082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

624 Chapter 20 Ordinary differential equations II
Table20.6
Comparison of the Runge--Kutta method
with exact solutions.
i x i y i y(x i )
0 0.0 2.0 2.0
1 0.25 1.8823 1.8824
2 0.5 1.5999 1.6000
3 0.75 1.2799 1.2800
4 1.0 1.0000 1.0000
Table20.7
The solution to
Example 20.9.
x i
y i
0.0 0.0
0.2 0.0213
0.4 0.0897
0.6 0.2112
Example20.9 Usethe fourth-order Runge--Kutta method toobtain a solution of
Solution We have
dy
dx =x2 +x−y
subject toy = 0 whenx = 0, for 0 x 0.6 withh = 0.2. Work throughout to four
decimal places.
where
y i+1
=y i
+ h 6 (k 1 +2k 2 +2k 3 +k 4 )
k 1
= f(x i
,y i
)
(
k 2
=f x i
+ h )
2 ,y i +h 2 k 1
(
k 3
=f x i
+ h )
2 ,y i +h 2 k 2
k 4
=f(x i
+h,y i
+hk 3
)
Inthisexample, f (x,y) =x 2 +x−yandx 0
= 0,y 0
= 0.Thefirststageinthesolution
isgiven by
k 1
=f(x 0
,y 0
)=0
k 2
= f(0.1,0) = 0.11
k 3
= f(0.1,0 + (0.1)(0.11)) = f(0.1,0.011) = 0.099
k 4
= f (0.2,0 + (0.2)(0.099)) = f (0.2,0.0198) = 0.2202
Therefore,
y 1
=0+ 0.2 (0 +2(0.11) +2(0.099) +0.2202) = 0.0213
6
which, in fact, is correct to four decimal places. Check the next stage for yourself. The
complete solution isshown inTable 20.7.