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Laplace (IVP): Having Trouble Understanding Solution Manual's Answer

  1. Jul 26, 2012 #1
    1. The problem statement, all variables and given/known data
    Use Laplace Transform to Solve IVP
    y"-2y'+2y=e^(-t) y(0)=0 y'(0)=1

    2. Relevant equations
    [s^2*L{y}-sy(0)-y'(0)]-[2sL{y}-y(0)]+2L{y}=0 (for Homogeneous eq)


    3. The attempt at a solution
    Homogeneous Attempt:
    L{y}=1/(s^2-2s+2) Not sure if correct

    There soluton to this point is:
    Y(s)=1/((s-1)^2+1) + 1/((s+1)[(s-1)^2+1])

    a) Not sure where second term came from
    b) Not sure how to handle the NonHomogeneous part
     
  2. jcsd
  3. Jul 26, 2012 #2
    I'm not going to attempt to read something like this that is not in tex (right now). But, why are you trying to solve the homogeneous eq with laplace transforms? Just take the Laplace Transform of the rhs of the ode.

    edit:
    you didn't multiply all of L{y'} by 2.
     
  4. Jul 26, 2012 #3
    Thank you for the comment. But the problem is what it is.
    Besides, I'm beginning to like Laplace.
    Seems like there are all kinds of possibilities.

    Multiply RHS by 2? Not sure I need to?
    The homogeneous part seems to be correct, which is the first term.
    I think the second term has to do with e^(-t).
    If someone could verify/explain I'd appreciate it. Thank you.
     
    Last edited: Jul 26, 2012
  5. Jul 26, 2012 #4
    Sorry - I was busy before. I didn't mean for my post to make me sound like suck an A-Hole, my apologies.

    To solve an ODE using laplace, you don't find the solution to the homogeneous and then a particular solution to the inhomogeneous part. Rather, you take the laplace transform of the whole ODE, solve for L{y} and then get the inverse laplace transform of that.
    But, your laplace transform of the homogeneous part is correct.

    I got a different (well, seemingly different) Laplace transform than your book did. (I think they probably just took the Laplace transform of the whole equation.) However, to get the inverse of my Laplace transform, I had to use "exponential shift" with a couple of hyperbolic functions. Perhaps that is a hint, I don't know.

    EDIT:
    I use [itex]\hat{y}[/itex] to be the Laplace transform of y.

    As for multiplying by 2, you have: [itex](s^2\hat{y}-sy(0)-y'(0))-(2s\hat{y}-y(0))+2 \hat{y}=0[/itex] it should be: [itex](s^2 \hat{y}-sy(0)-y'(0))-(2s\hat{y}-2y(0))+2\hat{y}=0[/itex]. However, it doesn't matter since [itex]y(0)=0[/itex].
     
    Last edited: Jul 26, 2012
  6. Jul 26, 2012 #5
    WHOOPS! I made a mistake in my work somewhere. I have

    EDIT:
    I made ANOTHER mistake. OK, so, I can't seem to do simple algebra tonight. Here is what I have:

    [tex]\hat{y} = \frac{1}{5}\big( \frac{s+2}{(s-1)^2+1} - \frac{1}{s+1}\big)[/tex]

    This is close to what you have as the book's solution. Are you sure you copied correctly? Anyway, my transform is the transform of the whole ODE, which is what your book did as well. (That is, not just the homogeneous part.) Hyperbolic functions aren't part of the solution - my apologies.

    You are going to have to play with the numerators to get this to something you can use.

    EDIT:
    I used the above solution to get the inverse laplace transform, and it satisfied the ODE. Sorry if there was any confusion. I quickly did it the first time, checked my solution, and it worked - then I realised I had made an error. THen I made a type fixing my error. Oh well, sorry if there was confusion.
     
    Last edited: Jul 26, 2012
  7. Jul 26, 2012 #6

    HallsofIvy

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    This is incorrect but probably just a typo- the middle term should be 2[sL{y}- y(0)]
    However, why did you do the homogeneous equation and not include [itex]e^{-t}[/itex]?

    If you did your equation would be
    [itex]s^2L{y}- 1- 2[sL{y}]+ 2L{y}= \frac{1}{1+ s}[/itex]


    That is where the second term came from. Including the nonhomogeneous part, the equation for the Laplace transform becomes
    [itex]s^2L- 1- 2sL+ 2y= \frac{1}{1+ s}[/itex]
    [itex](s^2- 2s+ 2)L= ((s+1)^2+ 1)L= 1+ \frac{1}{1+ s}= \frac{s+ 2}{s+1}[/itex]
    so that
    [tex]L= \frac{s+2}{((s+1)^2+ 1)(s+ 1)}[/tex]
    Of course,
    [tex]\frac{1}{((s+1)^2+ 1}+ \frac{1}{(s+1)((s+1)^2+1)}= \frac{s+1}{((s+1)^2+1)(s+1)}+ \frac{1}{(s+1)((s+1)^2+1)}= \frac{s+ 2}{(s+1)((s+1)^2+ 1)}[/tex]
     
  8. Jul 26, 2012 #7
    Hey! You were able to do this with less than 29 edits! I should take notes.
     
  9. Jul 26, 2012 #8

    HallsofIvy

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    I would suspect sarcasm- from anyone else!:tongue:
     
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