Thomas Calculus 13th [Solutions]

Section 17.1 Second-Order Linear Equations 1007
αx αx αx
63. Let r1 œ α i " and r2
œ α i " be complex roots. If e cos " x and e sin " x are linearly independent, then e cos " x
is not a constant multiple of e α xsin x (and vice versa). Assume that e α xcos x is a constant multiple of e α x
" " sin " x, then
for some nonzero constant c, e α xcos x c e α xsin x e α xcos x c e α xsin x 0 e α x
" œ " Ê " " œ Ê cos " x c sin " x
œ 0
αx
αx
Ê e œ 0 or cos " x c sin " x œ 0. Since e Á 0 Ê c œ cot " x, thus c is not a constant, which is a contridiction.
αx
αx
Thus e cos " x and e sin " x are linearly independent.
ww w w w w
1 2 3 1 2 3 1 2
64. Let y and y be linearly independent solutions of Pxy Qxy Rxy œ 0. Let y œ y y Ê y œ y y
ww ww ww ww w ww ww w w
Ê y3 œ y1 y 2. Then Pxy3 Qxy3
Rxy3 œ Pxy1 y2Qxy1 y2Rxy1 y2
ww ww w w
œ Pxy1 Pxy2 Qxy1 Qxy2 Rxy1 Rxy2
ww w ww w w w w
œ Pxy1 Qxy1 Rxy1 Pxy2 Qxy2 Rxy2 œ 00 œ 0. Let y4 œ y1 y2 Ê y4
œ y1 y2
ww ww ww ww w ww ww w w
Ê y4 œ y1 y 2. Then Pxy4
Qxy4 Rxy4 œ Pxy1 y2Qxy1 y2Rxy1 y2
ww ww w w
œ Pxy1 Pxy2 Qxy1 Qxy2 Rxy1 Rxy2
ww w ww w
œ Pxy1 Qxy1 Rxy1 cPxy2 Qxy2
Rxy2d
œ 0 0 œ 0. Thus y 3 and y 4 are both solutions.
Suppose that y 3 is a constant multiple of y 4, then there is a nonzero constant c such that y3 œ c y 4.
Ê y y œ c y y . If we solve this equation for y we obtain y œ 1
c y Ê y is a constant multiple of y ,
1 2 1 2 1 1 1 c 2 "
2
which is a contradiction since y 1 and y 2 are linearly independent Ê y1 y 2 and y1 y 2 are linearly independent.
ww 2 0†
x
1
65. (a) y 4y œ 0, y0 œ 0, y œ 1 Ê r 4 œ 0 Ê r œ !„ 2i Ê y œ e c1cos 2x c2sin 2x
Ê y œ c1cos 2x c2sin 2x; y0
œ 0 Ê c1 œ 0, and y1
œ 1 Ê c1
œ 1 Ê no solution
ww 2 0†
x
(b) y 4y œ 0, y0 œ 0, y1 œ 0 Ê r 4 œ 0 Ê r œ !„ 2i Ê y œ e c1cos 2x c2sin 2x
Ê y œ c1cos 2x c2sin 2x; y0
œ 0 Ê c1 œ 0, and y1
œ 0 Ê c1 œ 0 Ê c1 œ 0, c 2 can be any real number
Ê y œ c 2 sin2x
b„ Èb 4ac b Èb 4ac
2a 2a 2a
# #
66. Let a, b, and c be positive constants, then r œ œ „ . There are three cases to consider.
#
Case I: Two distinct real solutions Ê b 4ac 0. Since a, b, and c are positive Ê 4ac 0 and b 0
Ê 4ac 0 and b 0 Ê 0 b # 4ac b # Ê 0 Èb # 4ac b Ê b Èb # 4ac
!
È # b„ Èb 4ac bÈb 4ac
2a
1 2a
# #
and b b 4ac !. Since 2a 0 Ê 0. Thus r œ 0 and
È #
b
b 4ac
1 2
r2 œ
2a
0. The general solution to the differential equation is y œ c1e c2e .
rx 1 rx 2
lim ce ce œ 00œ
0.
1 2
xÄ∞ #
Case II: One repeated real solution Ê b 4ac œ 0. Since a, b, and c are positive Ê # a 0 and b 0
b
Ê r œ 2a 0. The general solution to the differential equation is y œ c1e c2x e . Since
r 0Ê r 0. lim rx rx
ce cxe œ lim ˆ rx cx 2 rx
ce ‰ cx
œ lim ce lim ˆ 2 ‰
xÄ∞ 1 2
xÄ∞ 1 e rx
xÄ∞ 1
xÄ∞
e rx
œ 0 lim ˆ c ‰ œ 0, using L'hopital's rule to evaluate the limit of the second expression.
xÄ∞ re
2 rx
# #
Case III: Two complex, nonreal solutions Ê b 4ac 0 Ê 4ac b 0. Since a, b, and c are positive Ê # a 0 and
b b È4acb b È4acb
b
2a 1 2a 2a 2 2a 2a 2a
# #
b 0 Ê 0. The roots are r œ i and r œ i, thus α œ 0 and
È 4ac b
" œ 0. The general solution to the differential equation is y œ e α #
2a
c"
cos " x c2sin " x . Since
1 Ÿ sin " x Ÿ 1 and 1 Ÿ cos " x Ÿ 1 Ê kc# k Ÿ c# sin " x Ÿ kc # k and kc" k Ÿ c" cos " x Ÿ kc"
k
Ê kc" kkc# k Ÿ c" cos" xc# sin"
x Ÿ kc" kkc#
k
αx αx αx
Ê e kc" kkc# k Ÿ e c" cos " x c# sin " x Ÿ e kc" kkc#
k
lim ce αx kc kkc kd œ kc kkc klim e αx œ 0 and lim ce αx kc kkc kd œ kc kkc klim
e αx
œ 0
" # " # " # " #
xÄ∞ xÄ∞ xÄ∞ xÄ∞
α x
thus by the Sandwich theorem, lim ce c" cos " x c#
sin " xd
œ 0.
xÄ∞
x
rx
rx
rx
rx
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