466 DIFFERENTIAL EQUATIONS
Table 49.8
xy y′
- 0 2 2
- 0.1 2.205 2.105
- 0.2 2.421025 2.221025
- 0.3 2.649232625 2.349232625
- 0.4 2.89090205 2.49090205
- 0.5 3.147446765
For line 3,x 1 = 0. 2
yP 1 =y 0 +h(y′) 0 = 2. 205 +(0.1)(2.105)
= 2. 4155
yC 1 =y 0 +^12 h[(y′) 0 +f(x 1 ,yP 1 )]
= 2. 205 +^12 (0.1)[2. 105 +(2. 4155 − 0 .2)]
=2.421025
(y′) 0 =yC 1 −x 1 = 2. 421025 − 0. 2 = 2. 221025
For line 4,x 1 = 0. 3
yP 1 =y 0 +h(y′) 0
= 2. 421025 +(0.1)(2.221025)
= 2. 6431275
yC 1 =y 0 +^12 h[(y′) 0 +f(x 1 ,yP 1 )]
= 2. 421025 +^12 (0.1)[2. 221025
+(2. 6431275 − 0 .3)]
=2.649232625
(y′) 0 =yC 1 −x 1 = 2. 649232625 − 0. 3
= 2. 349232625
For line 5,x 1 = 0. 4
yP 1 =y 0 +h(y′) 0
= 2. 649232625 +(0.1)(2.349232625)
= 2. 884155887
Table 49.9
Euler method Euler-Cauchy method Exact value
xy y y=x+ 1 +ex
- 0 2 2 2
- 0.1 2.2 2.205 2.205170918
- 0.2 2.41 2.421025 2.421402758
- 0.3 2.631 2.649232625 2.649858808
- 0.4 2.8641 2.89090205 2.891824698
- 0.5 3.11051 3.147446765 3.148721271
yC 1 =y 0 +^12 h[(y′) 0 +f(x 1 ,yP 1 )]
= 2. 649232625 +^12 (0.1)[2. 349232625
+(2. 884155887 − 0 .4)]
=2.89090205
(y′) 0 =yC 1 −x 1 = 2. 89090205 − 0. 4
= 2. 49090205
For line 6,x 1 = 0. 5
yP 1 =y 0 +h(y′) 0
= 2. 89090205 +(0.1)(2.49090205)
= 3. 139992255
yC 1 =y 0 +^12 h[(y′) 0 +f(x 1 ,yP 1 )]
= 2. 89090205 +^12 (0.1)[2. 49090205
+(3. 139992255 − 0 .5)]
=3.147446765
Problem 4 is the same example as Problem 3 and
Table 49.9 shows a comparison of the results, i.e.
it compares the results of Tables 49.3 and 49.8.
dy
dx
=y−xmay be solved analytically by the inte-
grating factor method of Chapter 48 with the solution
y=x+ 1 +ex. Substituting values ofxof 0, 0.1,
0 .2,...give the exact values shown in Table 49.9.
The percentage error for each method for each
value ofxis shown in Table 49.10. For example
whenx= 0 .3,
% error with Euler method
=
(
actual−estimated
actual
)
×100%
=
(
2. 649858808 − 2. 631
2. 649858808
)
×100%
=0.712%