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Help with Laplace Transformations and 2nd order ODEs

  1. Nov 23, 2008 #1

    TFM

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    1. The problem statement, all variables and given/known data

    Solve the following problems using Laplace Transforms:


    [tex] y' - y = 2e^t, y_0 = 3 [/tex]

    [tex] y'' + 4y' + 4y = e^{-2t}, y_0 = 0, y_0' = 4 [/tex]

    [tex] y'' + y = sin(t), y_0 = 1, y_0' = 0 [/tex]

    [tex] y'' + y = sin(t), y_0 = 1, y_0' = -\frac{1}{2} [/tex]

    2. Relevant equations

    N/A

    3. The attempt at a solution

    I don't seem to understand how to use Laplace Transformation to solve ODE's! If anyone can help, I will be most grateful.

    TFM
     
  2. jcsd
  3. Nov 23, 2008 #2
    lets look at the first one: let L(F(t)) denote the laplace of F(t)
    in general, you say that:
    L(y'(t))-L(y(t))=L(2e^t) call this (1)

    L(y(t))=y (in terms of s)
    L(y'(t))=sL(y(t) - y(0) (This is one of the formulas from the table of laplace transforms)
    =sy-3
    so, substituting these back into (1): sy-3-y=2/(s-1) (also from the tables)

    make y the subject, then you should get an expression for y in terms of s.
    Then to find y(t) (which is what you are looking for), you must find the inverse of y.

    I hope all is clear.
     
  4. Nov 23, 2008 #3

    TFM

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    Where did the 'in terms of S' come from?

    TFM
     
  5. Nov 23, 2008 #4
    It might be confusing to use the same label y for the function und its Laplace transform. Better come up with a new letter like Y:smile:

    You then have to solve (according to Sara)
    [tex]sY(s)-3-Y(s)=\frac{2}{s-1}[/tex]
    for Y(s) and find the inverse Laplace transform of this function.
     
  6. Nov 23, 2008 #5

    TFM

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    Okay lets see if this makes sense:

    you have the orginal equation, F(t)

    [tex] y' - y = 2e^t [/tex]

    then you make L(f(t)):

    [tex] L(y'(t)) - L(y(t)) = L(2e^t) [/tex] ---(1)

    I still can't see how you have gone from here to here:

    ???

    TFM
     
  7. Nov 23, 2008 #6
    that is not from one step to another.
    L(y'(t))-L(y(t))=L(2e^t) this is something
    and
    L(y(t))=Y is something else.
    In other words, LET L(y(t))=Y
    so therefore, L(y'(t))=sY-3
     
  8. Nov 23, 2008 #7

    TFM

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    Let's see then...:

    [tex] L(F(t): L(y'(t)) - L(y(t)) = L(2e^t) [/tex] ---(1)

    and also:

    [tex] L(y(t)) = Y [/tex]

    so inserting into (1)

    [tex] L(F(t): L(y'(t)) - Y = L(2e^t) [/tex]

    [tex] L(y'(t)) = Y + L(2e^t) [/tex]

    I see now where the 3 comes from, [tex] y_0 = 3 [/tex]

    [tex] L(y'(t)) = Y - y_0 [/tex]

    [tex] L(y'(t)) = Y - 3 [/tex]

    substitute back into (1):

    [tex] Y - 3 - L(y(t)) = L(2e^t) [/tex]

    does this look right so far?

    TFM
     
  9. Nov 23, 2008 #8
    you made a small mistake,
    L(y'(t))=sY-3
    because you must use the formula:
    L(F'(t))=sL(F(t))-F(0)
    Im not sure but I think you are expected to know this general formula.
    so this gives:
    sY-3-L(y(t))=L(2e^t)
     
  10. Nov 23, 2008 #9

    TFM

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    Yes, the formula:

    [tex] L(F(t)) = sL(F(t)) - F(0) [/tex]

    looks useful. I assume the [tex] sL(F(t)) [/tex] is inequivalent to: [tex] s*L(y(t)) = Y [/tex]

    ???

    so:

    [tex] L(y'(t)) = sY - 3 [/tex]

    putting back into (1)

    [tex] L(y'(t)) - L(y(t)) = L(2e^t) [/tex]

    [tex] sY - 3 - sY - 3 = L(2e^t) [/tex]

    does this look right?

    TFM
     
  11. Nov 23, 2008 #10
    The star means convolution (in Laplace language) so it is inequivalent
    You have:
    L(y'(t))-L(y(t))=L(2e^t) this is (1)
    we said let L(y(t))=Y
    and then we know that L(y'(t))=sY-3

    so putting into (1) gives:
    sY-3-Y=L(2e^t)

    Why are you putting sY-3 twice?
     
  12. Nov 23, 2008 #11

    TFM

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    I meant the star as in times, Sorry.

    I am not sure, how am I putting in sY-3 twice? sY-3-Y=L(2e^t) only has sY - 3 in once?

    TFM
     
  13. Nov 23, 2008 #12
    exactly so again you have to remember:
    L(y'(t))-L(y(t))=L(2e^t) this is (1)
    (A) we said let L(y(t))=Y
    (B) and then we know that L(y'(t))=sY-3
    so putting into (1) gives:
    (C) sY-3-Y=L(2e^t)
    These are the three main steps that should give you the equation of Y in terms of s.
    So, what would you do next?
     
  14. Nov 23, 2008 #13

    TFM

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    Well, judging from your first post, rearrange into Y?

    so this would be:

    [tex] sY-3-Y=L(2e^t) [/tex]

    put 3 to one side:

    [tex] sY-Y=L(2e^t) + 3 [/tex]

    factorise out Y:

    [tex] Y(s - 1) = L(2e^t) + 3 [/tex]

    divide by (s - 1):

    [tex] Y = \frac{L(2e^t) + 3}{s - 1} [/tex]

    Does this look right?

    ???

    TFM
     
  15. Nov 23, 2008 #14
    no, before you 'put to one side' on your second step, you have to find the laplace of
    2e^t
    sY-3-Y=L(2e^t)
    so,
    sY-3-Y=2/(s-1)
    (do you know that L(2e^t)=2/(s-1) ?)
    now, you make Y the subject.
     
  16. Nov 23, 2008 #15

    TFM

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    I have a table with Laplace conversions, and it gives:

    [tex] e^{-\alpha t} = \frac{1}{p + \alpha} [/tex]

    so

    [tex] 2e^{t} = \frac{2}{p + t} [/tex]

    I'm assuming p and s are equivilant

    so now:


    [tex] sY-3-Y = L(2e^t) [/tex]

    [tex] sY-3-Y = \frac{2}{s + t} [/tex]

    now take over the 3:

    [tex] sY-Y = \frac{2}{s + t} + 3 [/tex]

    factorise Y:

    [tex] Y(s - 1) = \frac{2}{s + t} + 3 [/tex]

    thus giving Y

    [tex] Y = \frac{\frac{2}{s + t} + 3}{s - 1} [/tex]

    Does this look better?

    TFM
     
  17. Nov 23, 2008 #16
    e^(-at)=1/(s-a)
    so: (the right hand side is the laplace so it has nothing to do with t)
    2e^(t)=1/(s+1) (since in our case, -a=1)

    Now, make Y the subject from:

    sY-3-Y=1/(s+1) (do you agree?)
     
  18. Nov 23, 2008 #17

    TFM

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    I feel slightly sill, I meant to do that, but didn't :redface:

    So:

    [tex] sY-3-Y = \frac{2}{s + 1} [/tex]

    now:

    [tex] sY-Y = \frac{2}{s + 1} + 3 [/tex]

    [tex] Y(s - 1) = \frac{2}{s + 1} + 3 [/tex]

    [tex] Y = \frac{\frac{2}{s + 1} + 3}{s - 1} [/tex]

    Is this right?

    TFM
     
  19. Nov 23, 2008 #18
    Im sorry, i misread; the L(2e^t) means that our a=1 so that means L(2e^t)=2/(s-1)
    not 2/(s+1)
    but your steps are correct and so your Y would be:
    sY-3-Y=2/(s-1)

    Y(s-1)=2/(s-1) +3
    so Y=2/(s-1)^2 + 3/(s-1) (by dividing both sides by s-1)
    is this clear?
    so, what would you do next?
     
  20. Nov 23, 2008 #19

    TFM

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    so:

    [tex] Y = \frac{\frac{2}{s - 1} + 3}{s - 1} [/tex]

    is the same as:

    [tex] Y = \frac{2}{(s - 1)^2}+\frac{3}{s - 1} [/tex]

    so now, from your first post again, I need to find the inverse:

    thus:

    [tex] \frac{1}{Y} = \frac{(s - 1)^2}{2}+\frac{s - 1}{3} [/tex]

    ???

    TFM
     
  21. Nov 23, 2008 #20
    no, carefull.... when i say inverse i dont mean 1/Y;
    Y=2/(s-1)^2 + 3/(s-1) call this (2) (This is you Laplace of the function you are looking for)

    So you want to find the function in terms of t that would give you (2) when you find its laplace. so you want to find a function of t.
    so, Y(t)= something that when you find its laplace you would get (2)

    so, first, what is the function in terms of t that when you find its laplace you would get 2/(s-1)^2 ? (you would need to use your tables)
     
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