Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

cos a sin a
constant k or --b = k = ---:--b
'cos
SID
=>
sina
cosa
sinb
cosb
SECTION 17.2 NONHOMOGENEOUS LINEAR EQUATIONS 0 347
=> tan a = tan b => b - a = mr, n any integer. (Note that
none of cos a, cos b, sin a, sin b are zero.) If the lines are both horizontal then cos a = cos b = 0 ::::? b - a = mr, and
similarly vertical lines means sin a = sin b = 0 =>
b- a = mr. Thus the systerit has a unique solution if b - a t= mr.
(b) The linear system has no solution if the lines are parallel but not identical. From part (a) the lines are parallel if
b - a = mr. If the lines are not horizontal, they are identical if ce-a = kde-b ::::?
ce-a =k=~
de-b cosb
c b cos a . sin a .
- = ea- --b. (If d = 0 then c = 0 also.) If they are honzontal then cos b = 0, but k = ---:--b also (and SID b t= 0) so
d ~ ~
. c a bsina h th h I . 'fb d c ..J. a- bcosa .
we reqmre -d = e - ---:--b. T us . e system as no so ullon 1 - a = n1r an -d r e --b unless cos b = 0, m
SID COS '
. h c _J. a- bsina
w h IC case d r e sin b .
(c) The linear system has infinitely many solution if the lines are identical (and necessarily parallel). From part (b) this occurs
b d
. c a-bcosa ni b 0 0 hi
cos
h c a- bsina
sm
when - a = n1r an -d = e --b u ess cos = , m w c case -d = e ---:--b.
=>
17.2 Nonhomogeneous Linear Equations
1. The auxiliary equation is r 2 - 2r - 3 = (r- 3)(r + 1) = 0 ::::? r = 3, r = - 1, so the complementary solution is
Yc(x) = c1e 3 "' + c2e-"'. We try the particular sc;>lution yp(x) = Acos2x + Bsin2x, so
y~ = - 2A sin 2x + 2B cos 2x andy~ = - 4A cos 2x - 4B sin 2x. Substitution into the differential equation gives
( -4Acos2x - 4Bsin2x)- 2( -2Asin2x + 2B~os2x)- 3(Acos2x + Bsin2x) = cos2x =>
(-7A- 4B) cos2x + (4A -7B) sin2x = cos2x. Then -7A - 4B = 1 and 4A - 7B = 0 => A =- :,., and
B = - 6 ~. Thus the general solution is y( x) = Yc( x) + YP ( x) = c1 e 3 "' + c2e - x -
7
65 cos 2x - 6
; sin 2x.
3. The auxiliary equation is r 2 + 9 = 0 with roots r = ± 3i, so the complementary solution is Yc(x) = c1 cos 3x + c 2 sin 3x.
Try the particular solution yp(x) = Ae- 2 "', soy~= - 2Ae- 2 "' andy~ = 4Ae- 2 "' . Substitution into the differential equation
gives 4Ae- 2 "' + 9(Ae- 2 "') = e- 2 "' or 13Ae- 2 "' = e- 2 "'. Thus 13A = 1 ;> A= f3 and the general solution is
5. The auxiliary equation is r 2 - 4r + 5 = 0 with roots r = 2 ± i, so the complementary ·solution is
Yc(x) = e 2 "'(c1 cosx + c2 sin x). Try yp (x) = Ae-"', soy~= - Ae-"' andy~ = Ae-"'. Substitution gives
Ae-:r- 4( -Ae-"') + 5(Ae-x) = e-"' => lOAe-"' = e-"' ::::? A = • 1
Thus the general solution is
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