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Homework Help: Solving a Differential Eq using Laplace and Unit-Step function

  1. Apr 6, 2008 #1
    !!!Solving a Differential Eq using Laplace and Unit-Step function

    1. The problem statement, all variables and given/known data
    I am having a terrible time just starting this. I need some explanation here. I have given up on the text I am using...

    I need to solve the following piecewise function using Laplace and Unit-Step function:

    y"+4y=f(t) y(0)=-1, y'(0)=0

    where f(t)= t, t<2
    5, t>2


    3. The attempt at a solution

    Now I condensed it to one line using the Unit Step function giving:

    [itex]y''+4y=t-t*U(t-2)+5*U(t-2)[/itex]

    Applying Laplace to the LHS is easy enough, but what is the Laplace of the unit step function? I am having a hard tome extrapolating it from the text.

    So far I have

    [itex]s^2Y(s)-sy(0)-y'(0)+4Y(s)=\frac{1}{s^2}-???[/itex]

    Could somebody help me out with the "U" terms here?

    Thank you
     
  2. jcsd
  3. Apr 6, 2008 #2
    SKIP AHEAD TO POST #10 That is where I need help
     
    Last edited: Apr 7, 2008
  4. Apr 6, 2008 #3
    Can somebody please help me with the Laplace Transform

    [tex]L(-t*U(t-2))[/tex]

    I cannot figure this out!!!!
     
  5. Apr 6, 2008 #4
    Is this even "Laplacable"? I guess I do not understand the rules
     
  6. Apr 7, 2008 #5
    nobody has any advice?
     
  7. Apr 7, 2008 #6

    HallsofIvy

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    Science Advisor

    Since the U(t-2) just changes the limits of integration to 2 to infinity, rather than 0 to infinity, you need to change the variable from x to u= x- 2 to get back the 0 limit.

    Doing that gives a general rule that the Laplace transform of f(x)U(x-a) is e-asF(s) where F(s) is the Laplace transform of f(x).
     
  8. Apr 7, 2008 #7
    So for [itex]-t*U(t-2)[/itex] I should get [itex]e^{-2s}*-\frac{1}{s^2}[/itex]?
     
  9. Apr 7, 2008 #8
    I don't think this is correct...
     
  10. Apr 7, 2008 #9
    Halls, I just found this in my text as a side note. It would appear that the Laplace of f(t)*U(t-a)=e^{-as}*L{g(t+a)} ........ not just +L{g(t)}
     
  11. Apr 7, 2008 #10
    Okay so I have got it down to this point:

    [tex]Y(s)[s^2+4]=\frac{1}{s^2}-e^{-2s}*\frac{1}{(s+2)^2}+5e^{-2s}*\frac{1}{s+2}-s[/tex]

    I was going to write the RHS over the LCD and then divide out by [s^2+4] .....sound good? .... there must be a neater way to so this... isn't there? This looks like it will get very sloppy very quick!
     
    Last edited: Apr 7, 2008
  12. Apr 7, 2008 #11
    This equation is B.S. F#$k this.
     
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