Higher Engineering Mathematics

NUMERICAL METHODS FOR FIRST ORDER DIFFERENTIAL EQUATIONS 471

I

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

It is seen from Table 49.16 thatthe Runge-Kutta

method is exact, correct to 5 decimal places.

Problem 8. Obtain a numerical solution of

the differential equation:

dy

dx

=3(1+x)−yin

the range 1.0(0.2)2.0, using the Runge-Kutta

method, given the initial conditions thatx=1.0

wheny=4.0

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 tox=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