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+--- Thread: (15C) Runge-Kutta 4th order method (/thread-4210.html)
(15C) Runge-Kutta 4th order method - Thomas Klemm - 06-23-2015 06:25 PM
This program is from a post by Valentin Albillo in the old forum:
Small HP calc versus MATHEMATICA ! [LONG]
Code:
001 - 42,21,11 LBL A 013 - 44 3 STO 3 025 - 45 1 RCL 1 037 - 32 2 GSB 2
002 - 44 0 STO 0 014 - 32 0 GSB 0 026 - 42 31 PSE 038 - 44,40, 3 STO+ 3
003 - 33 R↓ 015 - 32 0 GSB 0 027 - 45 2 RCL 2 039 - 43 32 RTN
004 - 43 32 RTN 016 - 45,40, 2 RCL+ 2 028 - 42 31 PSE 040 - 42,21, 2 LBL 2
005 - 42,21,12 LBL B 017 - 45 0 RCL 0 029 - 22 1 GTO 1 041 - 45,40, 1 RCL+ 1
006 - 44 1 STO 1 018 - 32 2 GSB 2 030 - 42,21, 0 LBL 0 042 - 32 13 GSB C
007 - 34 x<>y 019 - 45 3 RCL 3 031 - 2 2 043 - 45,20, 0 RCL× 0
008 - 44 2 STO 2 020 - 6 6 032 - 10 ÷ 044 - 44,40, 3 STO+ 3
009 - 42,21, 1 LBL 1 021 - 10 ÷ 033 - 45,40, 2 RCL+ 2 045 - 43 32 RTN
010 - 34 x<>y 022 - 44,40, 2 STO+ 2 034 - 45 0 RCL 0 046 - 42,21,13 LBL C
011 - 32 13 GSB C 023 - 45 0 RCL 0 035 - 2 2 047 - 43 32 RTN
012 - 45,20, 0 RCL× 0 024 - 44,40, 1 STO+ 1 036 - 10 ÷Example:
Quote:Given the equation \({y}' = x^2 + y^2\), with \(y(0) = 0\), find an accurate value for \(y(2)\).
It turns out that there is a singularity near \(x = 2\) which can be removed using a change of variable: \(y = \tan(z)\)
This leads to:
\({z}' = \frac{x^2 + \tan^2(z)}{1 + \tan^2(z)}\)
The initial condition is: \(z(0) = 0\)
This differential equation is programmed under LBL C:
Code:
046 - 42,21,13 LBL C
047 - 43 11 x²
048 - 34 x<>y
049 - 25 TAN
050 - 43 11 x²
051 - 40 +
052 - 1 1
053 - 43 36 LSTx
054 - 40 +
055 - 10 ÷
056 - 43 32 RTNIn RAD mode, using \(h = 0.1\), then \(h = 0.05\), we get:
\(\begin{array}{l l l}
\hline\hline
x & y (h=0.1) & y (h=0.05) \\
\hline
0 & 0 & 0 \\
1 & 0.3368811 & 0.3368813 \\
1.7 & 1.2463033 & 1.2463031 \\
1.8 & 1.3606417 & 1.3606412 \\
1.9 & 1.4666491 & 1.4666485 \\
2.0 & 1.5676491 & 1.5676489 \\
\hline
\end{array}
\)
This clearly shows that the singularity has been effectively removed, so we've got:
z(2) = 1.5676489
and thus:
y(2) = tan(z(2)) = 317.7225457