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To classify the critical points of the given second-order nonlinear differential equation, let's break down the problem.

### Step 1: Convert the Equation into a System of First-Order Equations

The equation given is:
\[
\ddot{x} + 2(1 - x^3)\dot{x} - 3x = 0
\]

Define:
\[
x_1 = x \quad \text{and} \quad x_2 = \dot{x}
\]

Then, we have:
\[
\dot{x}_1 = x_2
\]
and
\[
\dot{x}_2 = \ddot{x} = -2(1 - x_1^3)x_2 + 3x_1
\]

This gives us the first-order system:
\[
\begin{cases}
\dot{x}_1 = x_2 \\
\dot{x}_2 = -2(1 - x_1^3)x_2 + 3x_1
\end{cases}
\]

### Step 2: Find the Critical Points

To find the critical points, we set \(\dot{x}_1 = 0\) and \(\dot{x}_2 = 0\):
\[
x_2 = 0
\]
and
\[
-2(1 - x_1^3)x_2 + 3x_1 = 0
\]

Substitute \(x_2 = 0\) into the second equation:
\[
3x_1 = 0
\]
This implies \(x_1 = 0\).

Therefore, the only critical point is:
\[
(x_1, x_2) = (0, 0)
\]

### Step 3: Linearize the System Around the Critical Point

To classify the critical point, we need to linearize the system at \((0, 0)\). Calculate the Jacobian matrix \(J\) of the system at \((x_1, x_2) = (0, 0)\).

The system is:
\[
\begin{cases}
\dot{x}_1 = x_2 \\
\dot{x}_2 = -2(1 - x_1^3)x_2 + 3x_1
\end{cases}
\]

The partial derivatives are:
\[
\frac{\partial \dot{x}_1}{\partial x_1} = 0, \quad \frac{\partial \dot{x}_1}{\partial x_2} = 1
\]
\[
\frac{\partial \dot{x}_2}{\partial x_1} = 3, \quad \frac{\partial \dot{x}_2}{\partial x_2} = -2(1 - x_1^3)
\]

At \((0, 0)\), the Jacobian becomes:
\[
J = \begin{pmatrix} 0 & 1 \\ 3 & -2 \end{pmatrix}
\]

### Step 4: Analyze the Eigenvalues of the Jacobian

To classify the critical point, we find the eigenvalues of the Jacobian matrix \(J\):
\[
\det\begin{pmatrix} 0 - \lambda & 1 \\ 3 & -2 - \lambda \end{pmatrix} = 0
\]

Expanding the determinant:
\[
\lambda^2 + 2\lambda + 3 = 0
\]

Solve for \(\lambda\):
\[
\lambda = \frac{-2 \pm \sqrt{4 - 12}}{2} = -1 \pm i\sqrt{2}
\]

The eigenvalues are complex with a negative real part, which implies a **stable spiral** at the critical point \((0, 0)\).

### Conclusion

The critical point \((0, 0)\) is a **stable spiral**.

Would you like more details or have any questions?

### Related Questions:

1. How do you interpret the stability of a spiral in a dynamical system?
2. What other types of critical points exist in nonlinear systems?
3. How does the behavior change if the real part of the eigenvalues were positive?
4. Why does the Jacobian matrix help in linearizing a nonlinear system?
5. Can we classify critical points for systems of higher-order differential equations in a similar way?

### Tip

When classifying critical points, always check the signs and types of eigenvalues, as they directly indicate the nature of the critical point.

Classify the critical points of the plane autonomous system corresponding to the second-order nonlinear differential equation: \( \ddot{x} + 2(1 - x^3)\dot{x} - 3x = 0 \).

This solution provides a detailed approach to classifying critical points of a second-order nonlinear differential equation system. It covers the transformation to a first-order system, finding critical points, linearization, and stability analysis using eigenvalues. Suitable for undergraduate students studying differential equations.

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