numerical solution of differential equations

[Ether]

Originally Posted by sspoldi

one forward Euler integration, one backward Euler integration

V_n+1 = V_n + A_n∙dt;
X_n+1 = X_n + V_n+1∙dt;

Maybe provides some insight: the above is algebraically equivalent to

V_n+1 = V_n + A_n∙dt;
X_n+1 = X_n + dt∙(V_n+V_n+1)/2 + ½∙A_n∙dt²

[GeeTwo]

Originally Posted by Ether

Maybe provides some insight: the above is algebraically equivalent to

V_n+1 = V_n + A_n∙dt;
X_n+1 = X_n + dt∙(V_n+V_n+1)/2 + ½∙A_n∙dt²

..so counting the constant acceleration term twice in the position calculation mostly offsets not counting the jerk in the velocity calculation..at least in this case.