Using Laplace Transform to Solve Non-Zero Initial Condition PDEs

[mike1111]

Homework Statement

Help, I don't know how to do the following question:

Using Laplace to solve
x' -y =1
2x' +x +y' = (t²-2t+1)e^-(t-1)

Homework Equations

x(1)=0
y(3)=0

The Attempt at a Solution

The problem I'm having is the initial conditions aren't at zero, and I'm not sure how to approach the question

so far I have:
X(s) -x(0) -Y(s) = 1/s
3X(s) -2x(0) +Y(s)-y(0)= F{(t²-2t+1)e^-(t-1)}

 

[LCKurtz]

This isn't a pde. It is a system of ordinary DE's. And you have mistakes in your transforms.

L(x') is not X(s) - x(0) it is sX(s) - x(0), and ditto for y'. And, of course, you need to transform (t²-2t+1)e^-(t-1).

Just call x(0) = a and y(0) = b and leave them in there. Once you get the equations right and solve for X(s) and Y(s), you can take the inverse transforms. Your answers will have a and b in them. Finally, plug in your given initial conditions and choose a and b to make them work.

 

[mike1111]

Thanks a lot LCKurtz, I didn't see that mistake. and the solution makes more sense now. I was getting weird answer from other questions too and could work out why.