1. ## ODE order change

Hi, I would to check if my transformation from an ODE 2nd degree to ODE 1st degree is fine:

X''(t)+X(t)=0

I set:
X1(t)=X(t)
X2(t)=X'(t)
this implies that:
X1'(t)=X2(t)
so the original equation becomes:
X2'(t)=-X1(t)

PLease tell me if I am right?

B

Hi, I would to check if my transformation from an ODE 2nd degree to ODE 1st degree is fine:

X''(t)+X(t)=0

I set:
X1(t)=X(t)
X2(t)=X'(t)
this implies that:
X1'(t)=X2(t)
so the original equation becomes:
X2'(t)=-X1(t)

PLease tell me if I am right?

B
Yes this is the standard way of reducing a second order ODE
to a first order system. The only comment that I would make on
this is that the original equation becomes the system:

$\left[ \begin{array}{cc}x_1(t) \\ x_2(t)\end{array} \right]'\ =\ \left[{ \begin{array}{cc}x_2(t) \\ -x_1(t)\end{array} \right]$

RonL