My Math Forum Quick question: Coupled first order differentials, correct?

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 September 17th, 2013, 06:37 AM #1 Member   Joined: Nov 2010 Posts: 36 Thanks: 0 Quick question: Coupled first order differentials, correct? I am given the following second order differential equation. I am wondering whether my final x'=ax+f(t) equation is correct? $Mz''+B(z'-u'+K(z-u)=0" /> Putting all u terms on the right side: $Mz''+Bz'+Kz=Bu#39;+Ku$ Substituting using: $x=z#39;$ $Mx'+Bx+Kz=Bu#39;+Ku$ This gives me two first order differential equations: $x'=\frac{Bu#39;+Ku-Bx-Kz}{M}$ $x=z#39;$ This can ultimately be written in the form: $\begin{pmatrix} x'\\ z' \end{pmatrix} = \begin{pmatrix} -B/M & -K/M\\ 1 & 0 \end{pmatrix}\begin{pmatrix} x\\ z \end{pmatrix}+\begin{pmatrix} \frac{B}{M}u'+\frac{K}{M}u\\ 0 \end{pmatrix}$ Is it correct to put u' and u in the f(t) part of the equation? Despite knowing that a second order D.E. can only be replaced by two first order D.E's, I have my doubts that I should also substitute u' with another term to have a third first order D.E.
 September 17th, 2013, 07:31 AM #2 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Quick question: Coupled first order differentials, corre The question is really about your first equation, "Mz''+ B(z'- u')+ K(z- u)= 0". What is "u"? Is it a given function or an unknown function you have to determine? If it is a given function then "Bu'+ Ku" is just another function and what you have done is perfectly valid.

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