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Laplace Transforms to solve First Order ODE's.

  1. Apr 24, 2010 #1
    1. The problem statement, all variables and given/known data

    Solve the system of first order differential equations using Laplace Transforms:

    dx/dt = x - 4y

    dy/dt = x + y,
    subject to the initial conditions x(0)=3 and y(0)=-4.

    2. Relevant equations

    So far I've used the limited knowledge of Laplace Transforms for first order ODE's to get this far:

    L[x`] = L[x] - L[4y]
    s*L[x] - x(0) = L[x] - L[4y]
    s*L[x] - L[x] - 3 = -L[4y]
    (s-1)*L[x] = 3 + L[4y] <--------- Equation 1

    L[y'] = L[x] + L[y]
    s*L[y] - y(0) = L[x] + L[y]
    s*L[y] + 4 = L[x] + L[y]
    (s-1)*L[y] = L[x] - 4 <----------Equation 2
    or (s-1)*L[y] + 4 = L[x]


    Up to this stage I am kind of confident I have been using Laplace Transforms right (from the couple of examples I have in a text book I got from the library).

    3. The attempt at a solution

    The step where I become very confused is substituting equations 1 and 2 into one another to evaluate y(t) and x(t).

    When I substitute (2) into (1) i get the following:
    (s-1)[(s-1)*L[y] + 4] = 3 + L[4y]

    From here I have probably tried 20 different ways of getting a solution for y(s) but every single one is very complicated and leads to a dead end for me (they are way too long to type). Because of this I suspect I am doing something wrong here [potentially I am even applying Laplace Transforms completely wrong!].

    I am hoping someone knows where I am going wrong or what I'm doing wrong. Any help and advice would be greatly appreciated, thanks!
     
  2. jcsd
  3. Apr 24, 2010 #2

    vela

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    Looks good so far. Remember that the Laplace transform is linear, so you can say L[4y]=4 L[y]. Then it's just basic algebra to isolate L[y]. You should get

    [tex]L[y] = \frac{3-4(s-1)}{(s-1)^2-4}[/tex]
     
  4. Apr 24, 2010 #3

    gabbagabbahey

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    Keep in mind that L[4y]=4L[y]:wink:

    EDIT: Vela beat me to it.:smile:
     
  5. Apr 24, 2010 #4
    Thanks a lot for the replies guys, really appreciate it. Hmm I just made a small mistake in isolating L[y], I got that bit now thanks to your help!

    I have worked at it and did get:
    [3-4(s-1)] / [(s-1)^2 - 4]

    It's what I do from here to get y(t) thats confusing me now. I can't seem to find a way to isolate y and get a general solution for it. Sorry I'm not great with maths yet, just starting and keen.
     
    Last edited: Apr 24, 2010
  6. Apr 24, 2010 #5

    gabbagabbahey

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    Just take the inverse Laplace transform of both sides of your equation now.

    [tex]L^{-1}[L[y]]=y(t)=L^{-1}\left[\frac{3-4(s-1)}{(s-1)^2-4}\right][/tex]
     
  7. Apr 24, 2010 #6
    I've simplified the numerator to get:

    (7-4s) / [(s-1)^2 - 4]

    From here I suspect I should break the expression up into 2 parts and compare them to the table of Laplace transformations (from other questions I've done). But I just cant find the right parts and Laplace transformations.
     
  8. Apr 24, 2010 #7

    gabbagabbahey

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    Rather than simplifying the numerator, you might start by applying the frequency shift rule:

    [tex]L^{-1}\left[\frac{3-4(s-1)}{(s-1)^2-4}\right]=e^{-t}L^{-1}\left[\frac{3-4s}{s^2-4}\right][/tex]
     
  9. Apr 24, 2010 #8
    Thanks, I didn't see that at all!

    From there I have tried to now split the expression into 2 parts to use the Laplace Transformation table, by doing:

    (e^-t)*(L^-1)[3/(s^2 - 2^2)*(L^-1)[-4s/(s^2 - 2^2)]
    (e^-t)*[3/2](L^-1)[2/(s^2 - 2^2)*-4(L^-1)[s/(s^2 - 2^2)]

    From here I used the two following Laplace transformations:
    a/(s^2 + a^2) = sin(at) and s/(s^2 + a^2) = cos(at) to get:

    (e^-t)*[3/2]sin(-2t)*-4cos(-2t)
    -6(e^-t)*sin(-2t)*cos(-2t)

    Not sure if this is right, but it's the best application for the Laplace transformation table for this question I could see.
     
  10. Apr 25, 2010 #9

    gabbagabbahey

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    Why are you multiplying your fractions together like this?

    [tex]L^{-1}\left[f(s)+g(s)\right]=L^{-1}\left[f(s)\right]+L^{-1}\left[g(s)\right]\neq L^{-1}\left[f(s)\right]*L^{-1}\left[g(s)\right][/tex]
     
  11. Apr 25, 2010 #10
    Sorry didn't even pick up on that, just a careless error! So it should be:

    (e^-t)*[(L^-1)[3/(s^2 - 2^2)-(L^-1)[4s/(s^2 - 2^2)]]
    (e^-t)*([3/2](L^-1)[2/(s^2 - 2^2)*-4(L^-1)[s/(s^2 - 2^2)])

    From here I used the two following Laplace transformations:
    a/(s^2 + a^2) = sin(at) and s/(s^2 + a^2) = cos(at) to get:

    (e^-t)*([3/2]sin(-2t)-4cos(-2t))
     
  12. Apr 25, 2010 #11

    gabbagabbahey

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    There are a couple of errors that I missed...

    (1) You should have [tex]L[y] = \frac{3-4(s-1)}{(s-1)^2+4}[/tex] not [tex]L[y] = \frac{3-4(s-1)}{(s-1)^2-4}[/tex]

    (2) The frequency shift rule does not give exactly what I had posted earlier...can you figure out what it should be?:wink:
     
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