Homework 1 (4 points possible) - due Thursday (1/16/2025) 1. (Short Answer) Consider the following differential equation $$ \frac{d y}{d t}=(2 y-1)(y-3) $$ Which of the following are the equilibrium solutions? (write down eadh correct answer) $$ \begin{array}{cccc} x & x & (y(3)-1)(3-3) \ y(t)-0, & y(t)-\frac{1}{2}+y(t)-2, y(t)-3 & 5 \cdot 0=0 \ (a(0)-1)(0-3) & (2(5)-1 y(1-3) & (2(2)-1)(2-3) & 0 \end{array} $$ 2. In question 1, for which values of $$ y $$ will a solution $$ y(t) $$ be increasing? A $$ y3 $$ only $$ (2(4)-1)(4-3) \rightarrow(7)(1) \rightarrow++\rightarrow V $$ (2x $$ \frac{1}{2}3 $$

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Homework 1 (4 points possible) - due Thursday (1/16/2025) 1. (Short Answer) Consider the following differential equation $$ \frac{d y}{d t}=(2 y-1)(y-3) $$ Which of the following are the equilibrium solutions? (write down eadh correct answer) $$ \begin{array}{cccc} x & x & (y(3)-1)(3-3) \ y(t)-0, & y(t)-\frac{1}{2}+y(t)-2, y(t)-3 & 5 \cdot 0=0 \ (a(0)-1)(0-3) & (2(5)-1 y(1-3) & (2(2)-1)(2-3) & 0 \end{array} $$ 2. In question 1, for which values of $$ y $$ will a solution $$ y(t) $$ be increasing? A $$ y<\frac{1}{2} $$ only $$ \left(2\left(\frac{1}{4}-1\right)\left(\left(\frac{1}{4}\right)-3\right) \rightarrow(-0.5)(-2.75) \rightarrow-\cdots\right. $$ (b) $$ y>3 $$ only $$ (2(4)-1)(4-3) \rightarrow(7)(1) \rightarrow++\rightarrow V $$ (2x $$ \frac{1}{2}3 $$

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Sure, let's tackle each part of the homework step by step. --- ### **Problem 1: Equilibrium Solutions** **Given Differential Equation:** $$ \frac{dy}{dt} = (2y - 1)(y - 3) $$ **Equilibrium Solutions:** Equilibrium solutions occur where $$\frac{dy}{dt} = 0$$. This happens when: $$ (2y - 1)(y - 3) = 0 $$ **Solving for $$ y $$:** 1. **First Factor:** $$ 2y - 1 = 0 \quad \Rightarrow \quad y = \frac{1}{2} $$ 2. **Second Factor:** $$ y - 3 = 0 \quad \Rightarrow \quad y = 3 $$ **Thus, the equilibrium solutions are:** $$ y(t) = \frac{1}{2} \quad \text{and} \quad y(t) = 3 $$ --- ### **Problem 2: Increasing Solutions** **Determine Where $$\frac{dy}{dt} > 0$$:** $$ \frac{dy}{dt} = (2y - 1)(y - 3) > 0 $$ **Critical Points:** $$ y = \frac{1}{2} \quad \text{and} \quad y = 3 $$ These critical points divide the real number line into three intervals: 1. **Interval 1:** $$ y < \frac{1}{2} $$ 2. **Interval 2:** $$ \frac{1}{2} < y < 3 $$ 3. **Interval 3:** $$ y > 3 $$ **Test Each Interval:** 1. **For $$ y < \frac{1}{2} $$:** - Let $$ y = 0 $$: $$ (2(0) - 1)(0 - 3) = (-1)(-3) = 3 > 0 $$ - **Result:** $$\frac{dy}{dt} > 0$$ 2. **For $$ \frac{1}{2} < y < 3 $$:** - Let $$ y = 2 $$: $$ (2(2) - 1)(2 - 3) = (4 - 1)(-1) = (3)(-1) = -3 < 0 - $$ - **Result:** $$\frac{dy}{dt} < 0$$ 3. **For $$ y > 3 $$:** - Let $$ y = 4 $$: $$ (2(4) - 1)(4 - 3) = (8 - 1)(1) = (7)(1) = 7 > 0 $$ - **Result:** $$\frac{dy}{dt} > 0$$ **Conclusion:** $$ \frac{dy}{dt} > 0 \quad \text{when} \quad y < \frac{1}{2} \quad \text{or} \quad y > 3 $$ **Thus, the solution $$ y(t) $$ is increasing for:** $$ y < \frac{1}{2} \quad \text{and} \quad y > 3 $$ --- ### **Final Answers:** 1. **Equilibrium Solutions:** $$ y(t) = \frac{1}{2} $$ and $$ y(t) = 3 $$ 2. **Increasing Solutions:** $$ y < \frac{1}{2} $$ and $$ y > 3 $$ **Equilibrium Solutions:** $$ y = \frac{1}{2} $$ and $$ y = 3 $$ **Increasing Solutions:** $$ y < \frac{1}{2} $$ and $$ y > 3 $$

Homework 1 (4 points possible) - due Thursday (1/16/2025)

(Short Answer) Consider the following differential equation

Which of the following are the equilibrium solutions? (write down eadh correct answer)

In question 1, for which values of will a solution be increasing?

A only

(b) only
(2x
(x)

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Homework 1 (4 points possible) - due Thursday (1/16/2025) (Short Answer) Consider the following differential equation Which of the following are the equilibrium solutions? (write down eadh correct answer) In question 1, for which values of will a solution be increasing? A only (b) only (2x (x)