davidbenari said:Homework Statement
I keep getting the wrong answer, and wolphram seems to back me up.
Here's the system of equations
##(-10+s)X(s)-7Y(s)=\frac{10}{s}##
##X(s)+(-2+s)Y(s)=0##
Homework Equations
The Attempt at a Solution
Using Cramer's rule I've got
##X(s)=\frac{10}{(s-9)(s-3)}-\frac{20}{s(s-9)(s-3)}##
Using partial fraction decomposition I've got
##x(t)=e^{9t}(10/6-20/54)+e^{3t}(-10/6+20/18)-20/27##
Evaluating ##x(1)## I get ##10492.1##
davidbenari said:The correct answer is supposedly -1426.16 :/ Strangely enough, if I solve ##y## and evaluate ##y(1)## I get an answer close to that.
Laplace transform is a mathematical tool used to convert a differential equation into an algebraic equation, making it easier to solve. It transforms the equation from the time domain to the frequency domain, where the equations can be manipulated using algebraic techniques.
To solve a system of equations using Laplace transform, you first take the Laplace transform of each equation in the system. Then, you manipulate the transformed equations using algebraic techniques to eliminate variables and solve for the remaining variables. Finally, you take the inverse Laplace transform to get the solution in the time domain.
Evaluating x(1) means finding the value of the variable x at time t=1. This is important because it gives a specific solution for the system of equations at a specific time, which can be used to understand the behavior of the system over time.
Yes, there are some limitations to using Laplace transform. It is only applicable to linear systems of equations, and the equations must have constant coefficients. Additionally, it can only be used for initial value problems, where the initial conditions of the system are known.
Yes, Laplace transform can be used to solve real-world problems in various fields such as engineering, physics, and economics. It is particularly useful in solving differential equations that model real-world systems, as it provides an efficient and accurate method to find solutions.