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

618 Chapter 20 Ordinary differential equations II
Table20.1
ComparisonofnumericalsolutionbyEuler’s
method with exact solution.
i x i y i y(x i )
numerical exact
0 0.000 2.000 2.000
1 0.250 2.000 1.882
2 0.500 1.750 1.600
3 0.750 1.367 1.280
4 1.000 1.017 1.000
Table20.2
The solution to Example 20.5
byEuler’smethod.
i x i y i
0 1.00 1.00
1 1.20 1.20
2 1.40 1.40
3 1.60 1.60
4 1.80 1.80
5 2.00 2.00
Table 20.1 summarizes the numerical and exact solutions. In this example only one
correct significant figure is obtained. In practice, for most equations a very small step
size is necessary which means that computation is extremely time consuming. An improvement
to Euler’s method, which usually yields more accurate solutions, is given in
Section 20.7.
Example20.5 Obtain a solution forvalues ofxbetween 1 and 2 of
dy
dx = y x
subject toy = 1whenx = 1usingEuler’smethod.Useastepsizeofh = 0.2,working
throughouttotwodecimalplacesofaccuracy.Compareyouranswerwiththeanalytical
solution. Comment upon the approximate and exact solutions.
Solution Here we have f (x,y) = y x , y 0 = 1, x 0
= 1 and h = 0.20. With a step size of 0.2,
x 1
= 1.2,x 2
= 1.4,...,x 5
= 2.0. We need to calculatey 1
,y 2
,...,y 5
. Euler’s method,
y i+1
=y i
+hf (x i
,y i
), reduces to
( )
yi
y i+1
=y i
+0.2
x i
Therefore,
Similarly,
y 1
=y 0
+0.2
y 2
=y 1
+0.2
(
y0
(
y1
x 0
)
( 1
=1+0.2 = 1.20
1)
) ( ) 1.20
= 1.20 +0.2 = 1.40
x 1
1.20
ContinuinginasimilarfashionweobtaintheresultsshowninTable20.2.Toobtainthe
analytical solution we separate the variables togive
∫ ∫ dy dx
y = x
so that
lny=lnx+lnD=lnDx
Therefore,
y=Dx