find the critical points of y'=5y(x-3)(y^2-1)

To find the critical points of the differential equation \( y' = 5y(x-3)(y^2-1) \), we need to find the points where \( y' = 0 \). Critical points occur where the derivative is zero or undefined. Let's solve for these points step by step.

Given:
\[
y' = 5y(x-3)(y^2-1)
\]
For \( y' = 0 \), we set the right-hand side equal to zero:

\[
5y(x-3)(y^2-1) = 0
\]

This equation can be factored into three components:
\[
5 = 0, \quad y = 0, \quad x - 3 = 0, \quad y^2 - 1 = 0
\]

Clearly, \( 5 = 0 \) is false, so we can ignore that factor. Now we analyze the remaining factors:

1. \( y = 0 \): This gives us the solution where \( y = 0 \) for any \( x \).

2. \( x - 3 = 0 \): This gives \( x = 3 \). For \( x = 3 \), \( y \) can take any value since the factor \( y^2 - 1 \) or \( y \) can be non-zero.

3. \( y^2 - 1 = 0 \): Solving this gives:
\[
y^2 = 1 \implies y = \pm 1
\]

Thus, we have \( y = 1 \) or \( y = -1 \). For these values, \( x \) can be any number because \( x-3 \) does not affect the solution.

### Critical points:

- \( y = 0 \) for any \( x \) (horizontal line at \( y = 0 \)).
- \( x = 3 \), for any \( y \).
- \( y = 1 \) and \( y = -1 \) for any \( x \) (horizontal lines at \( y = 1 \) and \( y = -1 \)).

### Conclusion:
The critical points of the differential equation are:
- The line \( y = 0 \) for all \( x \).
- The line \( y = 1 \) for all \( x \).
- The line \( y = -1 \) for all \( x \).
- The vertical line \( x = 3 \) for any \( y \).

Would you like further details on the stability of these critical points or other aspects of this solution?

### 5 Relative Questions:
1. What are the different types of critical points in differential equations?
2. How do we analyze the stability of critical points in a differential equation?
3. Can we visualize these critical points on a phase plane?
4. How does changing the factor \( (x-3) \) affect the solution's critical points?
5. What happens to the system if we modify the function to \( y' = 5y(x-3)(y^2 + 1) \)?

### Tip:
In differential equations, critical points can often be classified as stable, unstable, or saddle points, depending on their behavior in nearby regions. Understanding their stability is key for interpreting solutions.

find the critical points of y' = 5y(x-3)(y^2-1)

Explore the method for finding critical points of the differential equation y' = 5y(x-3)(y^2-1). This solution covers key concepts in differential equations, including the critical points of the system and their behavior. Suitable for high school and college-level students.

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