Laplace transform with step function

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Homework Help Overview

The discussion revolves around solving a first-order differential equation using Laplace transforms, specifically focusing on the behavior of a piecewise function defined as f(t). The original poster expresses uncertainty about representing f(t) as a step function.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand how to express the piecewise function f(t) in terms of a step function and questions whether it should be represented as t*u(t-1). Other participants question the accuracy of the initial definition of f(t) and clarify its correct form.

Discussion Status

The discussion is active, with participants exploring the correct representation of the function f(t) and its implications for the Laplace transform. Guidance has been offered regarding the integration limits for the Laplace transform based on the piecewise nature of f(t).

Contextual Notes

There is a focus on the piecewise definition of f(t), which is critical for applying the Laplace transform correctly. The original poster's confusion about the step function representation indicates a need for clarification on piecewise functions in this context.

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Homework Statement



y'+2y=f(t), y(0)=0, where f(t)=\left\{t, 0\leq t<0 \;and \; 0, t \geq 0 (I don't know how to do piecewise)

Homework Equations



Equation for initial conditions; step function: u(t-a)

The Attempt at a Solution



I know how to solve a differential equation with initial conditions using Laplace transforms, I just don't know what f(t) looks like as a step function, or how it would behave. would it be t*u(t-1)?
 
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f(t) looks like 0 for all values of t, are you sure your definition of f(t) is correctly copied down?
 
Opps, it's:
t, \; 0 \leq t < 1
0, \; t \geq 1
 
OKay, then when taking the Laplace transform of the RHS, split the domain of integration to (0,1) and [1, infintity], the part where you integrate over 0 to infinity will vanish and you will be left with:
<br /> \int_{0}^{1}te^{-st}dt<br />
 

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