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

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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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