Laplace transform with step function

  • Thread starter CINA
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  • #1
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Homework Statement



[tex]y'+2y=f(t), y(0)=0, where f(t)=\left\{t, 0\leq t<0 \;and \; 0, t \geq 0[/tex] (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)?
 

Answers and Replies

  • #2
hunt_mat
Homework Helper
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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?
 
  • #3
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Opps, it's:
[tex]t, \; 0 \leq t < 1[/tex]
[tex]0, \; t \geq 1[/tex]
 
  • #4
hunt_mat
Homework Helper
1,745
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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:
[tex]
\int_{0}^{1}te^{-st}dt
[/tex]
 

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