Determine the characteristic polynomial
The characteristic polynomial is 
$$s^2 + 2s -8.$$

Explanation: Characteristic polynomials are defined in linear algebra as the following: For an \$n \times n\$ matrix \$A\$ the characteristic polynomial is
$$p_A(t) = det(A-tI)$$
where \$I\$ is the \$n \times n\$ identity matrix.

Your original differential equation is the same as the system of differential equations
$$y_1'(t) + 2y_1(t) - 8y_0(t) =  6 f(t)$$
$$y_1(t) = y_0'(t)$$

with initial conditions \$y_0(0^{-}) = 0\$ and  \$y_1(0^{-}) =1\$.

Writing \$\vec{y}(t) = \begin{bmatrix} y_1(t)\\ y_0(t) \end{bmatrix}\$ we end up with the matrix differential equation
$$\vec{y}(t)' = \begin{bmatrix} -2 &amp; 8 \\ 1 &amp; 0 \end{bmatrix} \vec{y}(t) + \begin{bmatrix} 6f(t) \\ 0 \end{bmatrix}.$$

The characteristic polynomial of the differential equation is defined as the characteristic polynomial of the coefficient matrix $$A = \begin{bmatrix} -2 &amp; 8 \\ 1 &amp; 0 \end{bmatrix}$$
which is $$p_A(t) = t^2 +2t-8.$$