Higher Engineering Mathematics, Sixth Edition

474 Higher Engineering Mathematics

Table 49.17

n xn k 1 k 2 k 3 k 4 yn

0 1.0 4.0

1 1.2 2.0 2.1 2.09 2.182 4.418733

2 1.4 2.181267 2.263140 2.254953 2.330276 4.870324

3 1.6 2.329676 2.396708 2.390005 2.451675 5.348817

4 1.8 2.451183 2.506065 2.500577 2.551068 5.849335

5 2.0 2.550665 2.595599 2.591105 2.632444 6.367886

5. k 4 =f(x 1 +h,y 1 +hk 3 )
   =f( 1. 2 + 0. 2 , 4. 418733 + 0. 2 ( 2. 254953 ))
   =f( 1. 4 , 4. 869724 )= 3 ( 1 + 1. 4 )− 4. 869724
   =2.330276


6. yn+ 1 =yn+

h

6

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

n=1:

y 2 =y 1 +

h

6

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

= 4. 418733 +

0. 2

6

{ 2. 181267 + 2 ( 2. 263140 )

+ 2 ( 2. 254953 )+ 2. 330276 }

= 4. 418733 +

0. 2

6

{ 13. 547729 }=4.870324

This completes the third row of Table 49.17. In a sim-

ilar mannery 3 , y 4 and y 5 can be calculated and the

results are as shown in Table 49.17. As in the previ-

ous problem such a table is best produced by using a

spreadsheet.

This problem is the same as Problem 1, page 462 which

used Euler’s method, and Problem 5, page 468 which

used the Euler-Cauchy method, and a comparison of

results can be made.

The differential equation

dy

dx

= 3 ( 1 +x)−y may be

solved analytically using the integrating factor method

of chapter 48, with the solution:

y= 3 x+e^1 −x

Substitutingvalues ofxof 1.0, 1.2, 1. 4 ,..., 2.0 willgive

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 49.18.

It is seen from Table 49.18 thatthe Runge-Kutta

method is exact, correct to 4 decimal places.

Table 49.18

Euler’s Euler-Cauchy Runge-Kutta

method method method Exact value

x y y y y= 3 x+e^1 −x

1.0 4 4 4 4

1.2 4.4 4.42 4.418733 4.418730753

1.4 4.84 4.8724 4.870324 4.870320046

1.6 5.312 5.351368 5.348817 5.348811636

1.8 5.8096 5.85212176 5.849335 5.849328964

2.0 6.32768 6.370739847 6.367886 6.367879441