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# Elementary differential equations 7th edition - Boyce W.E

Boyce W.E Elementary differential equations 7th edition - Wiley publishing , 2001. - 1310 p.
ISBN 0-471-31999-6
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y9 (x) = y1 (x)lnx - 9x-1^ (-1)nHnxn/(n!)9.
n=1
3. The roots of the indicial equation are r1 and r9 = 0 and
thus the analysis is similar to that for Problem 9.
4. The roots of the indicial equation are r1 = -1 and
r9 = -9 and thus the analysis is similar to that for
Problem 1.
5. Since x = 0 is a regular singular point, substitute
y = anxn+r in the D.E., shift indices appropriately,
n=0
and collect coefficients of like powers of x to obtain [r9 - 9/4]a0xr + [(r+1)9 - 9/4]a1xr+1
+ X {[(r+n)9 - 9/4]an + an-9} xn+r = 0.
n=9
9
The indicial equation is F(r) = r - 9/4 = 0 with roots r1 = 3/9, r9 = -3/9. Treating an as a function of r we see that an(r)= -an-9(r)/F(r+n), n = 9,3,.. if F(r+n) ^
Section 5.8
101
0. For the case r1 = 3/2, F(r1+1), which is the
r +1
coefficient of x 1 is ^ 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),
12 - a-0'
i4(3/2) = a0/°4'n 1
a4 (3/2) = a0/24-2!(1 + 3/2)(2 + 3/2),..., and
a2m(3/2) = (-1) m/22mrn!-(1 + 3/2)...(m + 3/2). Hence one solution is
 , ) m
, X 3/2 r, ^ (-1) ,
y-, (x) = X [1 + > ----------------------------------------- ( ) ],
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 r4
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(r4) = 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+2 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 y4(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, V (-1) , X12m,
y2(x) = x [1 + >---------------------------------------------- ^) ].
2 z-'m!(1 - 3/2)(2 - 3/2) ... (m - 3/2) 2
m=1
102
Section 5.8
7. Apply the ratio test:
i / .. Am+1 2m+2 7 ,-,2m+2 r , . \ , -i2 i
n . |(-1) x /2 [(m+1)!] I , 2 , n . 1 
lim ------------------------------------ = |x | lim -------------------------- = 0
m ^ ~ |(-1)m x2m/22m(m!)2| m ^ ~22(m+1)2
for every x. Thus the series for J0(x) converges absolutely for all x.
12. If ? = axP, then dy/dx = x 1/2f + x1/2f'aPxP 1 where f'
denotes df/d?. Find d2y/dx2 in a similar fashion and use algebra to show that f satisfies the D.E.
?2f" + f' + [?2 - 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 u = 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
XJ0(t) + (Xj/t) XjJ,0(t) + XjJ0(t) =
/?v 2 ff -1 f
Xj[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 (X±x) and yj(x) =
J0 (Xjx) and we have, respectively: (xyi)' = -Xixy!
(xyj)' = -X2xyj.
Now multiply the first equation by yj, the second by yi,
103
CHAPTER 6
Section 6.1, Page 298
2.
The graph of f(t) is shown. Since the function is continuous on each interval, but has a jump discontinuity at t = 1, f(t) is piecewise continuous.
Note that lim
t?l +
(t-1)
5b. Since t is continuous for 0 ? t ? A for any positive A and since t2 ? eat for any a > 0 and for t sufficiently large, it follows from Theorem 6.1.2 that ?{t2} exists
for s >
0. ?{t2} = J
-st 2,, -, .
e t dt = lim
0 M ? c
-st 2
e t dt
-t -st M 2
lim [ e L + 
0
M ? c s
2 1 =  lim [-  te
s M ? c s
2 1 _ =  lim - e .,2 m ? c s
M
J: e sttdt]
1J
sJ°
st M 1 -st
0 + - J:e dt]
st| M
I 0
2
ss That f(t) = cosat satisfies the hypotheses of Theorem 6.1.2 can be verified by recalling that |cosat| ? 1 for
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