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# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
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Section 5.8
101
0. For the case r1 = 3/2, F(r1+1), which is the
r +1
coefficient of x 1 is Ô 0 so we must set a1 = 0. It follows that a3 = a5 = ... = 0. For the even coefficients, set n = 2m so
2
a2m(3/2) = -a2m-2(3/2)/F(3/2 + 2m) = -a2m-2/2 m(m+3/2),
2
m = 1,2... . Thus a2(3/2) = - a0/2-1(1 + 3/2),
*2
³4(3/2) = a0/°4'n 1 ,ë
a4(3/2) = a0/24-2!(1 + 3/2)(2 + 3/2),..., and
a2m(3/2) = (-1) m/22mm!-(1 + 3/2)...(m + 3/2). Hence one solution is
 , ) m
, x 3/2 r, ^ (-1) ,
y 1 (x) = x [1 + X ----------------------------------------- ( ) ],
1 " m!(1 + 3/2)(2 + 3/2)...(m + 3/2) 2
m=1
where we have set a0 = 1. For this problem, the roots r1
and r2 of the indicial equation differ by an integer:
r1 - r2 = 3/2 - (-3/2) = 3. Hence we can anticipate that
there may be difficulty in calculating a second solution corresponding to r = r2. This difficulty will occur in
calculating a3(r) = - a1(r)/F(r+3) since when r = r2 = -3/2 we have F(r2+3) = F(r1) = 0. However, in this problem we are fortunate because a1 = 0 and it will not be necessary to use the theory described at the end of Section 5.7. Notice for r = r2 = -3/2 that the
coefficient of xr2+1 is [(r2 + 1)2 - 9/4]a1, which does not
vanish unless a1 = 0. Thus the recurrence relation for
the odd coefficients yields a5 = -a3/F(7/2),
a7 = -a5/F(11/2) = a3/F(11/2)F(7/2) and so forth.
Substituting these terms into the assumed form we see
that a multiple of y1(x) has been obtained and thus we
may take a3 = 0 without loss of generality. Hence
a3 = a5 = a7 = ... = 0. The even coefficients are given
by a2m(-3/2) = -a2m-2(-3/2)/F(2m - 3/2), m = 1,2... .
Thus a2 (-3/2) = -a0/2-1- (1 - 3/2),
a4 (-3/2) = a0/24-2!(1 - 3/2)(2 - 3/2),..., and
a2m(-3/2) = (-1) ma0/22mm!(1 - 3/2)(2 - 3/2) ... (m - 3/2).
Thus a second solution is
^ m
, , -3/2 r, ^ (-1) , x 42m,
y2(x) = x [1 + X-------------------------------------------- ( ) ].
2 z-'m!(1 - 3/2)(2 - 3/2) ... (m - 3/2) 2
m=1
102
Section 5.8
7. Apply the ratio test:
³ / 4 m+1 2m+2/,-..2m+2r/ . ³ , -i2 ³
n . l(-1) x /2 [(m+1)!] I , 2 , n . 1 
lim --------------------------------- = |x | lim --------------- = 0
m ^ to |(-1)m x2m/22m(m!)2| m ^ to22(m+1)2
for every x. Thus the series for J0(x) converges absolutely for all x.
12. If S = axe, then dy/dx = x 1/2f + x1/2f'aPxe 1 where f'
denotes df/dS. Find d2y/dx2 in a similar fashion and use algebra to show that f satisfies the D.E.
S2f" + Sf' + [S2 - U2]f = 0.
13. To compare y" - xy = 0 with the D.E. of Problem 12, we
must multiply by x2 to get x2y" - x3y = 0. Thus 2p = 3,
a2p2 = -1 and 1/4 - u2p2 = 0. Hence P = 3/2, a = 2i/3
2
and è = 1/9 which yields the desired result.
14. First we verify that J0(Xjx) satisfies the D.E. We know
that J0(t) is a solution of the Bessel equation of order
zero:
22 t J0(t) + tJ0(t) + t J0(t) = 0 or
0
0(
J^(t) + t 1J/0(t) + J0(t) = 0.
Let t = Xjx. Then
d d dt
 J0(Xjx) =  J0(t) = XjJ0(t)
dx dt dx
2
d d dt 2
 J0(Xjx) = Xj [J0(t)] = XjJ0 (t).
dx2 0 j j dt 0 dx j 0
Substituting y = J0 (Xjx) in the given D.E. and making use
of these results, we have
X2J'0(t) + (Xj/t) XjJ,0(t) + X2J0(t) =
2 -1 X[J0(t) + t 1J0(t) + J0(t)] = 0.
Thus y = J0 (Xjx) is a solution of the given D.E. For the
second part of the problem we follow the hint. First, rewrite the D.E. by multiplying by x to yield
xy" + y + Xjxy = 0, which can be written as (xy')' = -X2xy. Now let yi(x) = J0 (Xħx) and yj(x) =
J0 (Xjx) and we have, respectively: (xyi)' = -X2xyi
(xyj)' = -X2xyj.
Now multiply the first equation by yj, the second by Ó³,
103
CHAPTER 6
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