Re: Solving 2nd degree differntial equatation

[tmacchant]

>And initial condition x0=1, after rewritting it in the same manner will be
x0=[1;0]? 
You should have two values for initial condition for second order ODE.
Usually it is x(0) and x'(0). So that initial condition of the 1st order 2
simultaneous ODEs is essentially the same as that of the 2nd order ODE. 

As Juan suggested, validate your method with the ODE whose analytical
solution is known like  x(t)''+k*x(t)=0     

Tatsuro



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