Solve the initial value problem t2 y" + y = t < 2 y(0) = 0, y'(0) = 0 4t – 4 t> 2 |
Solve the initial value problem t2 y" + y = t < 2 y(0) = 0, y'(0) = 0 4t – 4 t> 2 |
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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Using Laplace Transforms
![**Title: Solving Initial Value Problems in Differential Equations**
**Introduction**
In this section, we'll explore how to solve an initial value problem (IVP) for a second-order linear differential equation with piecewise-defined forcing functions. Understanding these problems can help you apply mathematical concepts to real-world scenarios.
**Problem Statement**
Given the differential equation:
\[ y'' + y =
\begin{cases}
t^2 & \text{for } t < 2, \\
4t - 4 & \text{for } t \geq 2
\end{cases}
\]
with initial conditions:
\[ y(0) = 0, \quad y'(0) = 0 \]
**Objective**
Our goal is to find the function \( y(t) \) that satisfies both the differential equation and the initial conditions.
**Explanation**
The problem involves a piecewise function that changes its form based on the value of \( t \). For \( t < 2 \), the forcing function is \( t^2 \), while for \( t \geq 2 \), it changes to \( 4t - 4 \).
**Steps to Solve**
1. **Divide the Problem**: Solve the differential equation separately for each piece of the piecewise function, ensuring continuity at \( t = 2 \).
2. **Apply Initial Conditions**: Use the initial conditions \( y(0) = 0 \) and \( y'(0) = 0 \) to determine constants of integration for the solution on \( t < 2 \).
3. **Ensure Continuity**: At \( t = 2 \), ensure that both \( y(t) \) and \( y'(t) \) are continuous to find the constants for the \( t \geq 2 \) solution.
**Conclusion**
By solving each piece of the differential equation separately and ensuring continuity, you can find the function \( y(t) \) that satisfies the given initial value problem. Such methods are crucial for modeling systems that change behavior at specific points in time.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9cba6163-1bf7-4503-a8f6-4b3c743b9de1%2Fe4cf3172-1e68-48c0-ba67-a205f251fe84%2F89qlbw8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Solving Initial Value Problems in Differential Equations**
**Introduction**
In this section, we'll explore how to solve an initial value problem (IVP) for a second-order linear differential equation with piecewise-defined forcing functions. Understanding these problems can help you apply mathematical concepts to real-world scenarios.
**Problem Statement**
Given the differential equation:
\[ y'' + y =
\begin{cases}
t^2 & \text{for } t < 2, \\
4t - 4 & \text{for } t \geq 2
\end{cases}
\]
with initial conditions:
\[ y(0) = 0, \quad y'(0) = 0 \]
**Objective**
Our goal is to find the function \( y(t) \) that satisfies both the differential equation and the initial conditions.
**Explanation**
The problem involves a piecewise function that changes its form based on the value of \( t \). For \( t < 2 \), the forcing function is \( t^2 \), while for \( t \geq 2 \), it changes to \( 4t - 4 \).
**Steps to Solve**
1. **Divide the Problem**: Solve the differential equation separately for each piece of the piecewise function, ensuring continuity at \( t = 2 \).
2. **Apply Initial Conditions**: Use the initial conditions \( y(0) = 0 \) and \( y'(0) = 0 \) to determine constants of integration for the solution on \( t < 2 \).
3. **Ensure Continuity**: At \( t = 2 \), ensure that both \( y(t) \) and \( y'(t) \) are continuous to find the constants for the \( t \geq 2 \) solution.
**Conclusion**
By solving each piece of the differential equation separately and ensuring continuity, you can find the function \( y(t) \) that satisfies the given initial value problem. Such methods are crucial for modeling systems that change behavior at specific points in time.
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