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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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... = 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 ¦ æ, , .ã/, èÃ 2
------ = ---------- — = -f (x)sinô(x) = - f (x) v 1-x ,
dô dx dô
2 ,_______________________
d F(A) d , / 2 dx 2 ,
------ = -- [-f (x^ 1-x ] ---- = (1-x )f (x) - xf (x),
dô2 dx dô
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 Ô m the second equation can be subtracted from the first and the result integrated from -1 to 1 to obtain
1 2 2 1 J {Pm[(1-x )Pn] -Pn[(1-x )Pm] }dx = [m(m+1)-n(n+1)]J ^PnPmdx
The left side may be integrated by parts to yield
2 f 2 ' 1 ¥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 Ô 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
= x + — + ...,
6
which converges about x0 = 0 and thus xp(x) is analytic 2
at x0 = 0. xq(x), by similar steps, is also analytic at x0 = 0 and thus x0 = 0 is a regular singular point. For x0 = nn, we have
(x-nn)x (x-nn)[(x-nn) + nn]
(x-nn)p(x) =
sinx - (x-nn) 3
±(x-nn)+----------±
6
Section 5.4
85
19.
21.
(x-nn) 2
= [(x-nn)+nn][±1 ± ------------ ± ...], which
6
converges about x0 = nn and thus (x-nn)p(x) is analytic
at x= nn. Similarly (x+nn)p(x) and (x±nn) q(x) are analytic and thus x0 = ±nn are regular singular points.
Substituting y = anxn into the D.E. yields
n=0
2~X n(n-1)anxn 1 + 3^ nanxn 1 + anxn+1 = 0. The last sum
n=2 n=1 n=0
becomes Xan-2xn_1 üÓ replacing Ï+1 by n-1, the first term
n=2
of the middle sum is 3a1, and thus we have
3a1 + {[2n(n-1)+3n]an + an-2}xn 1 = 0. Hence a1 = 0 and
n=2
-an-2
an =-------------, which is the desired recurrance relation.
n = n(2n+1)
Thus all even coefficients are found in terms of a0 and all odd coefficients are zero, thereby yielding only one solution of the desired form.
If X = 1/x then
dy = dy _dX = -J1_ dy = x2 dy
dx dX dx x2 dX dX
d2y d ,-2dy dX dy ,-2d2y 1
—2 = ò³ã (-XSy> ^ = (-2X -y - X2^-) (--^
dx2 dX dX dx dX dX2 x2
= X4d2y + 2X3^y.
dX2 dX
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