We are given the differential equation:

$y' = y^3 - 9y$

with the initial condition $y(0) = -1.5$. We are tasked with finding $\lim_{t \to -\infty} y(t)$.

Step 1: Analyzing the differential equation

The equation can be factored as:

$y' = y(y^2 - 9) = y(y - 3)(y + 3)$

The critical points (equilibrium solutions) are obtained by solving:

$y(y - 3)(y + 3) = 0$

Thus, the equilibrium points are $y = -3$, $y = 0$, and $y = 3$.

Step 2: Stability of equilibrium points

To determine the stability of these points, we can perform a qualitative analysis by examining the sign of $y'$ in different intervals:

• For $y < -3$: $y(y - 3)(y + 3) > 0$, so $y' > 0$.
• For $-3 < y < 0$: $y(y - 3)(y + 3) < 0$, so $y' < 0$.
• For $0 < y < 3$: $y(y - 3)(y + 3) < 0$, so $y' < 0$.
• For $y > 3$: $y(y - 3)(y + 3) > 0$, so $y' > 0$.

From this analysis, we observe:

• $y = -3$ is a stable equilibrium (attractor).
• $y = 0$ and $y = 3$ are unstable equilibrium points.

Step 3: Behavior as $t \to -\infty$

Given the initial condition $y(0) = -1.5$, which lies between $y = -3$ and $y = 0$, the solution will move toward the stable equilibrium point $y = -3$ as $t \to -\infty$. This is because the slope $y'$ is negative in the interval $-3 < y < 0$, so $y(t)$ will decrease toward $-3$.

Conclusion

Thus, we conclude that:

$\lim_{t \to -\infty} y(t) = -3$

Would you like more details on any specific part of this explanation?

Here are 5 related questions:

1. How do you classify equilibrium points as stable or unstable?
2. Can you solve this differential equation explicitly?
3. What is the role of phase line analysis in understanding differential equations?
4. How would the solution behave if the initial condition were $y(0) = 2$?
5. How does the behavior of nonlinear differential equations differ from linear ones?

Tip: Understanding the stability of equilibrium points helps in predicting long-term behavior of solutions without solving the equation explicitly.