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

20.8 Runge--Kutta method of order 4 623
20.8 RUNGE--KUTTAMETHODOFORDER4
The‘Runge--Kutta’methods arealargefamilyofmethods usedforsolvingdifferential
equations. The Euler and improved Euler methods are special cases of this family. The
orderofthemethodreferstothehighestpowerofhincludedintheTaylorseriesexpansion.Weshallnowpresentthefourth-orderRunge--Kuttamethod.Thederivationofthis
isbeyond the scope ofthis book.
To solve the equation dy
dx = f (x,y) subject toy = y 0 whenx = x 0
we generate the
sequence ofvalues,y i
, from the formula
where
y i+1
=y i
+ h 6 (k 1 +2k 2 +2k 3 +k 4 ) (20.17)
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
)
Example20.8 UsetheRunge--Kuttamethodtosolve dy
dx = −xy2 ,for0 x 1,subjecttoy(0) = 2.
Useh = 0.25 and work tofour decimal places.
Solution We shall show the calculations required to findy 1
. You should follow this through and
then verifythe results inTable 20.6. The exact solution is shown forcomparison.
f(x,y)=−xy 2 h=0.25 x 0
=0 y 0
=2
Takingi = 0 inEquation (20.17), wehave
k 1
= f(x 0
,y 0
) = −0(2) 2 =0
k 2
= f (0.125,2) = −0.125(2) 2 = −0.5
(
k 3
=f 0.125,2 + 0.25 )
2 (−0.5) = f (0.125,1.9375)
= −0.125(1.9375) 2 = −0.4692
k 4
= f (0.25,2 +0.25(−0.4692)) = f (0.25,1.8827)
Therefore,
= −0.25(1.8827) 2 = −0.8861
y 1
=2+ 0.25 (0 +2(−0.5) +2(−0.4692) + (−0.8861)) = 1.8823
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