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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Question
![### Differential Equations Problem
**Problem Statement:**
Find a differential equation whose general solution is \( y(t) = c_1 e^{2t} + c_2 e^{-t} \).
**Explanation:**
The problem asks you to determine a differential equation for which the given function \( y(t) = c_1 e^{2t} + c_2 e^{-t} \) is a general solution. This type of problem is common in the study of differential equations, where you work backward from a known solution to find the corresponding differential equation.
**Approach:**
1. **Identify the form of the given solution:**
The given function \( y(t) = c_1 e^{2t} + c_2 e^{-t} \) suggests that the solution is a linear combination of exponential functions.
2. **Determine the characteristic equation:**
For a homogeneous linear differential equation with constant coefficients, the general solution can often be expressed as a combination of exponential functions. Assume a solution of the form \( y(t) = e^{rt} \). Substitute this into a general second-order linear differential equation \( ay'' + by' + cy = 0 \).
The characteristic equation associated with the given solution can be determined by finding the roots which in this case are \( r = 2 \) and \( r = -1 \). Therefore, the characteristic equation is:
\[
(r - 2)(r + 1) = 0 \Rightarrow r^2 - r - 2 = 0
\]
3. **Form the differential equation:**
Given the characteristic equation \( r^2 - r - 2 = 0 \), the corresponding differential equation is:
\[
y'' - y' - 2y = 0
\]
Therefore, the differential equation whose general solution is \( y(t) = c_1 e^{2t} + c_2 e^{-t} \) is:
\[
y'' - y' - 2y = 0
\]
This process demonstrates how to derive the differential equation starting from a given general solution.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc47a605b-1167-4b41-bc40-883b330bf596%2Fa8dc5c05-4c6e-422a-8b9a-cfede7ea8a0b%2Fi5ljck_processed.png&w=3840&q=75)
Transcribed Image Text:### Differential Equations Problem
**Problem Statement:**
Find a differential equation whose general solution is \( y(t) = c_1 e^{2t} + c_2 e^{-t} \).
**Explanation:**
The problem asks you to determine a differential equation for which the given function \( y(t) = c_1 e^{2t} + c_2 e^{-t} \) is a general solution. This type of problem is common in the study of differential equations, where you work backward from a known solution to find the corresponding differential equation.
**Approach:**
1. **Identify the form of the given solution:**
The given function \( y(t) = c_1 e^{2t} + c_2 e^{-t} \) suggests that the solution is a linear combination of exponential functions.
2. **Determine the characteristic equation:**
For a homogeneous linear differential equation with constant coefficients, the general solution can often be expressed as a combination of exponential functions. Assume a solution of the form \( y(t) = e^{rt} \). Substitute this into a general second-order linear differential equation \( ay'' + by' + cy = 0 \).
The characteristic equation associated with the given solution can be determined by finding the roots which in this case are \( r = 2 \) and \( r = -1 \). Therefore, the characteristic equation is:
\[
(r - 2)(r + 1) = 0 \Rightarrow r^2 - r - 2 = 0
\]
3. **Form the differential equation:**
Given the characteristic equation \( r^2 - r - 2 = 0 \), the corresponding differential equation is:
\[
y'' - y' - 2y = 0
\]
Therefore, the differential equation whose general solution is \( y(t) = c_1 e^{2t} + c_2 e^{-t} \) is:
\[
y'' - y' - 2y = 0
\]
This process demonstrates how to derive the differential equation starting from a given general solution.
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