To classify the equilibrium point at the origin for the system $\frac{dY}{dt} = AY$ where $A = \begin{bmatrix} -15 & 5 \\ 3 & -1 \end{bmatrix}$, we need to analyze the eigenvalues of matrix $A$. The type of equilibrium point depends on the nature (real or complex) and sign of the eigenvalues.

Step-by-Step Solution

1. Calculate the Eigenvalues of $A$: For matrix $A$, the eigenvalues $\lambda$ satisfy the characteristic polynomial: $\det(A - \lambda I) = 0$ where $I$ is the identity matrix. Thus, we have: $\det \begin{bmatrix} -15 - \lambda & 5 \\ 3 & -1 - \lambda \end{bmatrix} = 0$

2. Find the Determinant: Expanding the determinant: $(-15 - \lambda)(-1 - \lambda) - (5)(3) = 0$ Simplifying, we get: $\lambda^2 + 16\lambda + 20 = 0$

3. Solve for $\lambda$: Using the quadratic formula: $\lambda = \frac{-16 \pm \sqrt{(16)^2 - 4 \cdot 1 \cdot 20}}{2 \cdot 1}$ $= \frac{-16 \pm \sqrt{256 - 80}}{2}$ $= \frac{-16 \pm \sqrt{176}}{2}$ $= \frac{-16 \pm 4\sqrt{11}}{2}$ $= -8 \pm 2\sqrt{11}$ This results in two real eigenvalues: $\lambda_1 = -8 + 2\sqrt{11}$ and $\lambda_2 = -8 - 2\sqrt{11}$.

4. Determine the Type of Equilibrium Point:

   • Since both eigenvalues are real and negative, the equilibrium point is stable.
   • Specifically, when both eigenvalues are real and negative, the system is classified as a sink.

Answer:

The equilibrium point at the origin is a sink.

Would you like further details or have any questions?

Here are 5 related questions for further study:

1. How would the classification change if one eigenvalue was positive and the other negative?
2. What if both eigenvalues were complex with a negative real part?
3. How do we interpret the eigenvalues geometrically for stability analysis?
4. Can this classification be visualized on a phase portrait?
5. What are the implications of a saddle point in a system?

Tip: The eigenvalues of the system matrix provide insight into stability and the type of equilibrium point, with their real parts indicating growth or decay rates.