Understanding the Laplace System of Equations for Initial Value Problems

[CarbonWater]

Homework Statement

Solve the initial value problem

| x'=x+y
| y'=-x+y

x(0)=y(0)=1

Homework Equations

N/A

The Attempt at a Solution

This is actually a solved problem with a too many details missing. You have to set up the equation as

| sX(s)-x(0)=X(s)+Y(s)
| sY(s)-y(0)= -X(s)+Y(s)

The next step is then to solve for X and Y respectively and the book jumps to

| X(s)=s/((s-1)^2+1)
| Y(s)=(s-2)/((s-1)^2 +1)

How did they do this? Where did the Y(s) go in the first line of the system and where did the X(s) go in the second line of the system? I'm not clear how they got rid of those variables.

 

[Wingeer]

From the first equations you get (and be aware of the notation X(s)=X):
X(s-1) - 1 = Y
Y(s-1) = 1 - X

So by using basic algebra (solving a system of linear equations), you'll get to the next step.
Note that the reason why we transform the equation is because it is way easier to solve an algebraic problem than it is to solve a system of differential equations.

 

[LCKurtz]

From the first equations you get (and be aware of the notation X(s)=X):
X(s-1) - 1 = Y
Y(s-1) = 1 - X

So by using basic algebra (solving a system of linear equations), you'll get to the next step.
Note that the reason why we transform the equation is because it is way easier to solve an algebraic problem than it is to solve a system of differential equations.

I would think given the nature of the question, that notation would confuse the poster. Better to write the system:

(s-1)X(s) - Y(s) = 1

X(s) + (s-1)Y(s) = 1

Two equations in the two unknowns X(s) and Y(s).

 

[Wingeer]

Perhaps, but as I am lazy of nature I prefer the shortest notation. :-)