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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
Download (direct link): elementarydifferentialequat2001.pdf
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n—2
3 4 5
+ 20a5x + 30a6x + 42a7x + ...
3 . 5 . 2 3 4
+ (x-x /3!+x /5!-...)(a0+a!x+a2x +a3x +a4x +...)
23
— 2a2 + (6a3+a0)x + (12a4+a1)x + (20a5+a2-a0/6)x +
45
(30a6+ a3-a1/6)x + (42a7+a4+a0/5!)x + ... — 0. Hence
a2 — 0, a3 — -a0/6, a4 — -a^/12, a5 — a0/120, a6 — (a1+a0)/180, a7 = -a0/7! + a1/504, ... . We set
a0 — 1 and a1 — 0 and obtain
y1(x) — (1 - x3/6 + x5/120 + x6/180 + ...). Next we set
a0 — 0 and a1 — 1 and obtain
4 6 7
y2(x) — (x - x /12 + x /180 + x /504 +...). Since
82
Section 5.3
p(x) = 1 and q(x) = sinx both have p = ro, the solution in this case converges for all x, that is, p = ro
x 2 3
18. We know that e = 1 + x + x/2! + X/3! + ..., and
x2 2 4 . 6 .
therefore e = 1 + x + x/2! + x/3! + ... . Hence, if
y = Xanxn, we have
2 2 4 2
a! + 2a2x + 3a3x + ... = (1+x + x /2+...)(a0+a1x+a2x +...'
2
= a0 + a1x + (a0+a2)x + ...:
Thus, a1 = a0 2a2 = a1 and 3a3 = a0 + a2, which yield the desired solution.
ro
20. Substituting y = anxn into the D.E. we obtain
n=2
n-1 n 2
> nanx - > anx = x . Shifting indices in the summation
n2
nx
n=2 n=2
n2
yields > [(n+1)an+1 - an]x = x . Equating coefficients of
n=2
both sides then gives: a1 - a0 = 0, 2a2 - a1 = 0, 3a3 - a2 = 1
and (n+1)an+1 = an for n = 3,4,... . Thus a1 = a0,
a2 = a1/2 = a0/2, a3 = 1/3 + a2/3 = 1/3 + a0/2-3,
a4 = a3/4 = 1/3-4 + a0/2-3-4, ..., an = an-1/n = 2/n! + a0/n! and
hence
2 n 3 4 n
x x x x x
y(x) = a0(1 + x + — + ...+— ...) + 2( +— + ...+— + ...).
0 2! n! 3! 4! n!
Using the power series for ex, the first and second sums
x x 2
can be rewritten as a0e + 2(e - 1 - x - x /2).
22. Substituting y = anxn into the Legendre equation,
n=2
shifting indices, and collecting coefficients of like powers of x yields
[2-1-a2 + a(a+1)a0]x° + {3-2-a3 - [2-1 - a(a+1)]a1}x1 +
ro
{(n+2)(n+1)an+2 - [n(n+1) - a(a+1)]an}xn = 0. Thus
n=2
a2 = -a (a+1)a0/2!, a3 = [2-1 - a(a+1)]a1/3! = -(a-1)(a+2)a1/3! and the recurrence relation is
Section 5.4
83
(n+2)(n+1)an+2 = -[a(a+1) - n(n+1)]an = -(a-n)(a+n+1)an,
n = 2,3,... . Setting a1 = 0, a0 = 1 yields a solution
with a3 = a5 = a7 = ... = 0 and
a4 = a(a-2)(a+1)(a+3)/4!,..., a2m = (-1) ma(a-2)(a-4) ...
(a-2m+2)(a+1)(a+3) ... (a+2m-1)/(2m)!,... . The second
linearly independent solution is obtained by setting a0 = 0 and a1 = 1. The coefficients are a2 = a4 = a6 =
... = 0 and a3 = -(a-1)(a+2)/3!, a5 = -(a-3)(a+4)a3/5-4 = (a-1)(a-3)(a+2)(a+4)/5!,... .
26. Using the chain rule we have:
dF(^) dF[$(x)] dx ? ,w , 4T/, 2
= ----------------- — = -f (x)sin6(x) = - f (x)V1-x ,
d^ dx d^
2 ,______________________
d F(6) d , ./ 2 dx 2 ,
-------- = — [-f (x)V1-x ] — = (1-x )f (x) - xf (x),
d^2 dx d^
which when substituted into the D.E. yields the desired
result.
22 28. Since [(1-x )y']' = (1-x )y" - 2xy', the Legendre
Equation, from Problem 22, can be written as shown.
Thus, carrying out the steps indicated yields the two
equations:
2
Pm[(1-x )Pn] = -n(n+1)pnpm 2
Pn[(1-x )Pm] = -m(m+1)PnPm.
As long as n ^ m the second equation can be subtracted from the first and the result integrated from -1 to 1 to obtain
1 2 2 1 I {Pm[(1-x )Pn] -Pn[(1-x )Pm] }dx = [m(m+1)-n(n+1)]I ^PnPmdx
The left side may be integrated by parts to yield
2 f 2 ' 1 f 1 ' 2 f ' 2 '
[Pm(1-x )Pn - Pn(1-x )Pm]-1 + J_1[Pm (1-x )Pn - Pn(1-x )Pm]dX, which is zero. Thus J1 Pn(x)Pm(x)dx = 0 for n ^ m.
Section 5.4, Page 259
1. Since the coefficients of y, y' and y" have no common
factors and since P(x) vanishes only at x = 0 we conclude that x = 0 is a singular point. Writing the D.E. in the form y" + p(x)y' + q(x)y = 0, we obtain p(x) = (1-x)/x and q(x) = 1. Thus for the singular point we have
84
Section 5.4
12.
17.
2
lim x p(x) = lim 1-x = 1, lim x q(x) = 0 and thus x = 0
x—0 x—0 x—0
is a regular singular point.
Writing the D.E. in the form y" + p(x)y' + q(x)y = 0, we
22 find p(x) = x/(1-x)(1+x) and q(x) = 1/(1-x )(1+x).
Therefore x = ±1 are singular points. Since
lim (x-1)p(x) and lim (x-1) q(x) both exist, we conclude
x—1 x—1
x = 1 is a regular singular point. Finally, since lim (x+1)p(x) does not exist, we find that x = -1 is an
x—— -1
irregular singular point.
Writing the D.E. in the form y + p(x)y' + q(x)y = 0, we
see that p(x) = ex/x and q(x) = (3cosx)/x. Thus x = 0 is
a singular point. Since xp(x) = ex is analytic at x = 0 2
and x q(x) = 3xcosx is analytic at x = 0 the point x = 0 is a regular singular point.
Writing the D.E. in the form y" + p(x)y' + q(x)y = 0, we
x4
see that p(x) = ----- and q(x) = -----. Since lim q(x)
sinx sinx x—0
does not exist, the point x0 = 0 is a singular point and since neither lim p(x) nor lim q(x) exist either the
x—±nn x—±nn
points x0 = ±nn are also singular points. To determine whether the singular points are regular or irregular we must use Eq.(8) and the result #7 of multiplication and division of power series from Section 5.1. For x0 = 0, we have
2 2 2 x x x
xp(x) = ----- = = x[1 + — + ...]
sinx x3 6
x- — +...
6
3
x
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