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470 DIFFERENTIAL EQUATIONS<br />

6. y n+1 = y n + h 6 {k 1 + 2k 2 + 2k 3 + k 4 } and when<br />

n = 0:<br />

y 1 = y 0 + h 6 {k 1 + 2k 2 + 2k 3 + k 4 }<br />

= 2 + 0.1 {2 + 2(2.05) + 2(2.0525)<br />

6<br />

+ 2.10525}<br />

= 2 + 0.1 {12.31025} =2.205171<br />

6<br />

A table of values may be constructed as shown in<br />

Table 49.15. The working has been shown for the<br />

first two rows.<br />

Let n = 1 to determine y 2 :<br />

2. k 1 = f (x 1 , y 1 ) = f (0.1, 2.205171); since<br />

dy<br />

= y − x, f (0.1, 2.205171)<br />

dx<br />

= 2.205171 − 0.1 = 2.105171<br />

(<br />

3. k 2 = f x 1 + h 2 , y 1 + h )<br />

2 k 1<br />

(<br />

= f 0.1 + 0.1<br />

)<br />

0.1<br />

,2.205171 +<br />

2 2 (2.105171)<br />

= f (0.15, 2.31042955)<br />

= 2.31042955 − 0.15 = 2.160430<br />

(<br />

4. k 3 = f x 1 + h 2 , y 1 + h )<br />

2 k 2<br />

(<br />

= f 0.1 + 0.1<br />

)<br />

0.1<br />

,2.205171 +<br />

2 2 (2.160430)<br />

= f (0.15, 2.3131925) = 2.3131925 − 0.15<br />

= 2.163193<br />

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

= f (0.1 + 0.1, 2.205171 + 0.1(2.163193))<br />

= f (0.2, 2.421490)<br />

= 2.421490 − 0.2 = 2.221490<br />

6. y n+1 = y n + h 6 {k 1 + 2k 2 + 2k 3 + k 4 }<br />

and when n = 1:<br />

y 2 = y 1 + h 6 {k 1 + 2k 2 + 2k 3 + k 4 }<br />

= 2.205171+ 0.1<br />

6 {2.105171+2(2.160430)<br />

+ 2(2.163193) + 2.221490}<br />

= 2.205171 + 0.1 {12.973907} =2.421403<br />

6<br />

This completes the third row of Table 49.15. In a<br />

similar manner y 3 , y 4 and y 5 can be calculated and<br />

the results are as shown in Table 49.15. Such a table<br />

is best produced by using a spreadsheet, such as<br />

Microsoft Excel.<br />

This problem is the same as problem 3, page 463<br />

which used Euler’s method, and problem 4, page 465<br />

which used the improved Euler’s method, and a comparison<br />

of results can be made.<br />

The <strong>differential</strong> <strong>equation</strong> dy = y − x may be solved<br />

dx<br />

analytically using the integrating factor method of<br />

chapter 48, with the solution:<br />

y = x + 1 + e x<br />

Substituting values of x of 0, 0.1, 0.2, ..., 0.5 will<br />

give the exact values. A comparison of the results<br />

obtained by Euler’s method, the Euler-Cauchy<br />

method and the Runga-Kutta method, together with<br />

the exact values is shown in Table 49.16 below.<br />

Table 49.15<br />

n x n k 1 k 2 k 3 k 4 y n<br />

0 0 2<br />

1 0.1 2.0 2.05 2.0525 2.10525 2.205171<br />

2 0.2 2.105171 2.160430 2.163193 2.221490 2.421403<br />

3 0.3 2.221403 2.282473 2.285527 2.349956 2.649859<br />

4 0.4 2.349859 2.417339 2.420726 2.491932 2.891824<br />

5 0.5 2.491824 2.566415 2.570145 2.648838 3.148720

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