11—Numerical Analysis 333
Substitute this into the equation fory(0):
y(0) =
∑N
k=1
αk
∑∞
n=0
(−kh)n
y(n)(0)
n!
+h
∑N
k=1
βk
∑∞
n=0
(−kh)n
y(n+1)(0)
n!
.
This should be an identity to as high an order as possible. The coefficient ofh^0 gives
1 =
∑N
k=1
αk. (40)
The next orders are
0 =
∑
k
αk(−kh) +h
∑
k
βk
0 =
∑
k
1
2
αk(−kh)^2 +h
∑
k
βk(−kh)
..
. (41)
N= 1is Euler’s method again.
N= 2gives
α 1 +α 2 = 1
α 1 + 4α 2 = 2(β 1 + 2β 2 )
α 1 + 2α 2 =β 1 +β 2
α 1 + 8α 2 = 3(β 1 + 4β 2 ).
The solution of these equations is
α 1 =− 4 α 2 = +5 β 1 = +4 β 2 = +2
y(0) =− 4 y(−h) + 5y(− 2 h) +h
[
4 y′(−h) + 2y′(− 2 h)
]
. (42)
To start this algorithm off, two pieces of information are needed: the values ofyat−hand at− 2 h. This is in
contrast to Runge-Kutta, which needs only one point.