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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.
Download (direct link): elementarydifferentialequations2001.pdf
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t
integration by parts yields the general solution.
6. The equation must be divided by t so that it is in the
form of Eq.(3): y' + (2/t)y = (sint)/t. Thus
r2dt 2 2
m(t) = exp(J------- = t , and (t y) = tsint. Integration
t
then yields t y = -tcost + sint + c.
t2 4tdt 2 2
7. m(t) = e . 8. m(t) = exp(J---------) = (1+t ) .
1+t2
11. m(t) = et so (ety)' = 5etsin2t. To integrate the right
side you can integrate by parts (twice), use an integral table, or use a symbolic computational software program to find ety = et(sin2t - 2cos2t) + c.
13. m(t) = e-t and y = 2(t-1)e2t + cet. To find the value
for c, set t = 0 in y and equate to 1, the initial value
of y. Thus -2+c = 1 and c = 3, which yields the solution
of the given initial value problem.
15. m(t) = exp(J dt) = t2 and y = t2/4 - t/3 + 1/2 + c/t2.
t
Setting t = 1 and y = 1/2 we have c = 1/12.
18. m(t) = t2. Thus (t2y)' = tsint and
t2y = -tcost + sint + c. Setting t = p/2 and
y = 1 yields c = p2/4 - 1.
20. m(t) = tet.
Section 2.1
7
21b. m(t) = e-t/2 so (e-t/2y) = 2e-t/2cost. Integrating (see comments in #11) and dividing by e-t/2 yields
4 8 4
y(t) = cost + —sint + cet/2. Thus y(0) =-------------------+ c = a,
5 5 5
4 4 8 4 t/2
or c = a + — and y(t) =---------cost + —sint + (a + — )et/2.
5 5 5 5
4
If (a + —) = 0, then the solution is oscillatory for all
5
t, while if (a + —) Ο 0, the solution is unbounded as
5
t ^ •. Thus a
0
21a.
4
5
24a.
24b. m (t) = exp
-cost
y(t) =
4c
f 2dt 2 J — = t2, so
(t2y)' = sint and
t
2
c p
+ — . Setting t =---------yields
t
2
ap2 ap2/4 - cost
— = a or c = ------------ and hence y(t) = ------------------------, which
p2 4 t2
is unbounded as t ^ 0 unless ap 2/4 = 1 or a0 = 4/p2.
2
24c. For a = 4/p2 y(t) =
1 - cost
. To find the limit as
t2
t ^ 0 L'Hopital's Rule must be used:
sint cost 1
limy(t) = lim------- = lim------- = — .
t^0 t^0 2t t^0 2 2
28. (e ty)/ = e t + 3e tsint so
-t
e y = -e
-t -t sint+cost
- 3e (-------------
2
) + c or
3 -1 t
y(t) = -1 - ( )e (sint+cost) + ce .
2
If y(t) is to remain bounded, we must have c = 0. Thus
3 5 5
y(0) = -1 - — + c = y0 or c = y0 + — = 0 and y0 =-----------------.
2 0 0 2 0 2
8
Sect ion
2 . 2
30. m(t) = eat so the D.E. can be written as
(eaty)' = beate-1t = be(a-1)t. If a Ο l, then integration and solution for y yields y = [b/(a-1)]e-1t + ce-at. Then
lim y is zero since both l and a are positive numbers.

If a = l, then the D.E. becomes (eaty)' = b, which yields y = (bt+c)/e1t as the solution. L'Hopital's Rule gives
(bt+c) b
lim y = lim ---------- = lim —— = 0.
t——• t——• e t——• le
32. There is no unique answer for this situation. One
—2t
possible response is to assume y(t) = ce + 3 — t, then y'(t) = —2ce—2t — 1 and thus y' + 2y = 5 — 2t.
35. This problem demonstrates the central idea of the method
of variation of parameters for the simplest case. The solution (ii) of the homogeneous D.E. is extended to the corresponding nonhomogeneous D.E. by replacing the constant A by a function A(t), as shown in (iii).
36. Assume y(t) = A(t)exp(-J (-2)dt) = A(t)e2t.
Differentiating y(t) and substituting into the D.E.
yields A' (t) = t2 since the terms involving A(t) add to zero. Thus A(t) = t3/3 + c, which substituted into y(t) yields the solution.
dt
37. y(t) = A(t)exp(-J-----) = A(t)/t.
t
Section 2.2, Page 45
Problems 1 through 20 follow the pattern of the examples worked in this section. The first eight problems, however, do not have I.C. so the integration constant, c, cannot be found.
2
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