While often the eigenvalues for the second order linear systems end up being negative in applications, the technique works just as well with positive eigenvalues. Take the system $${\vec{x}\,}'' = \begin{bmatrix} {\frac{29}{2}} & -{\frac{21}{2}} \\ -{\frac{21}{2}} & {\frac{29}{2}} \end{bmatrix} \vec{x}$$ Find the two eigenvalues, and let $\lambda_1$ be the smaller one.

$\lambda_1 = {}$

$\lambda_2 = {}$

help (numbers)

Guess a solution of the form $\vec{x} = \vec{v} e^{\omega t}$, plug it into ${\vec{x}\,}' = A \vec{x}$ as usual. Then guess what $\vec{v}$ and $\omega$ should be in terms of the eigenvalues and eigenvectors of $A$. You should collect 4 different linearly independent solutions. For this notice that a square root of a positive number has two answers, a positive and a negative.

Then find the solution of the initial value problem where: $$\vec{x}(0) = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \qquad {\vec{x}\,}'(0) = \begin{bmatrix} 8 \\ 0 \end{bmatrix}$$

$x_1(t) = {}$

$x_2(t) = {}$

help (formulas)