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

SECTION 17.2 NONHOMOGENEOUS LINEAR EQUATIONS D 349
.Note: Solving Equations (7) and (9) in The Mettiod of Variation of Parameters gives
and
We will use these equations rather than resolving the system in each of the remaining exercises in this section.
19. (a) Here4r 2 + 1 = 0 => r = ±tiand Yc(x) = Clcos(tx) + czsin(~x). Wetryaparticularsolution ofthe form
Yv(x) = Acosx + Bsinx =>
y~ = -Asinx + Bcosx and v;: = - Acosx - Bsin x. Then the equatio~
4y" + y = cosx becomes 4( - Acosx- Bsinx) + (Acosx + Bsinx) = cosx or
-3Acosx- 3Bsinx = cosx => A= - 1, B = 0. Thus, Yv(x) = - % cosx and the general solution is
(b) From (a) we know that Yc(x) = c1 cos~+ cz sin~· Setting Yl =cos~. Yz = sin~, we have
I cosxcos~ 1 "' "'1 ·2x "'
andu 2 =
1
= 2 cos (2 · 2 )cos 2 = 2 (1 -2sill 2 )cos 2 . Then
4 ·2
( ) J (} • X 2 X • "') dx X + 2 3 Z
U 1
d
X = 2 Sill 2 - COS 2 Sill 2 = - COS 2
'j COS 2 an
U2 () X =
l x · 2x
J( x) d · :r. 2 · 3x Tl
2 COS 2 - Sill 2 COS 2
X = SID 2 - ~ SID 2 . lUS
yp(x) = (-cos~ + ~ cos 3 ~)cos~ + (s in~ - i sin 3 ~)sin~ = - ( cos 2 ~ - sin 2 ~) + j ( cos 4 ~ - sin 4 ~)
= - cos {2 · ~) + i ( cos 2 ~ + sin 2 ~ ) ( cos 2 ~ - sin 2 ~ ) = -cos x + j cos x = - i cos x
and the general solution is y(x) = Yc(x) + y 1 ,(x) = c1 c os~ . + Cz sin~- t cos x.
21. (a) r 2 - 2r + 1 = (r -1) 2 = 0 => r· = 1, S? the complementary solution is Yc(x) = c1e'" + czxex. A particular solution
is of the form Yv(x) = Ae 2 "'. Thus 4Ae 2 "'- 4Ae 2 "' + Ae 2 "' = e 2 "' => Ae 2 "' = e 2 :r: => A= 1 => Yv(x) = e 2 "'.
So a general solution is y(x) = Yc(x) + Yv(x) = c1ex + c2xe"' + e 2 "'.
(b) From (a), Yc(x) = c1e"' + czxex, so set Yl = ex, Yz = xex. Then, Y1Y~ - YzY~ = e 2 "' (1 + x) - xe 2 "' = e 2 "' and so
u~ = - xe"' => u1 (x) = - J xe:r: dx = - (x - 1)e"' [by parts] and u~ = e"' => u2(x) = J e"' dx = e"'. Hence
yp (x ) = (1- x)e 2 "' + xe 2 "' = e 2 "' and the general solution is y(x) = Yc.(x) + yp(x) = c 1 e"' + czxe"' + e 2 "'.
23. As in Example 5, Yc(x) = c1 sin x + cz cosx, so set Y1 = sinx, Yz = cos x . Then Y1Y~ - YzYi =-sin 2 x- cos 2 x = - 1,
so ui =
sec 2 x cosx
_ 1
= secx => ul(x) = J sccxdx =ln(secx + tan x)for. O