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

20.6 Euler’s method 617
and we can therefore generate a whole sequence of approximate values ofy. Naturally,
theaccuracyofthesolutionwilldependuponthestepsize,h.Infact,forEuler’smethod,
theerrorincurredisroughlyproportionaltoh,sothatbyhalvingthestepsizeweroughly
halve the error.
An alternative way of deriving Euler’s method is to use a Taylor series expansion.
Recall from Chapter 18 that
y(x 0
+h) =y(x 0
) +hy ′ (x 0
) + h2
2! y′′ (x 0
)+···
If wetruncate afterthe second termwe find
thatis,
y(x 0
+h) ≈y(x 0
) +hy ′ (x 0
)
y 1
=y 0
+hy ′ (x 0
)=y 0
+hf(x 0
,y 0
)
sothatEuler’smethodisequivalenttotheTaylorseriestruncatedafterthesecondterm.
Example20.4 UseEuler’s methodwithh = 0.25 toobtain a numericalsolution of
dy
dx = −xy2
subject toy(0) = 2, giving approximate values ofyfor 0 x 1. Work throughout
tothreedecimal places and determine the exact solution for comparison.
Solution We need to calculatey 1
,y 2
,y 3
andy 4
. The correspondingxvalues arex 1
= 0.25,x 2
=
0.5,x 3
= 0.75 andx 4
= 1.0.Euler’s method becomes
We find
y i+1
=y i
+0.25(−x i
y 2 i ) with x 0 =0 y 0 =2
y 1
= 2 −0.25(0)(2 2 ) = 2.000
y 2
= 2 −0.25(0.25)(2 2 ) = 1.750
y 3
= 1.750 −0.25(0.5)(1.750 2 ) = 1.367
y 4
= 1.367 −0.25(0.75)(1.367 2 ) = 1.017
The exact solution can be found by separating the variables:
∫ ∫ dy
y = − xdx
2
so that
− 1 y = −x2 2 +C
Imposingy(0) = 2 givesC = − 1 so that
2
Finally,
− 1 y = −x2 2 − 1 2
y = 2
x 2 +1