# Using laplace transforms to solve system of (two) equations

• fufufu
In summary: So the final answer would be:x(t) = -8e^6t - 8/root8e^6tsin(root8t) + e^3ty(t) = 2e^3t + (2/root8)e^3tsin(root8t)
fufufu

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

## The Attempt at a Solution

Last edited:
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)

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)?

fufufu said:
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.

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

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

## 1. What are Laplace transforms and how do they work?

Laplace transforms are a mathematical tool used to convert a function of time into a function of complex frequency. They work by representing a function as a combination of complex exponential functions, making it easier to solve differential equations.

## 2. Can Laplace transforms be used to solve systems of two equations?

Yes, Laplace transforms can be used to solve systems of two equations. By converting the system into a set of algebraic equations in the frequency domain, it becomes easier to solve for the unknown variables.

## 3. What is the advantage of using Laplace transforms to solve systems of equations?

The advantage of using Laplace transforms is that it simplifies the process of solving complex systems of equations. It also allows for the use of algebraic techniques, making it easier to manipulate and solve the equations.

## 4. Are there any limitations to using Laplace transforms for solving systems of equations?

Yes, there are limitations to using Laplace transforms for solving systems of equations. The method is most effective for linear systems with constant coefficients. Non-linear systems or systems with time-varying coefficients may not be easily solvable using Laplace transforms.

## 5. How do I know if a system of equations can be solved using Laplace transforms?

A system of equations can be solved using Laplace transforms if it is linear, has constant coefficients, and is homogeneous. Additionally, the initial conditions must be given for each equation in the system.

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