Transformation from an ODE 2nd degree to ODE

[brad sue]

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

 

[HallsofIvy]

Yes, that is correct. So you have two first order equations:
$$\frac{dX_1}{dt}= X_2$$
and
$$\frac{dX_2}{dt}= -X_1$$
instead of the single second order equation
$$\frac{d^2X}{dt^2}+ X= 0$$