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Another laplace transform IVP

  1. Nov 1, 2015 #1
    1. The problem statement, all variables and given/known data
    Use laplace transforms to find following initial value problem -- there is no credit for partial fractions. (i assume my teacher is against using it..)

    y'' + 2y' + 2y = 2 ; y(0)= y'(0) = 0

    2. Relevant equations

    table.JPG
    Lf'' = ((s^2)*F) - s*f(0) - f'(0)
    Lf' = sF - f(0)
    Lf = F(s)
    3. The attempt at a solution
    My first attempt is of course realizing that the above equation is expressed in terms of 't'. So I must take the laplace transform on both sides.


    L(y'') + 2 L(y') + 2L(y) = L(2)

    [s^2*Y(s) - sy(0) - y'(0) ] + 2 [ sY(s) - y(0) ] + 2*Y(s) = 2/s

    [s^2*Y(s) - 0 - 0 ] + 2 [ sY(s) - 0 ] + 2*Y(s) = 2/s

    s^2*Y(s) + 2s*Y(s) + 2*Y(s) = 2/s

    Y(s) [s^2 + 2s + 2] = 2/s

    Y(s) = 2/s * (1/ [s^2 + 2s + 2] )

    = 2/s * (1 / [(s+1)^2 + 1]) (using the formula (s+(a/2)^2) + b - (a^2)/4 )

    At this final step, I am lost on how to break the equation apart to solve for Y(t) using inverse laplace transform without using "Partial Fractions".
    Can somebody guide me to the right direction? Thank you.



    The solution according to my professor is:

    y = -e^-t * sin(t) - e^-t*cos(t) + 1
     
    Last edited: Nov 1, 2015
  2. jcsd
  3. Nov 1, 2015 #2

    vela

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    Don't you have a property which relates f(t) to F(s)/s?
     
  4. Nov 1, 2015 #3
    Not according to the table..
     
  5. Nov 1, 2015 #4

    SteamKing

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    You started with this equation:
    y'' - 2y' + 2y = 2

    Then you took the LT of this equation:
    L(y'') + 2 L(y') + 2L(y) = L(2)

    Seems like you should check the signs of each term more carefully before launching into the algebra of finding F(s).
     
  6. Nov 1, 2015 #5
    Sorry I meant to write:

    y'' + 2y' + 2y = 2 ; y(0)= y'(0) = 0


    Still does not answer my question.
     
  7. Nov 1, 2015 #6

    vela

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    In your other thread, you used a property (the delay property) that wasn't listed in the table you posted. You want to find one of those properties.
     
  8. Nov 1, 2015 #7
    I also tried using the shift theorem, but it still doesn't break up my equation into two or more parts.
     
  9. Nov 1, 2015 #8

    vela

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    Yeah, the problem is the factor of ##s## in the denominator. You need to find the property that allows you get rid of that. Look into what the transform of the integral of f(t) is.
     
  10. Nov 1, 2015 #9
    When I do take account of the shift theorem...I get Y(s) = [ 2/(s-1) ]* (1/ [(s)^2 + 1])
     
  11. Nov 1, 2015 #10
    So here's what I have so far---am i on the right track?
    Y(t) =
    2 * 1/s * 1/[(s+1)^2 + 1]

    L^(-1){1/[(s+1)^2 + 1]} = sint * e^(-t)

    Using the integral for transform... I get:

    y = 2 ∫ [ (from 0 to t) (e^(-t)) * sin(t) dt ]

    This would mean I would have to consider using integration by parts correct?
     
  12. Nov 1, 2015 #11

    Ray Vickson

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    You can figure out that for yourself.

    However, if I were doing the problem I would look instead at the IVP for ##z(t) = y(t) - 1##, because it has a homogeneous DE but with nonzero initial value. The effect of that would be to replace the ##1/s## in the transform by ##s##; that is, it would move the ##s## from the denominator into the numerator, and that would mean we could replace integration by differentiation when getting the final result for ##y(t) = z(t) + 1##.
     
    Last edited: Nov 1, 2015
  13. Nov 1, 2015 #12

    HallsofIvy

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    I am going to jump in here with both feet to again vent my dislike for the "Laplace transform" method. It is far easier to solve this problem by observing that the characteristic equation is [itex]r^2+ 2r+ 2= 0[/itex] so [itex]r^2+ 2r+ 1= (r+ 1)^2= -1[/itex] and the roots are 1+i and 1- I. The general solution to the related homogeneous equation is [itex]e^t(C_1 cos(t)+ C_2 sin(t)[/itex]. And y= 1 satisfies the entire equation so that the general solution to the entire equation is [itex]e^t(C_1 cos(t)+ C_2 sin(t)+ 1[/itex]. [itex]y(0)= C_1+ 1[/itex] so [itex]C_1= -1[/itex] and [tex]y'(0)= -1+ C_2= 0[/itex] so C_2= 1. There wasn't that much easier?

    It looks to me like the "Laplace transform" method gives Engineering students the illusion of a purely mechanical method where the can look the answer up in a table!

    Again, I will ask, can anyone give an example of a differential equation that can be done more simply by using the Laplace transform?
     
  14. Nov 1, 2015 #13

    Ray Vickson

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    Why does this matter? The fellow was given an assignment to solve some IVPs using the Laplace Transform method. He was required to do it; he was not given a choice.
     
    Last edited: Nov 1, 2015
  15. Nov 1, 2015 #14

    HallsofIvy

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    For this thread, yes. But that raises the question "Why is the Laplace Transform method taught at all?" My opinion is what I said before- it gives engineers a "mechanical" way of treating differential equations.
     
  16. Nov 1, 2015 #15

    Ray Vickson

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    And this is wrong, why? The engineer's main job is to do engineering, not math, and convenient tools that can help are not necessarily a bad thing. Besides, I suspect it is used primarily in the non-homogeneous case as an alternative to the method of undetermined coefficients or Green's function methods.

    Anyway, even if we did not use Laplace transforms for solving constant-coefficient linear DEs (the only kind for which the method works at all well!) they would still be invaluable in many other fields of application, such as stochastic process/probability theory, where they allow solution of problems that are very likely impossible to solve any other way. The Queueing Theory and Markov process literature have Laplace transforms falling off every page, just about.
     
    Last edited: Nov 1, 2015
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