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

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y" + p(x)y' + q(x)y = 0 we see that p(x) = 1/x and 22

q(x) = (x -1)/x . Thus x = 0 is a singular point and

2

since xp(x) — 1 and x q(x) — -1 as x — 0, x = 0 is a

to

regular singular point. Substituting y = anxn+r into

n=0

22

x y" + xy' + (x -1)y = 0, shifting indices appropriately, and collecting coefficients of common powers of x we obtain [r(r-1) + r - 1]a0xr + [(1+r)r + 1 + r -1]a1xr+1

+ X {[(n+r)2 - 1]an + an-2}xn+r = 0.

n=2

2

The indicial equation is r -1 = 0 so the roots are r1 = 1

and r2 = -1. For either value of r it is necessary to

r+1

take a1 = 0 in order that the coefficient of x be zero.

The recurrence relation is [(n+r) - 1]an = -an-2,

n = 2,3,4... . For r = 1 we have an = -an-2/[n(n+2)],

n = 2,3,4,... . Since a1 = 0 it follows that a3 = a5 = a7

= ... = 0. Let n = 2m. Then a2m = -a2m-2/2 m(m+1), m =

oo

Section 5.7

93

22

1,2,..., so a2 = -a0/2 -1-2, a4 = -a2/2 -1-2-3 =

a0/22 22 1-2-2-3,..., and a2m = (-1)m a0/22nm!(m+1)!. Thus one solution (set a0 = 1/2) of the Bessel equation of

order one is J1(x) = (x/2) (-1) nx2n/(n + 1)!n!22n. The

n=0

ratio test shows that the series converges for all x.

Also note that J1(x) ^ 0 as x ^ 0.

16b. For r = -1 the recurrence relation is 2

[(n-1) - 1]an = -an-2, n = 2,3,... . Substituting n = 2

2

into the relation yields [(2-1) - 1]a2 = 0 a2 = -a0.

Hence it is impossible to determine a2 and consequently impossible to find a series solution of the form

x-1 X bnxR-

n=0

Secton 5.7, Page 278

1. The D.E. has the form P(x)y" + Q(x)y' + R(x)y = 0 with

P(x) = x, Q(x) = 2x, and R(x) = 6ex. From this we find p(x) = Q(x)/P(x) = 2 and q(x) = R(x)/P(x) = 6ex/x and thus x = 0 is a singular point. Since xp(x) = 2x and

2x

x q(x) = 6xe are analytic at x = 0 we conclude that x = 0 is a regular singular point. Next, we have

2

xp(x) ^ 0 = p0 and x q(x) ^ 0 = q0 as x ^ 0 and thus the

-2

indicial equation is r(r-1) + 0-r + 0 = r - r = 0, which has the roots r1 = 1 and r2 = 0.

3. The equation has the form P(x)y" + Q(x)y' + R(x)y = 0

2

with P(x) = x(x-1), Q(x) = 6x and R(x) = 3. Since P(x),

Q(x), and R(x) are polynomials with no common factors and

P(0) = 0 and P(1) = 0, we conclude that x = 0 and x = 1

are singular points. The first point, x = 0, can be shown

to be a regular singular point using steps similar to

those to shown in Problem 1. For x = 1, we must put the

D.E. in a form similar to Eq.(1) for this case. To do

this, divide the D.E. by x and multiply by (x-1) to

23 obtain (x-1) y + 6x(x-1)y + — (x-1)y = 0. Comparing this

x

to Eq.(1) we find that (x-1)p(x) = 6x and (x-1) q(x) = 3(x-1)/x which are both analytic at

94

Section 5.7

x = 1 and hence x = 1 is a regular singular point. These last two expressions approach p0 = 6 and q0 = 0

respectively as x ^ 1, and thus the indicial equation is

r(r-1) + 6r + 0 = r(r+5) = 0.

-(1+x) 2

9. For this D.E., p(x) = ------- and q(x) = ---- and thus

x2(1-x) x(1-x)

x = 0, -1 are singular points. Since xp(x) is not

analytic at x = 0, x = 0 is not a regular singular point.

1+x 2 2(1-x)

Looking at (x-1)p(x) = ----- and (x-1) q(x) = - we

x2 x

see that x = 1 is a regular singular point and that

p0 = 2 and q0 = 0.

sinx cosx

17a. We have p(x) = -- and q(x) = -------, so that x = 0 is

22

xx

a singular point. Note that xp(x) = (sinx)/x ^ 1 = p0 2

as x ^ 0 and x q(x) = -cosx ^ -1 = q0 as x ^ 0. In order to assert that x = 0 is a regular singular point we

2

must demonstrate that xp(x) and xq(x), with xp(x) = 1 at 2

x = 0 and x q(x) = -1 at x = 0, have convergent power series (are analytic) about x = 0. We know that cosx is analytic so we need only consider (sinx)/x. Now

sinx = (-1) nx2n+1/(2n+1)! for -

n 2n+1

^ < x < so

n=0

(sinx)/x = (-1) nx2n/(2n+1)! and hence is analytic.

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