# How to solve second-order matrix diffrential equation?

1. ### jahandideh

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 ...

2. ### trambolin

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,

3. ### HallsofIvy

40,395
Staff Emeritus
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
$$Z^.= \left(\begin{array}{c}X^. \\Y^.\end{array}\right)= \left(\begin{array}{c}Y \\ -AY-AB\end{array}\right)$$

4. ### jahandideh

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.

5. ### jahandideh

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.

6. ### HallsofIvy

40,395
Staff Emeritus
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
$$Y= e^{\left(\begin{array}{cc}-5 & -2 \\ 2 & -2\end{array}\right)t}Y(0)$$.

7. ### jahandideh

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?