$$\begin{pmatrix} x_1'(t) \ x_2'(t) \end{pmatrix}=\begin{pmatrix} 1 & -2 \ 2 &-1 \end{pmatrix}\begin{pmatrix} x_1(t) \ x_2(t) \end{pmatrix}$$

Rewrite the system of first order ode’s into a second order differential equation for $x_1(t)$

Question

$$\begin{pmatrix}  x_1'(t) \ x_2'(t) \end{pmatrix}=\begin{pmatrix}  1 & -2 \ 2 &-1 \end{pmatrix}\begin{pmatrix}  x_1(t) \ x_2(t) \end{pmatrix}$$

Rewrite the system of first order ode’s into a second order differential equation for $x_1(t)$

unklerhaukus · Answer

i dont understand what the question wants

anonymous · Answer

we have  $$x_1'=x_1-2x_2$$$$x_2'=2x_1-x_2$$
what to do now?

unklerhaukus · Answer

$$x_1'=x_1-2(2x_1-x_2')$$$$x_1'=x_2'-3x_1$$

unklerhaukus · Answer

?

anonymous · Answer

@eliassaab

phi · Answer

they want a 2nd order in terms of x1.  so we need to eliminate x2 and x2'

one way (is there a quicker way?)
from the 2 equations put x2' in terms of x1 and x1'
multiply the 2nd equation by -2 and add to get
2*x2' = x1' + 3x1

take the derivative of the 1st equation
x1''= x1'-2*x2'
replace 2*x2' with x1'+3x1 to get a 2nd order

anonymous · Answer

ahh...misread....in terms of $x_1$ is given..

phi · Answer

it looks like you get
$$ x_1''= -3x_1 $$

anonymous · Answer

yeah...no problem with phi's work...$x_1''=-3x_1$

unklerhaukus · Answer

$$x_1'=x_1-2x_2$$$$x_2'=2x_1-x_2$$
$$2x_2=x_1-x_1'$$
$$x_2'=2x_1-\frac{x_1-x_1'}2$$$$-2x_2'=2x_2-4x_1$$
$$-x_2'=2x_2-4x_1+2x_1-\frac{x_1-x_1'}2$$$$-x_2'=2x_2-2x_1-\frac{x_1-x_1'}2$$$$2x_2'=4x_1-2x_2+{x_1-x_1'}$$$$2x_2'=5x_1-2x_2-x_1'$$

???

anonymous · Answer

$x_1′=x_1−2x_2 \rightarrow 2x_2=x_1-x_1'  \ \ \ (1)$
$x_1′=x_1−2x_2 \rightarrow x_1''=x_1'-2x_2'  \ \ \ (2)$

$x_2′=2x_1−x_2 \rightarrow 2x_2'=4x_1-2x_2 $   put this in $(2)$

$x_1''=x_1'-2x_2'=x_1'-(4x_1-2x_2)=x_1'-4x_1+2x_2  \ \ \ (3)$

put $(1)$  in  $(3)$$$x_1''=x_1'-4x_1+(x_1-x_1')=-3x_1$$

unklerhaukus · Answer

i think im with you now,
$$x_1''=-3x_1$$
$$x_1=\iint -3x_1\text dt^2=\frac{-3x_1t^2}2+ct+d$$ right?

anonymous · Answer

$$x_1''+3x_1=0$$ 
characteristic equation  $m^2+3=0$

unklerhaukus · Answer

ooh

phi · Answer

See http://www.wolframalpha.com/input/?i=x%27%27%3D+-3x click on "show steps" for an explanation

unklerhaukus · Answer

the choice of variables threw me off

unklerhaukus · Answer

so,
$$x_1=A\cos \sqrt 3t+B\sin\sqrt 3t$$


looking back at the matrix at in the question, i dont really understand what the solution says about the system

unklerhaukus · Answer

what is oscillating exactly?

phi · Answer

here is a plot of the solution, where A=2 and B=3 (just so we have an example) http://www.wolframalpha.com/input/?i=plot+2*sin%28sqrt%283%29+t%29%2B3*cos%28sqrt%283%29+t%29 the solution oscillates over time

unklerhaukus · Answer

what is oscillating, x_1, x_1'' or the relationship between x_1 and x_2

phi · Answer

you started with an equation x'' = -3x  (where x is x1 of the original 2 equation system)
x is a function of t
you found what x(t) satisfies this 2nd order differential equation (or what x1 satisfies the 2 1st order linear differential equations)

the solution x(t) oscillates as a function of t.  This equation corresponds to free undamped harmonic motion. It arises, for example, when describing an ideal spring that you stretch out, and then let it go.  Or, approximately (for small angles) the motion of a pendulum that you move to one side, and then let go.

unklerhaukus · Answer

so x is only oscillating in one dimension then ok,

phi · Answer

yes.  It probably makes more sense if you knew where the equation came from in the first place.  For example, in physics, if you write down the forces acting on a spring, using Hooke's law and f=ma (Newton's 2nd law), you end up with a 2nd order differential equation.  It's solution is an oscillation:  the spring moves back and forth, and its speed and position are a function of time.

unklerhaukus · Answer

so the original matrix is two measurements of a spring systems velocity/?