25 4 45 -es + 8 Yo 25 45

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
Question
### Initial Value Problem Exploration

Consider the initial value problem:

\[ y' + \frac{2}{5}y = 1 - \frac{1}{2}t, \quad y(0) = y_0. \]

---

#### Question 1: Touching the t-axis
**What equation expresses the requirement that the solution touches the t-axis?**

- [ ] \( y(t) = 0 \)
- [ ] \( y'(t) = 0 \)
- [x] \( y''(t) = 0 \)

---

#### Question 2: Not Crossing the t-axis
**What additional equation expresses the requirement that the solution does not cross the t-axis?**

- [ ] \( y''(t) = 0 \)
- [ ] \( y'(0) = 0 \)
- [x] \( y'(t) = 0 \)

---

#### Combining Equations to Solve for r
**The equations found in (a) and (b) form a system of two equations for \( t \) and \( y_0 \). Show how the DE can be used to solve directly for \( t \).**

1. \( y' + \frac{2}{5}y = 0 \) but also \( 1 - \frac{1}{2}t = 0 \), so \( t = 5 \).
   - [x] True
2. \( y'' + \frac{2}{5}y = 0 \) but also \( 1 - \frac{1}{5}t = 0 \), so \( t = 5 \).
   - [ ] True
3. \( y' + \frac{2}{5}y = 0 \) but also \( 1 - \frac{1}{2}t = 0 \), so \( t = 2 \).
   - [ ] True
4. \( y' + \frac{5}{2}y = 0 \) but also \( 1 - \frac{1}{3}t = 0 \), so \( t = 2 \).
   - [ ] True

---

#### Final Solution
**Find the solution for the initial value problem with initial condition \( y(0) = y_0 \).**

\[ y = \frac{5}{4} + \left( y_0 -
Transcribed Image Text:### Initial Value Problem Exploration Consider the initial value problem: \[ y' + \frac{2}{5}y = 1 - \frac{1}{2}t, \quad y(0) = y_0. \] --- #### Question 1: Touching the t-axis **What equation expresses the requirement that the solution touches the t-axis?** - [ ] \( y(t) = 0 \) - [ ] \( y'(t) = 0 \) - [x] \( y''(t) = 0 \) --- #### Question 2: Not Crossing the t-axis **What additional equation expresses the requirement that the solution does not cross the t-axis?** - [ ] \( y''(t) = 0 \) - [ ] \( y'(0) = 0 \) - [x] \( y'(t) = 0 \) --- #### Combining Equations to Solve for r **The equations found in (a) and (b) form a system of two equations for \( t \) and \( y_0 \). Show how the DE can be used to solve directly for \( t \).** 1. \( y' + \frac{2}{5}y = 0 \) but also \( 1 - \frac{1}{2}t = 0 \), so \( t = 5 \). - [x] True 2. \( y'' + \frac{2}{5}y = 0 \) but also \( 1 - \frac{1}{5}t = 0 \), so \( t = 5 \). - [ ] True 3. \( y' + \frac{2}{5}y = 0 \) but also \( 1 - \frac{1}{2}t = 0 \), so \( t = 2 \). - [ ] True 4. \( y' + \frac{5}{2}y = 0 \) but also \( 1 - \frac{1}{3}t = 0 \), so \( t = 2 \). - [ ] True --- #### Final Solution **Find the solution for the initial value problem with initial condition \( y(0) = y_0 \).** \[ y = \frac{5}{4} + \left( y_0 -
Finish solving the equations found in (a) and (b).

Select the correct value for \( y_0 \) from the following options:

1. \[ y_0 = \frac{45}{8} e^{\frac{25}{2}} + \frac{25}{8} \]
2. \[ y_0 = \frac{25}{8} e^{\frac{4}{3} \cdot 5} + \frac{45}{8} \]
3. \[ y_0 = \frac{25}{8} e^{\frac{4}{5}} - \frac{45}{8} \]
4. \[ y_0 = \frac{4}{25} e^{\frac{2}{3} \cdot 5} + \frac{45}{8} \]
5. \[ y_0 = \frac{45}{8} e^{\frac{4}{5}} + \frac{25}{8} \]

Please choose the correct option by evaluating the given equations.
Transcribed Image Text:Finish solving the equations found in (a) and (b). Select the correct value for \( y_0 \) from the following options: 1. \[ y_0 = \frac{45}{8} e^{\frac{25}{2}} + \frac{25}{8} \] 2. \[ y_0 = \frac{25}{8} e^{\frac{4}{3} \cdot 5} + \frac{45}{8} \] 3. \[ y_0 = \frac{25}{8} e^{\frac{4}{5}} - \frac{45}{8} \] 4. \[ y_0 = \frac{4}{25} e^{\frac{2}{3} \cdot 5} + \frac{45}{8} \] 5. \[ y_0 = \frac{45}{8} e^{\frac{4}{5}} + \frac{25}{8} \] Please choose the correct option by evaluating the given equations.
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