System of Differential equations using Laplace.

rellsun

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May 2, 2012
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Hey everyone i have a system of Differential equations that must be solved using Laplace transformation thats giving me trouble.
Heres the problem :9.3.jpg. all i have so far is this: x'=5x+5y+2z, y'=-6x-6y-5z, z'=6x+6y+5z.
 

CurveGuy

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Jul 22, 2020
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x'=5x+5y+2z, y'=-6x-6y-5z, z'=6x+6y+5z.
Its been many years, but the Laplace( x') = s * Laplace( x) if my memory serves me correctly. Thus, rewrite your equations like this: sX = 5X + 5Y + 2 Z, sY = -6X - 6Y - 5Z, and so on. Now collect terms on the left, e.g.
(s - 5)X = 5Y + 2Z ==> X = (5Y + 2Z) / (s -5). Do this same thing for Y & Z. Now you have a system of Laplace equations. Solve this system of 's' dependent variables and you'll have an answer.
 

LCKurtz

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May 3, 2019
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CurveGuy is correct about the method but assumed the initial conditions are all 0. For example \(\displaystyle \dot X = sX-x(0)\). This leads to the linear system:

\(\displaystyle \begin{pmatrix} sX-x(0)\\sY-y(0)\\sZ-z(0) \end{pmatrix} =
\begin{pmatrix} 5 & 5 & 2 \\ -6 & -6 & -5 \\ 6 & 6 & 5 \end{pmatrix} =
\begin{pmatrix} X \\ Y \\ Z \end{pmatrix} \)
where the initial conditions might not be \(\displaystyle 0\). You can solve the system by any favorite method you choose to get the transforms \(\displaystyle X,~Y,~Z\) after which you need to find their inverses.
That \(\displaystyle 6\) with a line above it should be \(\displaystyle -6\). There appears to be a problem with Latex rendering.
 
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