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ODE and Laplace transform

  • Thread starter Niles
  • Start date
  • #1
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Homework Statement


Hi

I am trying to solve the following system of ODE's by Laplace transforming:
[tex]
x' = 1 + 21y - 6x \\
y' = 6x-53y
[/tex]
with the initial conditions x(0)=y(0)=0. Laplace transforming gives me (X and Y denote the Laplace transformed variables)
[tex]
sX = 1 + 21y-6x \\
sY = 6x-53y
[/tex]
From these I find
[tex]
X(s) = \frac{1}{6+s-126/(s+53)}
[/tex]
The inverse Laplace transform is (I have checked this with Mathematica)
[tex]
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}}
[/tex]
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?

Best,
Niles.
 

Answers and Replies

  • #2
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,535
751

Homework Statement


Hi

I am trying to solve the following system of ODE's by Laplace transforming:
[tex]
x' = 1 + 21y - 6x \\
y' = 6x-53y
[/tex]
with the initial conditions x(0)=y(0)=0. Laplace transforming gives me (X and Y denote the Laplace transformed variables)
[tex]
sX = 1 + 21y-6x \\
sY = 6x-53y
[/tex]
Shouldn't these be$$
sX =\frac 1 s +21Y-6X$$ $$
sY = 6X-53Y$$
 
  • #3
1,868
0
You are right, thanks for that! I don't know why I thought it would just be constant.

Best,
Niles.
 

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