# How to solve second-order matrix diffrential equation?

by jahandideh
Tags: diffrential, equation, matrix, secondorder, solve
 P: 7 hi all this is the general problem X$$\ddot{}$$+AX$$\dot{}$$+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 ...
 P: 341 Can you define what do you mean by $\frac{d}{dt}X$. 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,
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,567 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 $$Z^.= \left(\begin{array}{c}X^. \\Y^.\end{array}\right)= \left(\begin{array}{c}Y \\ -AY-AB\end{array}\right)$$
 P: 7 How to solve second-order matrix diffrential equation? thanx yes $$X\dot{}$$ = dX / dt and t is time. for first-order system of differential equations like: $$X\dot{}=AX+BU$$ the solution is X(t) = $$e^{At}$$ X(0)+ $$\int e^{A(t-\tau)} BU(\tau) d\tau$$ for example I can solve this system : X\dot{} = {0 1 ; 2 3 } X + {0 1} u but I have problem with this $$Y\ddot{}$$ = {-5 -2 ; 2 -2} Y which $$Y\ddot{}$$ and Y are 2 by 2 matrices.
 P: 7 thanx yes $$X\dot{}$$ = dX / dt and t is time. for first-order system of differential equations like: $$X\dot{}=AX+BU$$ the solution is X(t) = $$e^{At}$$ X(0)+ $$\int e^{A(t-\tau)} BU(\tau) d\tau$$ for example I can solve this system : $$X\dot{}$$= {0 1 ; 2 3 } X + {0 1} u but I have problem with this $$Y\ddot{} = {-5 -2 ; 2 -2} Y$$ which $$Y\ddot{}$$ and $$Y$$ are 2 by 2 matrices.
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,567 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 $$Y= e^{\left(\begin{array}{cc}-5 & -2 \\ 2 & -2\end{array}\right)t}Y(0)$$.
 P: 7 no! no! that was not $$y\dot{}=\left(\begin{array}{cc}-5 & -2 \\ 2 & -2\end{array}\right) y$$ this is a second order system of differential equation $$y\ddot{}{}=\left(\begin{array}{cc}-5 & -2 \\ 2 & -2\end{array}\right)y$$ reducing the order of the system by assuming $$y\dot{} = p$$ 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?