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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.
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0t -5t -5t y = c1e + c2e = c1 + c2e .
2
7. The characteristic equation is r - 9r + 9 = 0 so that
r = (9±\J 81-36 )/2 = (9±3ä/"5)/2 using the quadratic
formula. Hence
y = c1exp[(9 + 3ä/5 )t/2] + c2exp[(9-3^/5 )t/2].
rt
10. Substituting y = e in the D.E.
we obtain the characteristic
2
equation r + 4r + 3 = 0, which has the roots r1 = -1, r2 = -3.
-t -3t Thus y = c1e + c2e and
-t -3t / ë
y = -c1e - 3c2e .
Substituting t = 0 we then have c1 + c2 = 2 and -c1 - 3c2 = -1,
yielding c1 = 5/2 and
5 -t 1 -3t
c2 = -1/2. Thus y = —e - —e
2 2 2
and hence y ^ 0 as t ^ •.
2
15. The characteristic equation is r + 8r - 9 = 0, so that r1 = 1 and r2 = -9 and the general solution is
t -9t
y = c1e + c2e . Since the I.C. are given at t = 1, it
is convenient to write the general solution in the form
(t-1) -9(t-1)
y = k1e + k2e . Note that
-1 9
c1 = k1e and c2 = k2e . The advantage of the latter form of the general solution becomes clear when we apply
36
Section 3 . 1
is y(t ) = ce + c e .
1 2
3
c - c = - --- , yielding
1 2 4
(t) = 1t - t
--- e - e = 0 or
4
the I.C. y(1) = 1 and y'(1) = 0. This latter form of y
(t-1) -9(t-1)
gives y = k1e - 9k2e and thus setting t = 1 in
y and y' yields the equations k1 + k2 = 1 and
k1 - 9k2 = 0. Solving for k1 and k2 we find that
(t-1) -9(t-1) (t-1)
y = (9e + e )/10. Since e has a positive
exponent for t > 1, y ^ • as t ^ • .
17. Comparing the given solution to Eq(17), we see that r = 2
1
and r = -3 are the two roots of the characteristic 2
2
equation. Thus we have (r-2)(r+3) = 0, or r + r - 6 = 0
as the characteristic equation. Hence the given solution
is for the D.E. y" + y' - 6y = 0.
19. The roots of the characteristic equation are r = 1, -1
t -t
5
y(0) = c + c = — and y'(0
124
1 t -t
y(t) = —e + e . From this
4 2t
e = 4 or t = ln2. The second derivative test or a graph of the solution indicates this is a minimum point.
-t 2t
21. The general solution is y = c1e + c2e . Using the I.C.
we obtain c1 + c2 = a and -c1 + 2c2 = 2, so adding the two
equations we find 3c2 = a + 2. If y is to approach zero
as t ^ •, c2 must be zero. Thus a = -2.
24. The roots of the characteristic equation are given by
-2t (a-1)t
r = -2, a - 1 and thus y(t) = c e + c e . Hence,
12
for a < 1, all solutions tend to zero as t ^ •. For a > 1, the second term becomes unbounded, but not the first, so there are no values of a for which all solutions become unbounded.
2
25a. The characteristic equation is 2r + 3r - 2 = 0, so
r1 = -2 and r2 = 1/2 and y = c1e 2t + c2et/2. The I.C.
1
yield c1 + c2 = 2 and -2c1 + —c2 = -p so that c1 = (1 + 2b)/5 and c2 = (4-2b)/5.
Section 3.1
37
25c.
27.
28.
30.
34.
37.
-2t
From part (a), if b = 2 then y(t) = e and the solution simply decays to zero. For b > 2, the solution becomes unbounded negatively, and again there is no minimum point.
-t
The second solution must decay faster than e , so choose -2t -3t
e or e etc. as the second solution. Then proceed as in Problem 17.
Let v = y', then v' = y" and thus the D.E. becomes 22 tv' + 2tv - 1 = 0 or tv' + 2tv = 1. The left side is 2
recognized as (t v)' and thus we may integrate to obtain 2
tv = t + c (otherwise, divide both sides of the D.E. by 22 t and find the integrating factor, which is just t in
this case). Solving for v = dy/dt we find
2
dy/dt = 1/t + c/t so that y = lnt + c /t + c .
1 2
Set v = y', then v' = y" and thus the D.E. becomes
v' + tv2 = 0. This equation is separable and has the -1 2 2
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