468 Higher Engineering Mathematics
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−x may be solved analytically by the inte-
grating factor method of Chapter 48 with the solution
y=x+ 1 +ex. Substitutingvalues ofxof 0, 0.1, 0. 2 ,...
give the exact values shown in Table 49.9.
The percentage error for each method for each value of
xis 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%
% error with Euler-Cauchy method
=
(
2. 649858808 − 2. 649232625
2. 649858808
)
×100%
=0.0236%
This calculation and the others listed in Table 49.10
show the Euler-Cauchy method to be more accurate than
the Euler method.
Table 49.10
x Error in Error in
Euler method Euler-Cauchy method
0 0 0
0.1 0.234% 0.00775%
0.2 0.472% 0.0156%
0.3 0.712% 0.0236%
0.4 0.959% 0.0319%
0.5 1.214% 0.0405%
Problem 5. Obtain a numerical solution of the
differential equation
dy
dx
= 3 ( 1 +x)−y
in the range 1.0(0.2)2.0, using the Euler-Cauchy
method, given the initial conditions thatx=1when
y=4.
This is the same as Problem 1 on page 462, and a
comparison of values may be made.
dy
dx
=y′= 3 ( 1 +x)−y i.e.y′= 3 + 3 x−y
x 0 = 1. 0 ,y 0 =4andh= 0. 2
(y′) 0 = 3 + 3 x 0 −y 0 = 3 + 3 ( 1. 0 )− 4 = 2
x 1 = 1 .2 and from equation (3),
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