Higher Engineering Mathematics, Sixth Edition

Numerical methods for first order differential equations 473

Table 49.16

Euler’s Euler-Cauchy Runge-Kutta

method method method Exact value

x y y y y=x+ 1 +ex

0 2 2 2 2

0.1 2.2 2.205 2.205171 2.205170918

0.2 2.41 2.421025 2.421403 2.421402758

0.3 2.631 2.649232625 2.649859 2.649858808

0.4 2.8641 2.89090205 2.891824 2.891824698

0.5 3.11051 3.147446765 3.148720 3.148721271

Using the above procedure:

1. x 0 = 1. 0 ,y 0 = 4 .0 and since h= 0 .2, and the
   range is fromx=1.0 to x=2.0, thenx 1 = 1. 2 ,
   x 2 = 1. 4 ,x 3 = 1. 6 ,x 4 = 1. 8 ,andx 5 = 2. 0

Letn=0 to determiney 1 :

2. k 1 = f(x 0 ,y 0 )=f(^1.^0 ,^4.^0 );since
   dy
   dx

= 3 ( 1 +x)−y,

f( 1. 0 , 4. 0 )= 3 ( 1 + 1. 0 )− 4. 0 =2.0

3. k 2 =f

(

x 0 +

h

2

,y 0 +

h

2

k 1

)

=f

(

1. 0 +

0. 2

2

, 4. 0 +

0. 2

2

( 2 )

)

=f( 1. 1 , 4. 2 )= 3 ( 1 + 1. 1 )− 4. 2 =2.1

4. k 3 =f

(

x 0 +

h

2

,y 0 +

h

2

k 2

)

=f

(

1. 0 +

0. 2

2

, 4. 0 +

0. 2

2

( 2. 1 )

)

=f( 1. 1 , 4. 21 )

= 3 ( 1 + 1. 1 )− 4. 21 =2.09

5. k 4 =f(x 0 +h,y 0 +hk 3 )

=f( 1. 0 + 0. 2 , 4. 1 + 0. 2 ( 2. 09 ))

=f( 1. 2 , 4. 418 )

= 3 ( 1 + 1. 2 )− 4. 418 =2.182

6. yn+ 1 =yn+

h

6

{k 1 + 2 k 2 + 2 k 3 +k 4 } and when

n=0:

y 1 =y 0 +

h

6

{k 1 + 2 k 2 + 2 k 3 +k 4 }

= 4. 0 +

0. 2

6

{ 2. 0 + 2 ( 2. 1 )+ 2 ( 2. 09 )+ 2. 182 }

= 4. 0 +

0. 2

6

{ 12. 562 }=4.418733

A table of values is compiled in Table 49.17. The

working has been shown for the first two rows.

Letn=1 to determiney 2 :

2. k 1 = f(x 1 ,y 1 )=f( 1. 2 , 4. 418733 );since
   dy
   dx

= 3 ( 1 +x)−y, f( 1. 2 , 4. 418733 )

= 3 ( 1 + 1. 2 )− 4. 418733 =2.181267

3. k 2 =f

(

x 1 +

h

2

,y 1 +

h

2

k 1

)

=f

(

1. 2 +

0. 2

2

, 4. 418733 +

0. 2

2

( 2. 181267 )

)

=f( 1. 3 , 4. 636860 )

= 3 ( 1 + 1. 3 )− 4. 636860 =2.263140

4. k 3 = f

(

x 1 +

h

2

,y 1 +

h

2

k 2

)

= f

(

1. 2 +

0. 2

2

, 4. 418733 +

0. 2

2

( 2. 263140 )

)

= f( 1. 3 , 4. 645047 )= 3 ( 1 + 1. 3 )− 4. 645047

=2.254953