Higher Engineering Mathematics

470 DIFFERENTIAL EQUATIONS

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 }

= 2 +

0. 1

6

{ 2 +2(2.05)+2(2.0525)

+ 2. 10525 }

= 2 +

0. 1

6

{ 12. 31025 }=2.205171

A table of values may be constructed as shown in
Table 49.15. The working has been shown for the
first two rows.

Letn=1 to determiney 2 :

2. k 1 =f(x 1 ,y 1 )=f(0.1, 2.205171); since

dy

dx

=y−x,f(0.1, 2.205171)

= 2. 205171 − 0. 1 =2.105171

3.k 2 =f

(

x 1 +

h

2

,y 1 +

h

2

k 1

)

=f

(

0. 1 +

0. 1

2

,2. 205171 +

0. 1

2

(2.105171)

)

=f(0.15, 2.31042955)

= 2. 31042955 − 0. 15 =2.160430

4.k 3 =f

(

x 1 +

h

2

,y 1 +

h

2

k 2

)

=f

(

0. 1 +

0. 1

2

,2. 205171 +

0. 1

2

(2.160430)

)

=f( 0 .15, 2. 3131925 )= 2. 3131925 − 0. 15

=2.163193

Table 49.15

n xn k 1 k 2 k 3 k 4 yn

0 0 2

1 0.1 2.0 2.05 2.0525 2.10525 2.205171

2 0.2 2.105171 2.160430 2.163193 2.221490 2.421403

3 0.3 2.221403 2.282473 2.285527 2.349956 2.649859

4 0.4 2.349859 2.417339 2.420726 2.491932 2.891824

5 0.5 2.491824 2.566415 2.570145 2.648838 3.148720

5.k 4 =f(x 1 +h,y 1 +hk 3 )

=f(0. 1 + 0 .1, 2. 205171 + 0 .1(2.163193))

=f(0.2, 2.421490)

= 2. 421490 − 0. 2 =2.221490

6 .yn+ 1 =yn+

h

6

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

and whenn=1:

y 2 =y 1 +

h

6

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

= 2. 205171 +

0. 1

6

{ 2. 105171 +2(2.160430)

+2(2.163193)+ 2. 221490 }

= 2. 205171 +

0. 1

6

{ 12. 973907 }=2.421403

This completes the third row of Table 49.15. In a

similar mannery 3 ,y 4 andy 5 can be calculated and

the results are as shown in Table 49.15. Such a table

is best produced by using aspreadsheet, such as

Microsoft Excel.

This problem is the same as problem 3, page 463

which used Euler’s method, and problem 4, page 465

which used the improved Euler’s method, and a com-

parison of results can be made.

The differential equation

dy

dx

=y−xmay be solved

analytically using the integrating factor method of

chapter 48, with the solution:

y=x+ 1 +ex

Substituting values ofxof 0, 0.1, 0.2,..., 0.5 will

give the exact values. A comparison of the results

obtained by Euler’s method, the Euler-Cauchy

method and the Runga-Kutta method, together with

the exact values is shown in Table 49.16 below.