Higher Engineering Mathematics, Sixth Edition

Numerical methods for first order differential equations 467

corrected value,yC 1 in the improved Euler method is
given by:

yC 1 =y 0 +^12 h[(y′) 0 +f(x 1 ,yP 1 )] (4)

The following worked problems demonstrate how
equations (3) and (4) are used in the Euler-Cauchy
method.

Problem 4. Apply the Euler-Cauchy method to

solve the differential equation

dy

dx

=y−x

in the range 0(0.1)0.5, given the initial conditions

that atx=0,y=2.

dy

dx

=y′=y−x

Since the initial conditions are x 0 =0andy 0 = 2
then (y′) 0 = 2 − 0 =2. Interval h= 0 .1, hence
x 1 =x 0 +h= 0 + 0. 1 = 0 .1.
From equation (3),

yP 1 =y 0 +h(y′) 0 = 2 +( 0. 1 )( 2 )= 2. 2

From equation (4),

yC 1 =y 0 +^12 h[(y′) 0 +f(x 1 ,yP 1 )]

=y 0 +^12 h[(y′) 0 +(yP 1 −x 1 )],

in this case

= 2 +^12 ( 0. 1 )[2+( 2. 2 − 0. 1 )]=2.205

(y′) 1 =yC 1 −x 1 = 2. 205 − 0. 1 = 2. 105

If we produce a table of values, as in Euler’s method,
we have so far determined lines 1 and 2 of Table 49.8.
The results in line 2 are now taken asx 0 ,y 0 and(y′) 0
for the next interval and the process is repeated.

Table 49.8

x y y′

1. 0 2 2


2. 0.1 2.205 2.105


3. 0.2 2.421025 2.221025


4. 0.3 2.649232625 2.349232625


5. 0.4 2.89090205 2.49090205


6. 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

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