Using laplace transforms to solve system of (two) equations

[fufufu]

Homework Statement

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

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

Homework Equations

y' = sY(s) - y(0)

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 can't 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 relevant 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 a lot but didnt explain how to take the transforms when x's and y's are in 1 equation...thanks for any help

Homework Statement

Homework Equations

The Attempt at a Solution

 

[rock.freak667]

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)

 

[fufufu]

thanks for help... ok i don't 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)?

 

[rock.freak667]

thanks for help... ok i don't 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)?

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.

 

[fufufu]

thanks for the help..
What form should the solution be in?
i just need to know if i am combining the equations correctly.. here's 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

 

[rock.freak667]

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