How do I solve a system of ODE's using Laplace transform?

[Niles]

Homework Statement

Hi

I am trying to solve the following system of ODE's by Laplace transforming:
$$x' = 1 + 21y - 6x \\<br /> y' = 6x-53y$$
with the initial conditions x(0)=y(0)=0. Laplace transforming gives me (X and Y denote the Laplace transformed variables)
$$sX = 1 + 21y-6x \\<br /> sY = 6x-53y$$
From these I find
$$X(s) = \frac{1}{6+s-126/(s+53)}$$
The inverse Laplace transform is (I have checked this with Mathematica)
$$x(t) = \frac{e^{-\frac{1}{2} \left(59+\sqrt{2713}\right) t} \left(-47+\sqrt{2713}+\left(47+\sqrt{2713}\right) e^{\sqrt{2713} t}\right)}{2 \sqrt{2713}}$$
When I take t=0, then I get x(0)=1, not x(0)=0. I not quite sure where I have gone wrong, I have double-checked everything by doing it numerically too.

Is there something that I have forgotten to do?Niles.

 

[LCKurtz]

Homework Statement

Hi

I am trying to solve the following system of ODE's by Laplace transforming:
$$x' = 1 + 21y - 6x \\<br /> y' = 6x-53y$$
with the initial conditions x(0)=y(0)=0. Laplace transforming gives me (X and Y denote the Laplace transformed variables)
$$sX = 1 + 21y-6x \\<br /> sY = 6x-53y$$

Shouldn't these be$$
sX =\frac 1 s +21Y-6X$$ $$
sY = 6X-53Y$$

 

[Niles]

You are right, thanks for that! I don't know why I thought it would just be constant.Niles.