1. **Problem statement:** Given the differential equation $$\frac{d^2x}{dt^2} + 5 \frac{dx}{dt} + 6x = 0,$$ we are to rewrite it as a system of first order differential equations, find eigenvalues of the associated matrix, and determine the long-term velocity of the particle. 2. **Substitution and matrix form:** Let $$y = \frac{dx}{dt}.$$ Then, $$\frac{dy}{dt} = \frac{d^2x}{dt^2} = -5y - 6x$$ from the original equation. Thus, the system is: $$\frac{dx}{dt} = y,$$ $$\frac{dy}{dt} = -6x - 5y.$$ In matrix form: $$\begin{pmatrix} \frac{dx}{dt} \\ \frac{dy}{dt} \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -6 & -5 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}.$$ 3. **Finding eigenvalues:** The matrix is $$A = \begin{pmatrix} 0 & 1 \\ -6 & -5 \end{pmatrix}.$$ Eigenvalues $$\lambda$$ satisfy: $$\det(A - \lambda I) = 0,$$ which is: $$\det \begin{pmatrix} -\lambda & 1 \\ -6 & -5 - \lambda \end{pmatrix} = 0,$$ so $$(-\lambda)(-5 - \lambda) - (-6)(1) = 0,$$ $$\lambda(5 + \lambda) + 6 = 0,$$ $$\lambda^2 + 5\lambda + 6 = 0.$$ 4. **Solving quadratic:** $$\lambda = \frac{-5 \pm \sqrt{25 - 24}}{2} = \frac{-5 \pm 1}{2}.$$ So, $$\lambda_1 = \frac{-5 + 1}{2} = -2,$$ $$\lambda_2 = \frac{-5 - 1}{2} = -3.$$ 5. **Long-term velocity:** Since both eigenvalues are negative, the system's solutions decay to zero as $$t \to \infty$$. Therefore, the long-term velocity $$y = \frac{dx}{dt}$$ tends to $$0$$. **Final answer:** The eigenvalues are $$-2$$ and $$-3$$, and the long-term velocity of the particle is $$0$$.