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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?

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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.

Classify the equilibrium point at the origin for the system \(\frac{dY}{dt} = AY\) with \(A = \begin{bmatrix} -15 & 5 \\ 3 & -1 \end{bmatrix}\).

Learn how to classify the equilibrium point at the origin for the system \(\frac{dY}{dt} = AY\), where \(A\) is a 2x2 matrix. This problem covers eigenvalue analysis for stability, with step-by-step calculations. Suitable for college-level students studying differential equations and stability analysis.

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