Yes this is the standard way of reducing a second order ODEOriginally Posted by braddy
to a first order system. The only comment that I would make on
this is that the original equation becomes the system:
RonL
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