Questions: (d) On a different graph make a rough sketch of the solution y(t) to the initial value problem y' = 3y(y+3)(y-2), y(0)=1 (e) Use the graph in part (d) to find the lim t -> ∞ y(t), where y(t) is the solution found in part (d).

(d) On a different graph make a rough sketch of the solution y(t) to the initial value problem

y' = 3y(y+3)(y-2), y(0)=1

(e) Use the graph in part (d) to find the lim t -> ∞ y(t), where y(t) is the solution found in part (d).

Transcript text: (d) On a different graph make a rough sketch of the solution $y(t)$ to the initial value problem \[ y^{\prime}=3 y(y+3)(y-2), \quad y(0)=1 \] (e) Use the graph in part (d) to find the $\lim _{t \rightarrow \infty} y(t)$, where $y(t)$ is the solution found in part (d).

Solution

Solution Steps

Step 1: Identify the differential equation and initial condition

The given differential equation is: \[ y' = 3y(y+3)(y-2) \] with the initial condition: \[ y(0) = 1 \]

Step 2: Analyze the differential equation

The differential equation is a first-order autonomous differential equation. The equilibrium points are found by setting the right-hand side to zero: \[ 3y(y+3)(y-2) = 0 \] Thus, the equilibrium points are: \[ y = 0, \quad y = -3, \quad y = 2 \]

Step 3: Determine the behavior around equilibrium points

For $ y > 2 $, $ y' > 0 $, so $ y $ increases.
For $ 0 < y < 2 $, $ y' < 0 $, so $ y $ decreases.
For $ y < 0 $, $ y' < 0 $, so $ y $ decreases.

Given the initial condition $ y(0) = 1 $, the solution will move towards $ y = 0 $ as $ t \rightarrow \infty $.

Final Answer

The solution $ y(t) $ to the initial value problem will approach $ y = 0 $ as $ t \rightarrow \infty $.

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