Higher Engineering Mathematics, Sixth Edition

Numerical methods for first order differential equations 469

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

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

=y 0 +^12 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.

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

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

(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

For line 6,x 1 = 2. 0

yP 1 =y 0 +h(y′) 0

= 5. 85212176 + 0. 2 ( 2. 54787824 )

= 6. 361697408

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

= 5. 85212176 +^12 ( 0. 2 )[2. 54787824

+( 3 + 3 ( 2. 0 )− 6. 361697408 )]

=6.370739843

Problem 6. Using the integrating factor

method the solution of the differential equation

dy

dx

= 3 ( 1 +x)−yof Problem 5 isy= 3 x+e^1 −x.

Whenx= 1 .6, compare the accuracy, correct to 3

decimal places, of the Euler and the Euler-Cauchy

methods.

When x= 1 .6, y= 3 x+e^1 −x= 3 ( 1. 6 )+e^1 −^1.^6 =

4. 8 +e−^0.^6 = 5 .348811636.

From Table 49.1, page 463, by Euler’s method, when

x= 1 .6,y= 5. 312

% error in the Euler method

=

(

5. 348811636 − 5. 312

5. 348811636

)

×100%

=0.688%