Solve the initial value problem x'=Ax

Question

$A = \begin{array}{cc} -2 & 1 \\ 5 & -4 \\ \end{array}$

$x(0) = \begin{array}{cc} 1\\ 3\\ \end{array}$

I undertsand that this is an eigenvector problem, and I got the values of $ -3+ \sqrt{6}$ and $ -3- \sqrt{6}$. I am unable to calculate the eigenvectors from here.

Answer 1

Using the eigenvalue $\lambda = -3 - \sqrt{6}$, we find $[A-\lambda I]v_1 = 0$ and arrive at a RREF of:

$$\left( \begin{array}{cc} 1 & \frac{1}{5} \left(-1+\sqrt{6}\right) \\ 0 & 0 \\ \end{array} \right)v_1 = 0$$

We choose $b = 1, a = \dfrac{1}{5} \left(1-\sqrt{6}\right)$, so:

$$v_1 = \left(\dfrac{1}{5} \left(1-\sqrt{6}\right), 1\right)$$

Since we have conjugate eigenvalues, we can write the eigenvector for the second eigenvalue as:

$$v_2 = \left(\frac{1}{5} \left(1+\sqrt{6}\right),1\right)$$

You can now write:

$$x(t) = c_1 ~e^{\lambda_1 t}~v_1 + c_2~ e^{\lambda_2 t}~v_2$$

Use the IC to find the constants.

Your final solution should be:

$x(t) = \dfrac{-4 e^{\left(-3-\sqrt{6}\right) t}+\sqrt{6} e^{\left(-3-\sqrt{6}\right) t}+4 e^{\left(-3+\sqrt{6}\right) t}+\sqrt{6} e^{\left(-3+\sqrt{6}\right) t}}{2 \sqrt{6}} =\dfrac{1}{3} e^{-3 t} \left(2 \sqrt{6} \sinh \left(\sqrt{6} t\right)+3 \cosh \left(\sqrt{6} t\right)\right) \\y(t) =\dfrac{-2 e^{\left(-3-\sqrt{6}\right) t}+3 \sqrt{6} e^{\left(-3-\sqrt{6}\right) t}+2 e^{\left(-3+\sqrt{6}\right) t}+3 \sqrt{6} e^{\left(-3+\sqrt{6}\right) t}}{2 \sqrt{6}} = \dfrac{1}{3} e^{-3 t} \left(\sqrt{6} \sinh \left(\sqrt{6} t\right)+9 \cosh \left(\sqrt{6} t\right)\right)$