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Re: Solving 2nd degree differntial equatation


From: Maverick
Subject: Re: Solving 2nd degree differntial equatation
Date: Sat, 4 Jun 2016 07:29:46 -0700 (PDT)

@Juan Pablo Carbajal-2 I know that results may vary, but difference seems
just to much for me. 
Alright as you both suggested i tested it with something known:  
x''+x'=0
x0(0)=2
x0'(0)=2
So if I am not wrong solution should be x(t)=C1+C2*e^(-t) And individual
solution will be 
x(t)=4-2*e^(-t)
I tested it for T=1, in Wolfram x(T)=3.264241117657... and lsode and RK
gives 3.26424 which is great and my hopeless Verlet 3.76570. It is quite
close. But when I try with my first equatation diffrence is much bigger. Now
I've discovered that for small *k* like 0.5 or less Verlet and RK4 gives
almost identical results(for my primary equatation). I may assume that my
implementation is fine, and it's fault of algorithm?



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