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

CHAPTER 17 REVIEW 0 355
(b) A boundary-value problem consists. of finding a solution y of a second-order differential equation that al~o satisfies given
boundary conditions y(xo) =Yo and y(x1) = Yl·
3. (a) ay" +In/ + CIJ = G{x) where a, b, and care constan ts and G is a continuous function.
(b) The complementary equation is the related homogeneous equation ay" +by'+ cy = 0. If we find the general solution Yc
of the complementary equation and YP is any particular solution of the original differential equation, then the general
solution of the original differential equation is y(x) = Yr>(x) + Yc(x).
(c) See Examples 1-5 and the associated discussion in Section 17.2.
(d) See the discussion on pages 11 77- 1179 [ ET 1153-11 55].
4. Second-order linear differential equations can be used to describe the motion of a vibrating spring or to analyze an electric
circuit; see the discussion in Section 17 .3.
5. See Example I and the preceding discus.sion in Section 17 .4..
1. True. See Theorem 17.1.3.
TRUE-FALSE QUIZ
3. True. cosh x and sinh x are linearly independent solutions of this linear homogeneous equation.
EXERCISES
·1. The auxiliary equation is 4r 2 - 1 = 0 => (2r + 1}(2r - 1) = 0 => r = ±t. Then the general solution
3. The auxiliary equation is r 2 + 3 = 0 => r = ±J3 i. Then the general solution is y = c1 cos ( J3 x) + c 2 sin ( J3 x).
5. r 2 - 4r + 5 = 0 => r = 2 ± i, so Yc (x) = e 2 "'{c 1 cosx + c 2 sin x ). Try YP (x) = Ae 2 "' => y~ = 2Ae 2 "'
and y~ = 4Ae 2 "'. Substitution into the differential equation gives 4Ae 2 "' - 8Ae 2 "' + 5Ae 2 "' =-e 2 "' => A = 1 and
the general solution is y( x) = e 2 "' ( c1 cos x + c2 sin x) + e 2 "'.
7. r 2 -21·+ l=O => r =landyc(x) = cle"' + c2xe"'. Tryy 1 ,(x)=(Ax +B)cosx + (Cx+D) sin x =>
y~ = (C- Ax- B) sinx +(A+ Cx +D) cos x andy~= (2C - B- Ax)-cosx + ( -2A- D- Cx) sinx. Substitution
gives (-2Cx + 2C- 2A- 2D)cosx + (2Ax - 2A + 2B - 2C}sinx =xcosx => A= 0, B = C = D = - t.
The general solution is y( x) = c1 e"' + c2 xe"' - ! cos x - !(x + 1) sin x.
9. r 2 - r- 6 = 0 . => r = -2, r = 3 and Yc(x) = c 1e __:_ 2 "' + C2 e 3 "'. For y" - y'- 6y = 1, try Yr>t (x) = A. Then
y~ 1 (x) = y~ 1 (x) = 0 and substitution into the differential equation gives A= -i- For y"- y'- 6y = e- 2 :r try
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