Higher Engineering Mathematics

NUMERICAL METHODS FOR FIRST ORDER DIFFERENTIAL EQUATIONS 467

I

% 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

00 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= 1

wheny=4.

This is the same as Problem 1 on page 461, 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 =4 andh= 0. 2

(y′) 0 = 3 + 3 x 0 −y 0 = 3 +3(1.0)− 4 = 2

x 1 = 1 .2 and from equation (3),
yP 1 =y 0 +h(y′) 0 = 4 + 0 .2(2)= 4. 4

yC 1 =y 0 + 21 h[(y′) 0 +f(x 1 ,yP 1 )]

=y 0 + 21 h[(y′) 0 +(3+ 3 x 1 −yP 1 )]

= 4 +^12 (0.2)[2+(3+3(1.2)− 4 .4)]

=4.42

(y′) 1 = 3 + 3 x 1 −yP 1 = 3 +3(1.2)− 4. 42 = 2. 18

Thus the first two lines of Table 49.11 have been
completed.

Table 49.11

x 0 y 0 y′ 0

1. 1.0 4 2


2. 1.2 4.42 2.18


3. 1.4 4.8724 2.3276


4. 1.6 5.351368 2.448632


5. 1.8 5.85212176 2.54787824


6. 2.0 6.370739847

For line 3,x 1 = 1. 4

yP 1 =y 0 +h(y′) 0 = 4. 42 + 0 .2(2.18)= 4. 856

yC 1 =y 0 +^12 h[(y′) 0 +(3+ 3 x 1 −yP 1 )]

= 4. 42 +^12 (0.2)[2. 18

+(3+3(1.4)− 4 .856)]

=4.8724

(y′) 1 = 3 + 3 x 1 −yP 1 = 3 +3(1.4)− 4. 8724

= 2. 3276

For line 4,x 1 = 1. 6

yP 1 =y 0 +h(y′) 0 = 4. 8724 + 0 .2(2.3276)

= 5. 33792

yC 1 =y 0 +^12 h[(y′) 0 +(3+ 3 x 1 −yP 1 )]

= 4. 8724 +^12 (0.2)[2. 3276

+(3+3(1.6)− 5 .33792)]

=5.351368

(y′) 1 = 3 + 3 x 1 −yP 1

= 3 +3(1.6)− 5. 351368

= 2. 448632

For line 5,x 1 = 1. 8

yP 1 =y 0 +h(y′) 0 = 5. 351368 + 0 .2(2.448632)

= 5. 8410944

yC 1 =y 0 +^12 h[(y′) 0 +(3+ 3 x 1 −yP 1 )]

= 5. 351368 +^12 (0.2)[2. 448632

+(3+3(1.8)− 5 .8410944)]

=5.85212176

(y′) 1 = 3 + 3 x 1 −yP 1

= 3 +3(1.8)− 5. 85212176

= 2. 54787824