Finally, 
x
∗
 is asymptotically stable if it is both attracting and Liapunov stable.
For each of the following systems, decide whether the origin is attracting, Liapunov stable, asymptotically stable, or none of the above.
a)
x
˙
=
y
, 
y
˙
=
−
4
x


We are given the system of differential equations:

\[
\dot{x} = y
\]
\[
\dot{y} = -4x
\]

This system can be analyzed to determine the stability of the origin \((x^*, y^*) = (0,0)\) by checking if it is attracting, Liapunov stable, asymptotically stable, or none of the above.

### Step 1: Write the system in matrix form

This is a second-order linear system, which can be expressed in matrix form as:

\[
\frac{d}{dt} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -4 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}
\]

The corresponding system matrix is:

\[
A = \begin{pmatrix} 0 & 1 \\ -4 & 0 \end{pmatrix}
\]

### Step 2: Find the eigenvalues of the system matrix

The stability of the system can be analyzed by finding the eigenvalues of \( A \). The eigenvalues are obtained by solving the characteristic equation:

\[
\det(A - \lambda I) = 0
\]

where \( I \) is the identity matrix and \( \lambda \) represents the eigenvalues. The characteristic equation is:

\[
\det\begin{pmatrix} -\lambda & 1 \\ -4 & -\lambda \end{pmatrix} = 0
\]

The determinant is:

\[
\lambda^2 + 4 = 0
\]

Solving for \( \lambda \), we get:

\[
\lambda^2 = -4
\]
\[
\lambda = \pm 2i
\]

Thus, the eigenvalues are purely imaginary, \( \lambda = 2i \) and \( \lambda = -2i \).

### Step 3: Analyze the eigenvalues for stability

- **Liapunov Stability**: The origin is Liapunov stable if all eigenvalues have non-positive real parts. In this case, both eigenvalues are purely imaginary, meaning their real parts are zero. Therefore, the origin is **Liapunov stable**.
  
- **Attracting**: The origin is attracting if all eigenvalues have negative real parts (indicating solutions tend to the origin). Since the real parts of the eigenvalues are zero, the origin is **not attracting**.

- **Asymptotically Stable**: The origin is asymptotically stable if it is both Liapunov stable and attracting. Since the origin is not attracting, it is **not asymptotically stable**.

### Conclusion

For the system:

\[
\dot{x} = y, \quad \dot{y} = -4x
\]

The origin \((0,0)\) is **Liapunov stable** but **not attracting** and hence **not asymptotically stable**.

---

Do you want further details or have any questions?

### Related questions:
1. What is the definition of Liapunov stability in dynamical systems?
2. How do complex eigenvalues affect stability in linear systems?
3. What is the relationship between attracting and asymptotic stability?
4. Can a system be Liapunov stable but not asymptotically stable?
5. How would the stability change if the eigenvalues had negative real parts?

#### Tip:
When analyzing linear systems for stability, the eigenvalues of the system matrix provide crucial insight into the nature of the equilibrium points.

Finally, x* is asymptotically stable if it is both attracting and Liapunov stable. For each of the following systems, decide whether the origin is attracting, Liapunov stable, asymptotically stable, or none of the above. a) x˙ = y, y˙ = −4x

Analyze the stability of the system of differential equations x˙ = y, y˙ = -4x using Lyapunov stability and eigenvalue methods. This problem involves determining whether the origin is attracting, Liapunov stable, or asymptotically stable. Suitable for undergraduate students studying differential equations and stability analysis.

Math Problem Statement

Finally, x ∗ is asymptotically stable if it is both attracting and Liapunov stable. For each of the following systems, decide whether the origin is attracting, Liapunov stable, asymptotically stable, or none of the above. a) x ˙

y , y ˙

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