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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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n=2 n=1 n=1
(b) Substituting the series representation for J0(x) in Eq. (i), show that
2 2 ^ 2 n ^ (1)n2nx2n
b1x + 2 b2x +J2 (nbn + bn2)x =~2? 22n ( !)2 . (ii)
^ 3 ^ 1 2 (n!)
(c) Note that only even powers of x appear on the right side of Eq. (ii). Show that b1 = b3 = b5 =    = 0, b2 = 1 /22 (1! )2, and that
(2n)2b2n + b2n_2 = 2(1)n(2n)/2ln(n!)2, n = 2, 3, 4,..
Deduce that
1 ( 1 \ 1 ( 11 b4 =------2Γ-2 I 1 +  I and be = 2 2 2 I 1 + ~ 
4 224^ V 2/ 6 22426^ V 2 3
The general solution of the recurrence relation is b2n = (1)n+1 Hn /22n (n!)2. Substituting for bn in the expression for y2(x) we obtain the solution given in Eq. (10).
11. Find a second solution of Bessels equation of order one by computing the cn (r2) and a ofEq. (24) of Section 5.7 according to the formulas (19) and (20) of that section. Some guidelines along the way of this calculation are the following. First, use Eq. (24) of this
V
5.8 Bessels Equation
291
section to show that a1 (1) and a\ (1) are 0. Then show that c1(1) = 0 and, from the recurrence relation, that cn (1) = 0 for n = 3, 5,.... Finally, use Eq. (5) to show that
( ) (1)ma0
am (Γ ) =
2m (r + 1)(r + 3)   (r + 2m  1)2(r + 2m + 1)
for m = 1, 2, 3,..., and calculate
cm (1) = (1)m+1(Hm + Hm-1)/mm !(m  1)!.
1. By a suitable change of variables it is sometimes possible to transform another differential equation into a Bessel equation. For example, show that a solution of
x2y" + (ΰ2β2xe + 1  v2e )y = 0, x > 0
is given by y = x1/2 f (axβ) where f (\$) is a solution of the Bessel equation of order v.
13. Using the result of Problem 12 show that the general solution of the Airy equation
y"  xy = 0, x > 0
is y = x 1/2[c1 f1(ixi/2) + cf2(3ixi/2)] where f1 (\$) and f2(\$) are linearly independent solutions of the Bessel equation of order one-third.
14. It can be shown that J0 has infinitely many zeros for x > 0. In particular, the first three zeros are approximately'2.405, 5.520, and 8.653 (see Figure 5.8.1). Let kj, j = 1, 2, 3,..., denote the zeros of J0; it follows that
) =| 1 x = L
Verify thaty = J0(kjx) satisfies the differential equation
1 * 
' 1 2,
y + y + kjy = 0, x > 0.
x
Hence show that
1
1
0
xJ0(kix)J^kx) dx = 0 if ki = kj.
This important property of J0(kix), known as the orthogonality property, is useful in solving boundary value problems.
Hint: Write the differential equation for J0(kix). Multiply it by xJ0(kjx) and subtract it from xJ0(kix) times the differential equation for J0(kjx). Then integrate from 0 to 1.
REFERENCES
Coddington, E. A., An Introduction to Ordinary Differential Equations (Englewood Cliffs, NJ: Prentice Hall, 1961; New York: Dover, 1989).
Copson, E. T., An Introduction to the Theory of Functions of a Complex Variable (Oxford: Oxford University, 1935).
Proofs of Theorems 5.3.1 and 5.7.1 can be found in intermediate or advanced books; for example, see Chapters 3 and 4 of Coddington, or Chapters 3 and 4 of:
Rainville, E. D., Intermediate Differential Equations (nd ed.) (New York: Macmillan, 1964).
Also see these texts for a discussion of the point at infinity, which was mentioned in Problem 1 of Section 5.4. The behavior of solutions near an irregular singular point is an even more advanced topic; a brief discussion can be found in Chapter 5 of:
292
Chapter 5. Series Solutions ofSecond Order Linear Equations
Coddington, E. A., and Levinson, N., Theory of Ordinary Differential Equations (New York: McGraw-Hill, 1955).
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