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

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Section 5.8

101

0. For the case r1 = 3/2, F(r1+1), which is the

r +1

coefficient of x 1 is Ô 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),

*2

³4(3/2) = a0/°4'n 1 ,ë

a4(3/2) = a0/24-2!(1 + 3/2)(2 + 3/2),..., and

a2m(3/2) = (-1) m/22mm!-(1 + 3/2)...(m + 3/2). Hence one solution is

, ) m

, x 3/2 r, ^ (-1) ,

y 1 (x) = x [1 + X ----------------------------------------- ( ) ],

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 r1

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(r1) = 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+1 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 y1(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, ^ (-1) , x 42m,

y2(x) = x [1 + X-------------------------------------------- ( ) ].

2 z-'m!(1 - 3/2)(2 - 3/2) ... (m - 3/2) 2

m=1

102

Section 5.8

7. Apply the ratio test:

³ / 4 m+1 2m+2/,-..2m+2r/ . ³ , -i2 ³

n . l(-1) x /2 [(m+1)!] I , 2 , n . 1

lim --------------------------------- = |x | lim --------------- = 0

m ^ to |(-1)m x2m/22m(m!)2| m ^ to22(m+1)2

for every x. Thus the series for J0(x) converges absolutely for all x.

12. If S = axe, then dy/dx = x 1/2f + x1/2f'aPxe 1 where f'

denotes df/dS. Find d2y/dx2 in a similar fashion and use algebra to show that f satisfies the D.E.

S2f" + Sf' + [S2 - 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 è = 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

X2J'0(t) + (Xj/t) XjJ,0(t) + X2J0(t) =

2 -1 X[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 (Xħx) and yj(x) =

J0 (Xjx) and we have, respectively: (xyi)' = -X2xyi

(xyj)' = -X2xyj.

Now multiply the first equation by yj, the second by Ó³,

103

CHAPTER 6

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