We are tasked with finding and classifying the equilibria of the differential equation:

$y' = y(3 - y)(25 - y^2)$

Step 1: Finding the equilibria

Equilibria occur where $y' = 0$, meaning the derivative of $y$ is zero. Set the right-hand side equal to zero:

$y(3 - y)(25 - y^2) = 0$

Now solve for $y$:

• $y = 0$
• $3 - y = 0 \quad \Rightarrow \quad y = 3$
• $25 - y^2 = 0 \quad \Rightarrow \quad y^2 = 25 \quad \Rightarrow \quad y = \pm 5$

Thus, the equilibria are $y = 0$, $y = 3$, $y = 5$, and $y = -5$.

Step 2: Classifying the equilibria

To classify the equilibria, we need to examine the stability of each equilibrium. This can be done by finding the derivative of the right-hand side with respect to $y$ (i.e., the function $f(y) = y(3 - y)(25 - y^2)$) and evaluating it at each equilibrium point.

Finding the derivative:

$f(y) = y(3 - y)(25 - y^2)$

Using the product rule, differentiate $f(y)$:

$f'(y) = \frac{d}{dy} \left[ y(3 - y)(25 - y^2) \right]$ $f'(y) = (3 - y)(25 - y^2) + y \cdot (-1)(25 - y^2) + y(3 - y) \cdot (-2y)$

Simplifying each term:

$f'(y) = (3 - y)(25 - y^2) - y(25 - y^2) - 2y^2(3 - y)$

Now, evaluate $f'(y)$ at the equilibrium points $y = 0$, $y = 3$, $y = 5$, and $y = -5$.

At $y = 0$:

$f'(0) = (3 - 0)(25 - 0^2) = 3 \times 25 = 75$

Since $f'(0) > 0$, the equilibrium at $y = 0$ is unstable.

At $y = 3$:

$f'(3) = (3 - 3)(25 - 3^2) - 3(25 - 3^2) - 2(3)^2(3 - 3)$ $f'(3) = 0 - 3(25 - 9) - 0 = -3(16) = -48$

Since $f'(3) < 0$, the equilibrium at $y = 3$ is stable.

At $y = 5$:

$f'(5) = (3 - 5)(25 - 5^2) - 5(25 - 5^2) - 2(5)^2(3 - 5)$ $f'(5) = (-2)(25 - 25) - 5(25 - 25) - 2(25)(-2) = 0 + 0 + 100 = 100$

Since $f'(5) > 0$, the equilibrium at $y = 5$ is unstable.

At $y = -5$:

$f'(-5) = (3 + 5)(25 - (-5)^2) - (-5)(25 - (-5)^2) - 2(-5)^2(3 + 5)$ $f'(-5) = (8)(25 - 25) + (-5)(25 - 25) - 2(25)(8) = 0 + 0 - 400 = -400$

Since $f'(-5) < 0$, the equilibrium at $y = -5$ is stable.

Conclusion:

• $y = 0$ is an unstable equilibrium.
• $y = 3$ is a stable equilibrium.
• $y = 5$ is an unstable equilibrium.
• $y = -5$ is a stable equilibrium.

Would you like to explore any further details on these classifications or their behavior?

Related questions:

1. How would you determine the stability of equilibria using a phase line?
2. Can you sketch the direction field of the differential equation to confirm the behavior of solutions?
3. How can we apply this knowledge to real-world systems (e.g., population models)?
4. What would happen if the equation had a higher-order term, such as $y^3$?
5. How would the classifications change if we included a damping term?

Tip:

When analyzing stability, remember that if the derivative $f'(y)$ at an equilibrium point is positive, the equilibrium is unstable (solutions move away), while if $f'(y)$ is negative, the equilibrium is stable (solutions move toward it).