Originally Posted by

**braddy** 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:

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

RonL