Solve the equation y"+y=f(t), y(0)=0, y'(0)=1 1 0
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Title: Solving Second-Order Linear Differential Equations with Piecewise Functions**
**Overview:**
In this exercise, we are tasked with solving a second-order linear differential equation with a piecewise function. The equation and initial conditions are provided along with multiple possible solutions.
**Problem Statement:**
Solve the equation \( y'' + y = f(t) \), subject to the initial conditions:
\[ y(0) = 0 \]
\[ y'(0) = 1 \]
**Piecewise Function:**
The function \( f(t) \) is defined as:
\[
f(t) = \begin{cases}
1 & \text{for } 0 < t \leq \pi/2 \\
0 & \text{for } \pi/2 \leq t < \infty
\end{cases}
\]
The Laplace transform of the piecewise function \( f(t) \), denoted \( F(s) \), is:
\[ F(s) = \frac{1 - e^{-(\pi/2)s}}{s} \]
**Possible Solutions:**
Evaluate which of the following functions \( y(t) \) satisfies the differential equation and initial conditions:
1. \( y(t) = 1 - \cos t + \sin t - \mu(t - \pi/2)(1 - \sin t) \)
2. \( y(t) = \cos t + \mu(t - \pi/2)(1 - \sin(t - \pi/2)) \)
3. \( y(t) = 1 - \sin t + \mu(t - \pi/2)(1 - \cos(t - \pi/2)) \)
4. \( y(t) = \sin t - \cos t - \mu(t - \pi/2)(1 - \sin(t - \pi/2) + \cos(t - \pi/2)) \)
**Steps to Solve:**
1. **Identify the Laplace Transform**:
The Laplace transform of \( f(t) \) has been provided, which is crucial for transforming the differential equation to an algebraic equation in terms of \( s \).
2. **Apply Initial Conditions**:
By applying the initial conditions \( y(0) = 0 \) and \( y'(0) = 1 \), we can determine the specific constants in the transformed equation](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F92c05ba6-e775-4ea8-9e55-a6761b98c61f%2Fb72bbecf-13e6-405c-8b15-70ec22cd1522%2Ffcix0ic_processed.png&w=3840&q=75)
Transcribed Image Text:**Title: Solving Second-Order Linear Differential Equations with Piecewise Functions**
**Overview:**
In this exercise, we are tasked with solving a second-order linear differential equation with a piecewise function. The equation and initial conditions are provided along with multiple possible solutions.
**Problem Statement:**
Solve the equation \( y'' + y = f(t) \), subject to the initial conditions:
\[ y(0) = 0 \]
\[ y'(0) = 1 \]
**Piecewise Function:**
The function \( f(t) \) is defined as:
\[
f(t) = \begin{cases}
1 & \text{for } 0 < t \leq \pi/2 \\
0 & \text{for } \pi/2 \leq t < \infty
\end{cases}
\]
The Laplace transform of the piecewise function \( f(t) \), denoted \( F(s) \), is:
\[ F(s) = \frac{1 - e^{-(\pi/2)s}}{s} \]
**Possible Solutions:**
Evaluate which of the following functions \( y(t) \) satisfies the differential equation and initial conditions:
1. \( y(t) = 1 - \cos t + \sin t - \mu(t - \pi/2)(1 - \sin t) \)
2. \( y(t) = \cos t + \mu(t - \pi/2)(1 - \sin(t - \pi/2)) \)
3. \( y(t) = 1 - \sin t + \mu(t - \pi/2)(1 - \cos(t - \pi/2)) \)
4. \( y(t) = \sin t - \cos t - \mu(t - \pi/2)(1 - \sin(t - \pi/2) + \cos(t - \pi/2)) \)
**Steps to Solve:**
1. **Identify the Laplace Transform**:
The Laplace transform of \( f(t) \) has been provided, which is crucial for transforming the differential equation to an algebraic equation in terms of \( s \).
2. **Apply Initial Conditions**:
By applying the initial conditions \( y(0) = 0 \) and \( y'(0) = 1 \), we can determine the specific constants in the transformed equation
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