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
- 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 - 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