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## Homework Statement

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)

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