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How to solve secondorder matrix diffrential equation? 
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#1
Mar2808, 01:30 AM

P: 7

hi all
this is the general problem X[tex]\ddot{}[/tex]+AX[tex]\dot{}[/tex]+BX=0 let A, B,X be 2*2 matrices its application is in vibrations. any opinion will be great I can solve the firstorder but ... 


#2
Mar2808, 05:45 AM

P: 341

Can you define what do you mean by [itex]\frac{d}{dt}X[/itex]. Is it the time derivative of the entries or something else?
You can start diagonalizing the matrices and then solving the mode shapes, the trick as usual in vibtration analysis, 


#3
Mar2808, 06:44 AM

Math
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PF Gold
P: 39,338

A system of differential equations, of any order, can always be reduced to a (larger) system of first order differential equations.
Here, let Y= X^{.}. Then X^{..}= Y^{.} so your equation becomes Y^{.}+ AY+ BX= 0. That, together with X^{.}= Y gives two first order matrix equation. If X is an n by n matrix, then so is Y and letting Z be the matrix n by 2n matrix with the rows of X above the rows of Y, we have [tex]Z^.= \left(\begin{array}{c}X^. \\Y^.\end{array}\right)= \left(\begin{array}{c}Y \\ AYAB\end{array}\right)[/tex] 


#4
Mar2808, 07:36 AM

P: 7

How to solve secondorder matrix diffrential equation?
thanx
yes [tex]X\dot{}[/tex] = dX / dt and t is time. for firstorder system of differential equations like: [tex]X\dot{}=AX+BU[/tex] the solution is X(t) = [tex] e^{At} [/tex] X(0)+ [tex]\int e^{A(t\tau)} BU(\tau) d\tau [/tex] for example I can solve this system : X\dot{} = {0 1 ; 2 3 } X + {0 1} u but I have problem with this [tex]Y\ddot{}[/tex] = {5 2 ; 2 2} Y which [tex]Y\ddot{}[/tex] and Y are 2 by 2 matrices. 


#5
Mar2808, 07:39 AM

P: 7

thanx
yes [tex]X\dot{}[/tex] = dX / dt and t is time. for firstorder system of differential equations like: [tex]X\dot{}=AX+BU[/tex] the solution is X(t) = [tex] e^{At} [/tex] X(0)+ [tex]\int e^{A(t\tau)} BU(\tau) d\tau [/tex] for example I can solve this system : [tex]X\dot{} [/tex]= {0 1 ; 2 3 } X + {0 1} u but I have problem with this [tex]Y\ddot{} = {5 2 ; 2 2} Y[/tex] which [tex]Y\ddot{}[/tex] and [tex]Y[/tex] are 2 by 2 matrices. 


#6
Mar2908, 05:58 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,338

Why would you have a problem with that? That's just your general form with B= 0. According to your formu;a, the solution is
[tex]Y= e^{\left(\begin{array}{cc}5 & 2 \\ 2 & 2\end{array}\right)t}Y(0)[/tex]. 


#7
Mar3008, 01:14 AM

P: 7

no! no!
that was not [tex]y\dot{}=\left(\begin{array}{cc}5 & 2 \\ 2 & 2\end{array}\right) y[/tex] this is a second order system of differential equation [tex]y\ddot{}{}=\left(\begin{array}{cc}5 & 2 \\ 2 & 2\end{array}\right)y [/tex] reducing the order of the system by assuming [tex]y\dot{} = p[/tex] is an idea but it is very time consuming because we have to calculate e^At towice , i am searching for a better way . by the way to solve e^At (A is a squar matrix) there is several ways like: caleyhamilton theorium , using similarity tansform and Jordan matrix , but I found the Laplace transform a better way so is there any faster way to calculate the e^At? 


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