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

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SUMMARY

The discussion focuses on solving a system of ordinary differential equations (ODEs) using the Laplace transform method. The equations presented are x' = 1 + 21y - 6x and y' = 6x - 53y, with initial conditions x(0) = y(0) = 0. The participant initially misapplied the Laplace transform, leading to an incorrect expression for X(s). The correct transformed equations are sX = 1/s + 21Y - 6X and sY = 6X - 53Y, which was clarified by another participant.

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Niles
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Homework Statement


Hi

I am trying to solve the following system of ODE's by Laplace transforming:
<br /> x&#039; = 1 + 21y - 6x \\<br /> y&#039; = 6x-53y <br />
with the initial conditions x(0)=y(0)=0. Laplace transforming gives me (X and Y denote the Laplace transformed variables)
<br /> sX = 1 + 21y-6x \\<br /> sY = 6x-53y <br />
From these I find
<br /> X(s) = \frac{1}{6+s-126/(s+53)}<br />
The inverse Laplace transform is (I have checked this with Mathematica)
<br /> 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}} <br />
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.
 
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Niles said:

Homework Statement


Hi

I am trying to solve the following system of ODE's by Laplace transforming:
<br /> x&#039; = 1 + 21y - 6x \\<br /> y&#039; = 6x-53y <br />
with the initial conditions x(0)=y(0)=0. Laplace transforming gives me (X and Y denote the Laplace transformed variables)
<br /> sX = 1 + 21y-6x \\<br /> sY = 6x-53y <br />
Shouldn't these be$$
sX =\frac 1 s +21Y-6X$$ $$
sY = 6X-53Y$$
 
You are right, thanks for that! I don't know why I thought it would just be constant.Niles.
 

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