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System Of Laplace Transforms

  1. Apr 5, 2008 #1
    1. The problem statement, all variables and given/known data
    Solve using Laplace Transforms

    [tex]\frac{dx}{dt}=x-2y[/tex]
    [tex]\frac{dy}{dt}=5x-y[/tex]

    Subject to x(0)=0 y(0)=0


    Okay. So I know that there are more efficient ways to solve this, but I missed the class in systems of Laplace Transforms and would like to get this resolved.

    I guess my first question is what the first step is. Is it the same as any other kind of systems approach, that is, to find a multiplier that allows me to sum the two equations eliminating one of the variables?

    I must assume this is the approach, but for some reason I am confused.
    So I have:

    [itex]x'-x+2y=0[/itex] (1)
    [itex]y'+y-5x=0[/itex] (2)

    I am looking in my text, I will come back to edit...

    Okay. So all of the examples given only have one dependent variable, so bear with me.

    Taking Laplace of each equation yields:

    [itex]sX(s)-x(0)-X(s)+2Y(s)=0[/itex] (3)
    [itex]sY(s)-y(0)+Y(s)-5X(s)=0[/itex] (4)

    Applying initial values

    [itex]sX(s)-X(s)+2Y(s)=0[/itex]
    [itex]sY(s)+Y(s)-5X(s)=0[/itex]

    I am assuming that this is the point where I eliminate? I also assume I should sort these first. Giving:
    [itex]X(s)[s-1]+2Y(s)=0[/itex] (5)
    [itex]Y(s)[s+1]-5X(s)=0[/itex] (6)

    Multiplying
    5*(5) and [s-1]*(6) yields

    [itex]5X(s)[s-1]+10Y(s)=0[/itex] (7)
    [itex]Y(s)[s+1][s-1]-5X(s)[s-1]=0[/itex] (8)

    Adding (7) + (8) yields

    [itex]10Y(s)+Y(s)(s+1)(s-1)=0[/itex] (9)
     
    Last edited: Apr 5, 2008
  2. jcsd
  3. Apr 5, 2008 #2
    Can someone tell me if I am on the right path? I cannot see the next move. If I factor out Y(s) and divide it off, I am going to get Y(s)=0 !

    Did I approach this right?
     
  4. Apr 5, 2008 #3

    Tom Mattson

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    Staff Emeritus
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    Gold Member

    Yes, you're going to get Y(s)=0. This problem is a bit silly, as it only has the zero solution. Even if you don't understand Laplace transforms, you can still understand why this is the case. Let us separate the coupled differential equations.

    Differentiate the first equation to get:

    [tex]\frac{d^2x}{dt^2}=\frac{dx}{dt}-2\frac{dy}{dt}[/tex]

    Now use the original equations to sub in for [itex]\frac{dx}{dt}[/itex] and [itex]\frac{dy}{dt}[/itex]. You obtain the following.

    [tex]\frac{d^2x}{dt^2}+9x=0[/tex].

    Furthermore, [itex]x(0)=y(0)=0[/itex] implies that [itex]\frac{dx(0)}{dt}=0[/itex] as well. So you have a sourceless 2nd order DE with zero initial conditions.

    The only solution is the zero solution.
     
  5. Apr 5, 2008 #4
    Hey now! I see what you are saying. Thanks Tom. I should have gone with my gut!
     
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