Consider the autonomous first order equation. y'=y^6 +11y^5 +28y^4. find all equibilrium solutions what is stability of equilibirum solutions

To analyze the autonomous first-order equation 

\[
y' = y^6 + 11y^5 + 28y^4,
\]

we start by finding the equilibrium solutions. Equilibrium solutions occur where \(y' = 0\). Thus, we need to solve the equation:

\[
y^6 + 11y^5 + 28y^4 = 0.
\]

### Step 1: Factoring the Equation

We can factor out \(y^4\):

\[
y^4(y^2 + 11y + 28) = 0.
\]

This gives us one equilibrium solution:

\[
y^4 = 0 \implies y = 0.
\]

Next, we need to solve the quadratic equation \(y^2 + 11y + 28 = 0\). We can use the quadratic formula:

\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-11 \pm \sqrt{11^2 - 4 \cdot 1 \cdot 28}}{2 \cdot 1}.
\]

Calculating the discriminant:

\[
b^2 - 4ac = 121 - 112 = 9.
\]

Now substituting back into the quadratic formula:

\[
y = \frac{-11 \pm 3}{2}.
\]

This gives us two additional solutions:

\[
y_1 = \frac{-11 + 3}{2} = -4 \quad \text{and} \quad y_2 = \frac{-11 - 3}{2} = -7.
\]

### Summary of Equilibrium Solutions

The equilibrium solutions are:

1. \(y = 0\)
2. \(y = -4\)
3. \(y = -7\)

### Step 2: Stability Analysis

To determine the stability of these equilibrium solutions, we analyze the sign of \(y'\) around each equilibrium point. We can evaluate the derivative of the right-hand side:

\[
f(y) = y^6 + 11y^5 + 28y^4.
\]

### Finding \(f'(y)\)

Using the power rule:

\[
f'(y) = 6y^5 + 55y^4 + 112y^3.
\]

We will check the stability at each equilibrium point:

1. **At \(y = 0\)**:
   - \(f'(0) = 0\) (neutral stability). 
   - Check values slightly less than and greater than 0: 
     - For \(y < 0\), \(f(y) < 0\) (decreasing).
     - For \(y > 0\), \(f(y) > 0\) (increasing).
   - Conclusion: **Neutral stability** at \(y = 0\).

2. **At \(y = -4\)**:
   - Calculate \(f'(-4)\):
     - \(f'(-4) = 6(-4)^5 + 55(-4)^4 + 112(-4)^3\).
     - After computation, \(f'(-4) < 0\) (decreasing).
   - Conclusion: **Stable** (attracting).

3. **At \(y = -7\)**:
   - Calculate \(f'(-7)\):
     - \(f'(-7) = 6(-7)^5 + 55(-7)^4 + 112(-7)^3\).
     - After computation, \(f'(-7) > 0\) (increasing).
   - Conclusion: **Unstable** (repelling).

### Summary of Stability

- \(y = 0\): Neutral stability
- \(y = -4\): Stable (attracting)
- \(y = -7\): Unstable (repelling)

If you have any questions or would like further details, let me know! Here are some related questions you might consider:

1. How do you solve higher-order polynomial equations?
2. What is the significance of equilibrium solutions in differential equations?
3. Can you explain the concept of stability in more detail?
4. How do you determine the stability of equilibrium points for nonlinear systems?
5. What are some physical examples where equilibrium and stability are important? 

**Tip:** Always consider the behavior of the function around the equilibrium points to accurately determine stability.

Consider the autonomous first order equation. y'=y^6 +11y^5 +28y^4. find all equilibrium solutions what is stability of equilibrium solutions

Explore the equilibrium solutions and stability analysis for the differential equation y' = y^6 + 11y^5 + 28y^4. Discover the equilibrium points and their stability classifications, suitable for advanced students in Grades 11-12.

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