- #1
schattenjaeger
- 178
- 0
forever!
I missed a day of notes, I know for the second order
Y(n+1) =(approx.) Y(n)+K2, and I have the algorithm for finding k1 and then k2, how does this differ from the 4th order?
I missed a day of notes, I know for the second order
Y(n+1) =(approx.) Y(n)+K2, and I have the algorithm for finding k1 and then k2, how does this differ from the 4th order?