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{SECT 0 {PARA 256 "" 0 "" {TEXT -1 36 "Section 8.4 - The Runge-Kutta M
ethod" }}{PARA 257 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
84 "Here is a procedure for using the Runge-Kutta method.  It takes as
 input a function " }{XPPEDIT 18 0 "f(t,y);" "6#-%\"fG6$%\"tG%\"yG" }
{TEXT -1 0 "" }{TEXT -1 19 ", an initial value " }{XPPEDIT 18 0 "y[0];
" "6#&%\"yG6#\"\"!" }{TEXT -1 0 "" }{TEXT -1 5 " for " }{XPPEDIT 18 0 
"y;" "6#%\"yG" }{TEXT -1 0 "" }{TEXT -1 18 ", an initial time " }
{XPPEDIT 18 0 "t[0];" "6#&%\"tG6#\"\"!" }{TEXT -1 0 "" }{TEXT -1 5 " (
so " }{XPPEDIT 18 0 "y(t[0]) = y[0];" "6#/-%\"yG6#&%\"tG6#\"\"!&F%6#F*
" }{TEXT -1 0 "" }{TEXT -1 16 "), a final time " }{XPPEDIT 18 0 "t[1];
" "6#&%\"tG6#\"\"\"" }{TEXT -1 0 "" }{TEXT -1 14 ", an interval " }
{XPPEDIT 18 0 "L;" "6#%\"LG" }{TEXT -1 0 "" }{TEXT -1 87 " for the dis
tance between\npoints at which you want to know the results and a step
 size " }{XPPEDIT 18 0 "h;" "6#%\"hG" }{TEXT -1 0 "" }{TEXT -1 1 "." }
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "runge_kutta_method := proc(
f,y0,t0,t1,L,h);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "   n := (t1-t0)
/h;" }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "n;" "6#%\"nG" }{TEXT -1 0 "" }
{TEXT -1 30 " is the total number of steps." }{TEXT -1 0 "" }}{PARA 0 
"> " 0 "" {MPLTEXT 1 0 19 "   m := round(L/h);" }}{PARA 0 "" 0 "" 
{TEXT -1 22 "The results should be " }{TEXT -1 14 "printed every " }
{XPPEDIT 18 0 "m;" "6#%\"mG" }{TEXT -1 7 " steps." }}{PARA 0 "> " 0 "
" {MPLTEXT 1 0 15 "   y(t0) := y0;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 
14 "   t(0) := t0;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "   for i from
 1 to n " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "      do" }}{PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 24 "         t(i) := t0+i*h;" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 35 "         k1 := f(t(i-1),y(t(i-1)));" }}{PARA 0 "> " 
0 "" {MPLTEXT 1 0 50 "         k2 := f(t(i-1)+0.5*h,y(t(i-1))+0.5*h*k1
);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "         k3 := f(t(i-1)+0.5*h
,y(t(i-1))+0.5*h*k2);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "         k
4 := f(t(i),y(t(i-1))+h*k3);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "   \+
      y(t(i)):= y(t(i-1))+h*(k1+2*k2+2*k3+k4)/6; " }}{PARA 0 "> " 0 "
" {MPLTEXT 1 0 66 "         if type (i/m,integer)='true' then print(t(
i),y(t(i))) fi;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "       od;" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "end;" }}}{PARA 0 "" 0 "" {TEXT -1 
50 "Let's try it on the function in Problem 1, p. 438." }{TEXT -1 0 "
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "f := (t,y) -> 2*y-1;" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "runge_kutta_method(f,1,0,.4
,.1,.1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "runge_kutta_met
hod(f,1,0,.4,.1,.05);" }}}{PARA 0 "" 0 "" {TEXT -1 27 "Here is the exa
ct solution." }{TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
8 "restart:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "sol := dsolv
e(\{diff(y(t),t)=2*y(t)-1,y(0)=1\},y(t));" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 25 "Y := unapply(rhs(sol),t);" }}}{PARA 0 "" 0 "" 
{TEXT -1 66 "Here it is evaluated at the same points as the Runge Kutt
a method." }{TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "
evalf(Y(.1));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "evalf(Y(.2
)); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "evalf(Y(.3));" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "evalf(Y(.4));" }}}{PARA 0 "
" 0 "" {TEXT -1 23 "The error is less than " }{XPPEDIT 18 0 "5*10^(-6)
;" "6#*&\"\"&\"\"\")\"#5,$\"\"'!\"\"F%" }{TEXT -1 27 " with a step siz
e of .05.  " }{MPLTEXT 1 0 1 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 0 "" }}}}{MARK "19" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }
{PAGENUMBERS 0 1 2 33 1 1 }