Books in black and white
 Books Biology Business Chemistry Computers Culture Economics Fiction Games Guide History Management Mathematical Medicine Mental Fitnes Physics Psychology Scince Sport Technics

# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
Previous << 1 .. 417 418 419 420 421 422 < 423 > 424 425 426 427 428 429 .. 609 >> Next

= (.5) Ï(ó0-12) + 12. Mathematical induction can now be used to prove that this is the correct solution.
The governing equation is óÏ+1 = ðóÏ-Ü, which has the Ï 1-ðÏ
solution óÏ = ð y0 - ---------b (Eq.(14) with a negative b).
1-P
Setting y360 = 0 and solving for b we obtain
~\~360
(1-ð)ð Ó0
b = -------------, where ð = 1.0075 for part a.
1-ð360
You must solve Eq.(14) numerically for ð when n = 240, Ó240 = 0, b = -\$900 and y0 = \$95,000.
ð-1
Substituting un = ----- + vn into Eq.(21) we get
n ð n ð-1 ð-1 ð-1
---- + vn+1 = ð(---- + vn)(1 - - - vn) or
ð ð ð
ð-1 1
vn+1 = - --- + (ð-1 + PVÏ)( - - vn)
ð ð
1-ð ð-1 2 2
= ----- + --- - ^-1)vn + vn- PVÏ = (2-P)VÏ - ð^-
ð ð
15a. For u0 = .2 we have u1 = 3.2u0(1-u0) = .512 and u2 = 3.2u1(1-u1) = .7995392. Likewise u3 = .51288406, u4 = .7994688, u5 = .51301899, u6 = .7994576 and u7 = .5130404. Continuing in this fashion, u14 = u16 = .79945549 and u15 = u17 = .51304451.
For both parts of this problem a computer spreadsheet was used and an initial value of u0 = .2 was chosen. Different initial values or different computer programs may need a slightly different number of iterations to reach the limiting value.
Miscellaneous Problems
33
17a.The limiting value of .65517 (to 5 decimal places) is
reached after approximately 100 iterations for p = 2.9. The limiting value of .66102 (to 5 decimal places) is reached after approximately 200 iterations for p = 2.95. The limiting value of .66555 (to 5 decimal places) is reached after approximately 910 iterations for p = 2.99.
17b. The solution oscillates between .63285 and .69938 after approximately 400 iterations for p = 3.01. The solution oscillates between.59016 and .73770 after approximately 130 iterations for p = 3.05. The solution oscillates between .55801 and .76457 after approximately 30 iterations for p = 3.1. For each of these cases additional iterations verified the oscillations were correct to five decimal places.
18. For an initial value of .2 and p = 3.448 we have the solution oscillating between .4403086 and .8497146.
After approximately 3570 iterations the eighth decimal place is still not fixed, though. For the same initial value and p = 3.45 the solution oscillates between the four values: .43399155, .84746795, .44596778 and
.85242779 after 3700 iterations.. For p = 3.449, the solution is still varying in the fourth decimal place after 3570 iterations, but there appear to be four values.
Miscellaneous Problems, Page 126
Before trying to find the solution of a D.E. it is necessary to know its type. The student should first classify the D.E. before reading this section, which indentifies the type of each equation in Problems 1 through 32.
1. Linear 2. Homogeneous
3. Exact 4. Linear equation in x(y)
5. Exact 6. Linear
2 dy dy
7. Letting u = x yields --------- = 2x— and thus
dx du
du 3
— - 2yu = 2y which is linear in u(y). dy
34 Miscellaneous Problems
8. Linear 9. Exact
10. Integrating factor 11. Exact
depends on x only
12. Linear 13. Homogeneous
14. Exact or homogeneous 15. Separable
16. Homogeneous 17. Linear
18. Linear or homogeneous 19. Integrating factor
depends on x only
20. Separable 21. Homogeneous
22. Separable 23. Bernoulli equation
24. Separable 25. Exact
26. Integrating factor 27. Integrating factor
depends on x only depends on x only
28. Exact 29. Homogeneous
30. Linear equation in x(y) 31. Separable
32. Integrating factor depends on ó only.
35
CHAPTER 3
Section 3.1, Page 136
•ë -Ë rt , . , / rt J // 2 rt
3. Assume y = e , which gives y = re and y = re .
2 rt
Substitution into the D.E. yields (6r -r-1)e = 0.
rt
Since e Ï0, we have the characteristic equation
2
6r2-r-1 = 0, or (3r+1)(2r-1) = 0. Thus r = -1/3, 1/2 and y = c1et/2 + c2e-t/3.
2
5. The characteristic equation is r + 5r = 0, so the roots
are r1 = 0, and r2 = -5. Thus
Previous << 1 .. 417 418 419 420 421 422 < 423 > 424 425 426 427 428 429 .. 609 >> Next