Consider the differential equation: x'=-(x+3)(x+1.5)^3(x-0.5)^2(x-1). x=-2 and solve

We are given the differential equation:

\[
x' = -(x+3)(x+1.5)^3(x-0.5)^2(x-1)
\]
and the initial condition \( x = -2 \).

To solve this differential equation qualitatively, let's break it down into several key parts:

### 1. **Equilibrium Points**
The right-hand side of the differential equation is a product of several terms. To find the equilibrium points, we set \( x' = 0 \), which corresponds to solving:

\[
-(x+3)(x+1.5)^3(x-0.5)^2(x-1) = 0
\]

The solutions are the points where each factor is zero:

\[
x+3 = 0 \quad \Rightarrow \quad x = -3
\]
\[
(x+1.5)^3 = 0 \quad \Rightarrow \quad x = -1.5
\]
\[
(x-0.5)^2 = 0 \quad \Rightarrow \quad x = 0.5
\]
\[
(x-1) = 0 \quad \Rightarrow \quad x = 1
\]

Thus, the equilibrium points are:

\[
x = -3, -1.5, 0.5, 1
\]

### 2. **Behavior Around Equilibrium Points**
The stability of each equilibrium point can be analyzed by considering the sign of \( x' \) around these points:

- **At \( x = -3 \)**: The factor \( (x+3) \) changes sign from negative to positive as \( x \) increases through \( -3 \). The other factors remain the same. This implies \( x = -3 \) is a **repelling** equilibrium point.

- **At \( x = -1.5 \)**: Since the factor \( (x+1.5)^3 \) is raised to an odd power, it changes sign as \( x \) passes through \( -1.5 \). This makes \( x = -1.5 \) a **repelling** equilibrium point.

- **At \( x = 0.5 \)**: The factor \( (x - 0.5)^2 \) is squared, so it doesn't change sign. As a result, \( x = 0.5 \) is a **neutral** equilibrium point (it behaves like a saddle or semi-stable point).

- **At \( x = 1 \)**: The factor \( (x - 1) \) changes sign from negative to positive as \( x \) increases through \( 1 \). Thus, \( x = 1 \) is a **repelling** equilibrium point.

### 3. **Initial Condition**
We are given the initial condition \( x = -2 \). To analyze the behavior starting from this point, note that:

- \( x = -2 \) is between the equilibrium points \( x = -3 \) and \( x = -1.5 \).
- Since \( x = -3 \) and \( x = -1.5 \) are both repelling, the solution will move away from these points.
  
By analyzing the differential equation's structure, it's evident that starting at \( x = -2 \), the solution will tend to move away from \( x = -3 \) and \( x = -1.5 \) and approach \( x = 0.5 \), the next closest equilibrium.

### Conclusion
Thus, the solution to the differential equation with the initial condition \( x = -2 \) qualitatively suggests that the trajectory will move toward the neutral equilibrium point at \( x = 0.5 \) over time.

Let me know if you want any details or have further questions. Here are five related questions:

1. How does the order of the factors in a differential equation affect stability?
2. What is the significance of equilibrium points in differential equations?
3. How can you classify equilibrium points as stable, unstable, or neutral?
4. How would the solution change if the initial condition was \( x = 0.8 \)?
5. What would happen if we linearized the equation around each equilibrium point?

**Tip:** Checking the sign changes of factors near equilibrium points is a useful way to determine stability.

Consider the differential equation: x' = -(x+3)(x+1.5)^3(x-0.5)^2(x-1). x=-2 and solve.

Explore how to solve the nonlinear differential equation x' = -(x+3)(x+1.5)^3(x-0.5)^2(x-1) with an initial condition x = -2. This solution covers key topics like equilibrium points, stability analysis, and the behavior of solutions near equilibrium. Suitable for undergraduate math students.

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