3. (Short Answer) Consider the following differential equation: $$ \frac{d y}{d t}=y\left(4-y^{2}\right) $$ Which of the following are the equilibrium solutions? (write down each correct answer) 4. In question 3 , for which values of $$ y $$ will a solution $$ y(t) $$ be decreasing? (A) $$ y

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3. (Short Answer) Consider the following differential equation: $$ \frac{d y}{d t}=y\left(4-y^{2}\right) $$ Which of the following are the equilibrium solutions? (write down each correct answer) 4. In question 3 , for which values of $$ y $$ will a solution $$ y(t) $$ be decreasing? (A) $$ y<-2,04 $$ (d) $$ -22 $$ B. $$ y=-5:-5\left(4-(-5)^{2}\right)=-5(-21)=1-\rightarrow $$ $$ y=1=t+ $$

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### **Question 3: Equilibrium Solutions** To find the equilibrium solutions of the differential equation: $$ \frac{dy}{dt} = y(4 - y^2) $$ **Equilibrium solutions** occur where $$\frac{dy}{dt} = 0$$. Setting the right-hand side to zero: $$ y(4 - y^2) = 0 $$ Solving for $$y$$: 1. $$y = 0$$ 2. $$4 - y^2 = 0 \Rightarrow y^2 = 4 \Rightarrow y = 2$$ or $$y = -2$$ **Therefore, the equilibrium solutions are:** $$ y = -2, \quad y = 0, \quad y = 2 $$ --- ### **Question 4: Values of $$ y $$ for Which $$ y(t) $$ is Decreasing** We need to determine for which values of $$ y $$ the solution $$ y(t) $$ is decreasing. This happens when $$\frac{dy}{dt} < 0$$: $$ y(4 - y^2) < 0 $$ Let's analyze the inequality: 1. **Critical Points:** $$ y = -2, 0, 2 $$ 2. **Intervals to Test:** - $$ y < -2 $$ - $$ -2 < y < 0 $$ - $$ 0 < y < 2 $$ - $$ y > 2 $$ 3. **Sign Analysis:** - **For $$ y < -2 $$:** - Example: $$ y = -3 $$ - $$\frac{dy}{dt} = (-3)(4 - 9) = (-3)(-5) = 15 > 0$$ (Increasing) - **For $$ -2 < y < 0 $$:** - Example: $$ y = -1 $$ - $$\frac{dy}{dt} = (-1)(4 - 1) = (-1)(3) = -3 < 0$$ (Decreasing) - **For $$ 0 < y < 2 $$:** - Example: $$ y = 1 $$ - $$\frac{dy}{dt} = (1)(4 - 1) = 3 > 0$$ (Increasing) - **For $$ y > 2 $$:** - Example: $$ y = 3 $$ - $$\frac{dy}{dt} = (3)(4 - 9) = (3)(-5) = -15 < 0$$ (Decreasing) **Conclusion:** The solution $$ y(t) $$ is **decreasing** in the intervals: $$ -2 < y < 0 \quad \text{and} \quad y > 2 $$ **Corresponding to Option (D):** $$ -2 < y < 0, \quad y > 2 $$ The equilibrium solutions are $$ y = -2 $$, $$ y = 0 $$, and $$ y = 2 $$. For the solution $$ y(t) $$ to be decreasing, $$ y $$ must be in the intervals $$ -2 < y < 0 $$ or $$ y > 2 $$.

(Short Answer) Consider the following differential equation:

Which of the following are the equilibrium solutions? (write down each correct answer)
4. In question 3 , for which values of will a solution be decreasing?
(A)
A.

(b)
(e)
(d)
B.

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(Short Answer) Consider the following differential equation: Which of the following are the equilibrium solutions? (write down each correct answer) 4. In question 3 , for which values of will a solution be decreasing? (A) A. (b) (e) (d) B.