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:

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