Chapter 4: Problem 4
Use the Laplace transform to solve the given differential equation subject to the indicated initial conditions. $$ y^{\prime \prime}+16 y=\delta(t-2 \pi), \quad y(0)=0, y^{\prime}(0)=0 $$
Short Answer
Expert verified
The solution is \( y(t) = u(t - 2\pi)\sin(4(t - 2\pi)) \).
Step by step solution
01
Apply Laplace Transform to the Equation
Begin by taking the Laplace transform of both sides of the differential equation. The equation is \( y'' + 16y = \delta(t - 2\pi) \). The Laplace transform of the left side is \( \mathcal{L}\{y''\} + 16\mathcal{L}\{y\} \) and the right side is \( \mathcal{L}\{\delta(t - 2\pi)\} \). Use the initial conditions \( y(0) = 0 \) and \( y'(0) = 0 \) to simplify \( \mathcal{L}\{y''\} \).
02
Simplify Using Initial Conditions
The Laplace transform of the second derivative with initial conditions is \( s^2Y(s) - sy(0) - y'(0) \). With \( y(0) = 0 \) and \( y'(0) = 0 \), this becomes \( s^2 Y(s) \). Thus, the transform of the equation becomes \( s^2 Y(s) + 16 Y(s) = e^{-2\pi s} \).
03
Solve for \( Y(s) \)
Factor out \( Y(s) \) from the left-hand side: \( (s^2 + 16) Y(s) = e^{-2\pi s} \). Solve for \( Y(s) \) by dividing both sides by \( (s^2 + 16) \): \( Y(s) = \frac{e^{-2\pi s}}{s^2 + 16} \).
04
Apply the Inverse Laplace Transform
To find \( y(t) \), apply the inverse Laplace transform to \( Y(s) = \frac{e^{-2\pi s}}{s^2 + 16} \). Use the shifting property of the Laplace transform, \( \mathcal{L}^{-1}\left\{ \frac{e^{-as}}{s^2 + \omega^2} \right\} = u(t-a)\sin(\omega(t-a)) \), yielding \( y(t) = u(t - 2\pi)\sin(4(t - 2\pi)) \).
05
Write the Final Solution
The solution to the differential equation is \( y(t) = u(t - 2\pi)\sin(4(t - 2\pi)) \), where \( u(t - 2\pi) \) is the Heaviside step function, ensuring \( y(t) = 0 \) for \( t < 2\pi \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differential Equations
Differential equations are mathematical equations that relate a function with its derivatives. They are used in various fields such as physics, engineering, and biology to model complex systems. In the context of this exercise, we have a second-order linear differential equation, which means it involves a second derivative and is linear in terms of the unknown function and its derivatives.
This specific equation is given by \[ y'' + 16y = \delta(t - 2\pi) \].
The equation includes a delta function, which often represents a sudden force acting at a specific point in time. Solving these equations often requires determining a function, \( y(t) \), that satisfies the equation for all values of \( t \) and adheres to any given initial conditions.
This specific equation is given by \[ y'' + 16y = \delta(t - 2\pi) \].
The equation includes a delta function, which often represents a sudden force acting at a specific point in time. Solving these equations often requires determining a function, \( y(t) \), that satisfies the equation for all values of \( t \) and adheres to any given initial conditions.
Initial Conditions
Initial conditions are crucial in solving differential equations because they allow us to find a unique solution. When we apply the Laplace transform to a differential equation, these conditions help us solve for the constants that arise.
In this problem, the initial conditions are \( y(0) = 0 \) and \( y'(0) = 0 \). These conditions specify the state of the system at time \( t = 0 \).
In this problem, the initial conditions are \( y(0) = 0 \) and \( y'(0) = 0 \). These conditions specify the state of the system at time \( t = 0 \).
- \( y(0) = 0 \) indicates the initial position is zero.
- \( y'(0) = 0 \) indicates the initial velocity is zero.
Heaviside Step Function
The Heaviside step function, denoted as \( u(t - a) \), is a simple mathematical function used to switch a signal on at a particular time. In our solution, it appears as part of the inverse Laplace transform. The function jumps from 0 to 1 at \( t = a \), effectively "turning on" the solution from this point forward.
In our problem, it is used to control the function \( y(t) = u(t - 2\pi)\sin(4(t - 2\pi)) \). This representation ensures that \( y(t) \) remains at zero for times before \( 2\pi \), which aligns with the initial impact occurring at \( t = 2\pi \) due to the delta function \( \delta(t - 2\pi) \).
The Heaviside function is integral in solutions involving sudden changes, acting as a switch to incorporate discontinuities within the solution seamlessly.
In our problem, it is used to control the function \( y(t) = u(t - 2\pi)\sin(4(t - 2\pi)) \). This representation ensures that \( y(t) \) remains at zero for times before \( 2\pi \), which aligns with the initial impact occurring at \( t = 2\pi \) due to the delta function \( \delta(t - 2\pi) \).
The Heaviside function is integral in solutions involving sudden changes, acting as a switch to incorporate discontinuities within the solution seamlessly.