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Laplace transforms of differential equations

  1. Mar 24, 2012 #1
    1. The problem statement, all variables and given/known data
    I've attached the problem.


    2. Relevant equations
    L(1) = 1 /s
    L(t^n) = n!/s^(n+1)


    3. The attempt at a solution
    because the question only asks for X(s) I only considered the x' + x + 4y=3 equation.
    I applied laplace tranforms and got:

    s X(s) - x(0) + 1/s^2 + 4/s^2 = 3/s. since x(0) = 0 it gets cancelled out so..
    s X(s) + 5/s^2 = 3/s
    s X(s) = 3/s - 5/s^2

    therfore
    X(s) = (3/s - 5/s^2) / s

    only problem is, i didn't use the other equation which makes me feel i've done it wrong.
     

    Attached Files:

  2. jcsd
  3. Mar 24, 2012 #2
    In

    [tex]sX(s) - x(0) + \frac{1}{s^2} + \frac{4}{s^2} = \frac{3}{s}[/tex]
    how are you getting
    [tex] \frac{1}{s^2} + \frac{4}{s^2} [/tex]
    ?
     
  4. Mar 24, 2012 #3

    LCKurtz

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    It's well you should feel you've done it wrong.:frown:

    You have two equations in two unknowns, x and y. When you transform, each equation will have an X(s) and Y(s). You left out the Y(s). You will get two equations in the unknowns X(s) and Y(s) which you can solve for X(s).
     
  5. Mar 24, 2012 #4
    Omg I feel so stupid right now, i cant believe i was trying to laplace transform a x and y variable >.<

    so it should be s X(s) - x(0) + X(s) + 4 Y(s) = 3/s and s Y(s) - y(0) + 9 X(s) + Y(s) = 0 simplify and it becomes

    s X(s) + X(s) + 4 Y(s) = 3/s and s Y(s)+ 9 X(s) + Y(s) = 0
     
  6. Mar 24, 2012 #5
    rearrange the second equation and it gets me Y(s) = -9 X(s) / (s+1)

    s X(s) + X(s) - 36 X(s) / s+1 = 3/s
    X(s) [s+1- 36/(s+1) ] = 3/s
    X(s) = 3(s+1) / x(s^2 + 2s - 35)

    not sure if calculations are correct.
     
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