The Initial Value Problem y' = l(x, y); y(e) = d 89
(18) kz = l(xn + ehn, Yn + dk1hn) = l(xn, Yn) + ehnlx + dk1hnly+
ezh2 d2h2 kz
T lxx + edh;kifxy + 2n^1 lyy + O(h^3 ).Substituting ki = l into (18), substituting the resulting equation into (16),
and rearranging in ascending powers of hn, we findComparing (17) with (19), we see that for the corresponding coefficients of
hn and h~ to agree, we must have(20) a+ b = 1,
1
be=
2
,1
and bd =
2
.Thus, we have three equations in four unknowns. Hence, we might hope to
be able to choose the constants in such a manner that the coefficients of h~
in (17) and (19) agree. However, for these coefficients to agree we must havebe^2
21
6'1
bed= -
3'bd^2
216'
and lxly + l;l = 0.
Obviously, the last equality is not satisfied by a ll functions f.
There are an infinite number of solutions to the simultaneous equations (20).The choice a = ~, b = ~, e = 1, and d = 1 yields the improved Euler's
method (14). Choosing a= 0, b = 1, e = ~' and d = ~results in the following
recursion which is known as the modified Euler's method:
(21)Fourth Order Runge-Kutta Method If one tries to develop a general
recursion of the form
(22) Yn+i = Yn + hn[aif(xn, Yn) + a2l(xn + bihn, Yn + bihnk1)+
a3l(xn + b2hn, Yn + b2hnk2) + a4l(xn + b3hn, Yn + b3hnk3)]where ki = l(xn, Yn) and ki = l(xn + bi-1hn, Yn + bi-1hnki-1) for i = 2, 3, 4
by determining the constants ai, az, a3, a4, bi, b2, and b3 in such a manner that
(22) will agree with a Taylor series expansion of as high an order as possible,
one obtains a system of algebraic equations in the constants. In this case, as
before, there are an infinite number of solutions to the system of equations.