = \bold{K}$

Or

$\displaystyle p_1 + p_2 = 4$
$\displaystyle -p_1 - p_2 = -4$

We can even choose $\displaystyle p_1 = 4$ then $\displaystyle p_2 = 0$ but for F U N we will choose $\displaystyle p_1 = 1$ so that $\displaystyle p_2 = 3$.

Then our second solution is:

$\displaystyle \bold{X}_2 = \begin{bmatrix}1 \\3 \end{bmatrix}e^{t} + \begin{bmatrix}4 \\-4 \end{bmatrix}te^{t}$

And the general solution is:

$\displaystyle \bold{X} = c_1\bold{X}_1 + c_2\bold{X}_2 = c_1\begin{bmatrix}4 \\-4 \end{bmatrix}e^{t} + c_2\left(\begin{bmatrix}1 \\3 \end{bmatrix}e^{t} + \begin{bmatrix}4 \\-4 \end{bmatrix}te^{t}\right)$