Higher Engineering Mathematics, Sixth Edition

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

1. 0 2 2 2


2. 0.1 2.2 2.205 2.205170918


3. 0.2 2.41 2.421025 2.421402758


4. 0.3 2.631 2.649232625 2.649858808


5. 0.4 2.8641 2.89090205 2.891824698


6. 0.5 3.11051 3.147446765 3.148721271