Re: Solving 2nd degree differntial equatation

[Maverick]

Thanks for reply.
Yes, I've rewritten it like that/(f=@(x,t) [x(2);-k*sin(x(1))];/) 
And initial condition x0=1, after rewritting it in the same manner will be
x0=[1;0]?
I know it works considerebly slower than build in function, but it's my
assigment to test both methods. 
Could you take a look at Verlet method? I read it like 10 times and I see no
error but it doesn't converges to same point as lsode() or RK. 
where:
x0 is initial condition
f is our function
T is argument for which we calculate x(T)
t0 is staring point
N is number of steps in method
k is our constant
/function x=Verlet(f,t0,T,x0,N,k,h=(T-t0)/N)
t=t0;
x=x0;
v=f(x,t);
for n=1:N

    v=v-(h/2)*k*sin(x);
    x=x+h*v;
    v=v-(h/2)*k*sin(x);
end

endfunction /



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