The rate of change of y with respect to x is one-half times the value of y. Find an equation for y, given that y = -7 when x = 0. You get: A. B. C. D. 2 = 1 y = −7(1)ex y = e05x_7 =-7₂0.5x y E. =-7(¹)*
The rate of change of y with respect to x is one-half times the value of y. Find an equation for y, given that y = -7 when x = 0. You get: A. B. C. D. 2 = 1 y = −7(1)ex y = e05x_7 =-7₂0.5x y E. =-7(¹)*
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...
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![### Problem Statement
The rate of change of \( y \) with respect to \( x \) is one-half times the value of \( y \). Find an equation for \( y \), given that \( y = -7 \) when \( x = 0 \).
### Options
A. \(\frac{dy}{dx} = \frac{1}{2}y \)
B. \( y = -7\left(\frac{1}{2}\right)e^x \)
C. \( y = e^{0.5x} - 7 \)
D. \( y = -7e^{0.5x} \)
E. \( y = -7\left(\frac{1}{2}\right)^x \)
### Explanation of the Problem and Options
The problem asks to find the function \( y(x) \) given a differential equation and initial condition. The differential equation given is:
\[ \frac{dy}{dx} = \frac{1}{2}y \]
Initial Condition: \( y = -7 \) when \( x = 0 \).
Here are the steps to solve for \( y(x) \):
1. **Solve the Differential Equation**:
Since the differential equation is linear and separable, we separate variables and integrate:
\[
\frac{1}{y}dy = \frac{1}{2}dx
\]
Integrating both sides:
\[
\ln|y| = \frac{1}{2}x + C
\]
2. **Solve for \( y \)**:
Exponentiating both sides to solve for \( y \):
\[
y = e^{\frac{1}{2}x + C} = e^C \cdot e^{\frac{1}{2}x}
\]
Let \( e^C = k \), which is a constant:
\[
y = ke^{\frac{1}{2}x}
\]
3. **Use Initial Condition to find the constant \( k \)**:
Given \( y = -7 \) when \( x = 0 \):
\[
-7 = ke^0 \implies k = -7
\]
4. **Final Solution**:
Substituting \( k \) back into](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2bf6a656-c902-430b-8919-3920c6e4552b%2F45be933f-0248-4a09-b8f3-3721e69b0bb1%2Fm9nf9fj_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
The rate of change of \( y \) with respect to \( x \) is one-half times the value of \( y \). Find an equation for \( y \), given that \( y = -7 \) when \( x = 0 \).
### Options
A. \(\frac{dy}{dx} = \frac{1}{2}y \)
B. \( y = -7\left(\frac{1}{2}\right)e^x \)
C. \( y = e^{0.5x} - 7 \)
D. \( y = -7e^{0.5x} \)
E. \( y = -7\left(\frac{1}{2}\right)^x \)
### Explanation of the Problem and Options
The problem asks to find the function \( y(x) \) given a differential equation and initial condition. The differential equation given is:
\[ \frac{dy}{dx} = \frac{1}{2}y \]
Initial Condition: \( y = -7 \) when \( x = 0 \).
Here are the steps to solve for \( y(x) \):
1. **Solve the Differential Equation**:
Since the differential equation is linear and separable, we separate variables and integrate:
\[
\frac{1}{y}dy = \frac{1}{2}dx
\]
Integrating both sides:
\[
\ln|y| = \frac{1}{2}x + C
\]
2. **Solve for \( y \)**:
Exponentiating both sides to solve for \( y \):
\[
y = e^{\frac{1}{2}x + C} = e^C \cdot e^{\frac{1}{2}x}
\]
Let \( e^C = k \), which is a constant:
\[
y = ke^{\frac{1}{2}x}
\]
3. **Use Initial Condition to find the constant \( k \)**:
Given \( y = -7 \) when \( x = 0 \):
\[
-7 = ke^0 \implies k = -7
\]
4. **Final Solution**:
Substituting \( k \) back into
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