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

354 0 CHAPTER 17 SECOND-ORDER DIFFERENTIAL EQUATIONS
nc,.. + Cn c,..(n + 1 ) . Cn ~ 0 1 2 ·o fth · d' · · {0) 1 8
Cn+2=(n+ 2
)(n + 1 )=(n+ 2
){n+ 1
)=n+ 2
.orn=, , , .. . . neo eg•vencon•tlons •sy =. ut
y(O) = n~o c,..(O)" = ~ + 0 + 0 + .. · =eo •. so Co= 1. Hence, C2 = ~ = ~· C4 = ~ = /
4 , C6 = ~ = 2 . ~ . 6 , ... ,
C2n = 2
! 1
• The other given condition is y' (0) = 0. But y' {0) = f ncn(O)n-l = c1 + 0 + 0 + · · · = C1, so C1 = 0.
n. n=l
By the recursion relation, ca = ~ = 0 ~ cs = 0, .. . '. C2n+l = 0 for n = 0, 1, 2, . . .. Thus, the solution to the initial-value
/
I
00 00 00 x2n 00 (x2/2)n 2
problem is y(x) = L; c,..xn = L; C2nX 2 n = L; - ,.. 2 1 = L; - - 1 - = e"' / 2 .
n =O . n=O n=O n. n = O n.
00 00 00 00 00
11. Assuming that y(x) = L; cnx", we have xy = x L; c,..xn = L; c,..xn+l, x 2 y' = x 2 L; nc,,xn- l = L; ncnxn+t,
n=O n=O n=O n=l n=O
y"(x)= f: n(n-1)c,..x"- 2 = f: (n+3){n + 2)Cn+ax"+ 1 [replace n with n + 3]
n=2 n=- 1
= 2c2 + f (n + 3){n + 2)Cn+ax"+l,
n =O
00
and the equation y" + x 2 y' + xy = 0 becomes 2c2 + L; [(n + 3)(n + 2)Cn+a + ncn + c,..J xn+l = 0. So c2 = 0 and the
n =O
(n + l)c,..
( )( ) , n = 0, 1, 2, .. . . But Co = y(O) = 0 = c2 and by the
n+3 n+2
recursion relation, Can = can+2 ~ 0 for n = 0, 1, 2, .... Also, c1 = y' (0) = 1, so C4 = - :~~ = - 4
~ 3
,
17 Review
CONCEPT CHECK
1. (a) ay" +by' + cy = 0 where a, b, and care constants.
(c) If the auxiliary equation has two distinct real roots r 1 and r·2, the solution is y = c 1 ert"' + c2er 2 "'. If the roots are real and
equal, the solution is'!/ = c 1 er"' + c2xer"' where r is the common root. If the roots are complex, we can write r1 = a+. i/3
and r2 = a- i/3, and the solution is y = ea"'(c1 cos/3x + c2 sin/3x).
2. (a) An initial-value problem consists of finding a solution y of a second-order differential equation that also satisfies given
conditions y(xo) =Yo andy' (xo) = '!/I. where yo and '!/1 are constants.
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