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

20.7 Improved Euler method 621
Table20.3
Applying the improved Eulermethod to Example 20.6.
i x i y i f(x i ,y i ) y i +hf(x i ,y i ) f(x i+1 ,y i y i+1
+hf(x i ,y i ))
0 0 2 0 2 −1 1.8750
1 0.25 1.8750 −0.8789 1.6553 −1.3700 1.5939
2 0.5 1.5939 −1.2703 1.2763 −1.2217 1.2824
3 0.75 1.2824 −1.2334 0.9741 −0.9489 1.0096
Table20.4
Comparison of the improved Euler
method with the exact solution.
i x i y i y(x i )
0 0.000 2.0000 2.0000
1 0.2500 1.8750 1.8824
2 0.5000 1.5939 1.6000
3 0.7500 1.2824 1.2800
4 1.0000 1.0096 1.0000
Example20.7 Apply both Euler’s method and the improved Euler method tothe solution of
dy
dx =2x y=1whenx=0
for 0 x 0.5 using h = 0.1. Compare your answers with the analytical solution.
Work throughout tothreedecimal places.
Solution We have dy
dx = 2x,x 0 = 0,y 0
= 1.Therefore, usingEuler’s method we find
y i+1
=y i
+hf(x i
,y i
)
Therefore,
=y i
+0.1(2x i
)
=y i
+0.2x i
y 1
=y 0
+0.2x 0
=1+ (0.2)(0) =1
y 2
=y 1
+0.2x 1
= 1 + (0.2)(0.1) = 1.02
The complete solution appears in Table 20.5. Check the values given in the table for
yourself.Usingthe improved Euler method we have
y i+1
=y i
+ h 2 {f(x i ,y i )+f(x i+1 ,y i +hf(x i ,y i ))}
=y i
+0.05(2x i
+2x i+1
)
=y i
+0.1(x i
+x i+1
)
Table20.5
ThesolutionofExample20.7usingEuler’smethod,
the improved Eulermethod and the exact solution.
x i y i (Euler) y i (improved y(x i )
Euler) (exact)
0 1.000 1.000 1.000
0.1 1.000 1.010 1.010
0.2 1.020 1.040 1.040
0.3 1.060 1.090 1.090
0.4 1.120 1.160 1.160
0.5 1.200 1.250 1.250