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Using laplace transforms to solve system of (two) equations

  1. Mar 18, 2012 #1
    1. The problem statement, all variables and given/known data
    x' = 3x - 4y
    y' = 2x + 3y

    x(0) = 1
    y(0) = 0


    2. Relevant equations
    y' = sY(s) - y(0)


    3. The attempt at a solution
    i am confused as to how to take laplace of either equation when i am left with a term i cant take the laplace of: specifically,
    x' = 3x - 4y
    sX(s) - 1 -3X(s) = -4Y(s)
    Y(s)[s-3] - 1 = -4Y(s)

    at this point i will usually solve for X(s) and then take the relevent inverse transforms. But what am i supposed to do with the Y(s)?? I know the general idea is to make this into an algebra prob of 2 equations and 2 unknowns but how do you take inv L.T. of an equation with X(s) and Y(s)?
    thanks for any help.

    (its same problem as here https://www.physicsforums.com/showthread.php?t=249363 but thread is closed otherwise i wouldv resurrected thread.. that worked problem helped alot but didnt explain how to take the transforms when x's and y's are in 1 equation...thanks for any help
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Mar 18, 2012
  2. jcsd
  3. Mar 18, 2012 #2

    rock.freak667

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    Homework Helper

    Taking the laplace transform of

    x' = 3x - 4y
    y' = 2x + 3y

    will give you two equations which only consist of X(s) and Y(s) (as it is there you have dx/dy,dy/dt,x and y in the equations which aren't too helpful).

    Taking the Laplace transform will give you the two equations which you can solve simultaneously to get X(s) to be in some function of s, which you can then get x(t) from. Similarly, you can get Y(s) to be some function of s and then get y(t)
     
  4. Mar 18, 2012 #3
    thanks for help... ok i dont know why i was having probs with Y(s)...anyway i get
    Y(s) = 1/s+1
    y(t) = e^-t

    next solve other equation + inv L.T., then solve 2 equations 2 unknowns to get y(t) = and x(t) =?
    thanks again

    (also, is this the method to use for all these kinds of problems? (system of equations)?
     
  5. Mar 18, 2012 #4

    rock.freak667

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    Right so then just put back in Y(s) = 1/(s+1) and get X(s) to get x(t).

    Usually this method will work as you have the initial conditions.
     
  6. Mar 18, 2012 #5
    thanks for the help..
    What form should the solution be in?
    i just need to know if i am combining the equations correctly.. heres what i did:

    x' = 3x - 4y...............1
    y' = 2x + 3y.............2

    solved for X(s) in 1
    X(s) = -4Y(s)(1/s-3) + 1/s-3

    took inverse L.T. of X(s)
    x(t) = -4Y(s)(e^3t) + e^3t

    inserted X(s) into #2
    Y(s) = 2/(s-3)(s-3)^2 + 8

    took inverse L.T. of Y(s)
    y(t) = 2e^3t + (2/root8)e^3tsin(root8t)

    plugged y(t) into x(t) equation:
    x(t) = -8e^6t - 8/root8e^6tsin(root8t) + e^3t

    so should the answer be in the form x(t) = and y(t) = ..or should it be one equation? thanks again
     
  7. Mar 18, 2012 #6

    rock.freak667

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    It would be in the form of two equations as that is how your equations were in the beginning.
     
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