(adsbygoogle = window.adsbygoogle || []).push({}); 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)

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# System Of Laplace Transforms

**Physics Forums | Science Articles, Homework Help, Discussion**