How to solve second-order matrix diffrential equation?

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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 first-order but ...
 

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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,
 
HallsofIvy
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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 \\ -AY-AB\end{array}\right)[/tex]
 
thanx
yes [tex]X\dot{}[/tex] = dX / dt and t is time.

for first-order 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.
 
thanx
yes [tex]X\dot{}[/tex] = dX / dt and t is time.

for first-order 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.
 
HallsofIvy
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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].
 
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: caley-hamilton 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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