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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.
Download (direct link): elementarydifferentialequations2001.pdf
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y = eXz[c1 cos(^z) + c2 sin(^z)] = xX[c1 cos(^ lnx) + c2 sin(^ lnx)].
In each of Problems 24 through 29 use the method of Problem 23 to solve the given equation for x > 0.
24. x2y" — 2y = 0 25. x2y" — 3xy' + 4y = lnx
26. x2yw + 7xy' + 5y = x 27. x2y" — 2xy' + 2y = 3x2 + 2 lnx
28. x2y" + xy' + 4y = sin(lnx) 29. 3x2y" + 12xy'+ 9y = 0
30. Show that if L [y ] = x 2y" + a xy' + Ðó , then
L [(—x ) ] = (—x )rF (r)
for all x < 0, where F (r) = r (r — 1) + ar + â. Hence conclude that if ^ = r2 are roots
of F (r) = 0, then linearly independent solutions of L [y] = 0 for x < 0 are (—x)r1 and
( —x )r2.
31. Suppose that xr1 and xr2 are solutions of an Euler equation for x > 0, where r1 = r2, and r1 is an integer. According to Eq. (24) the general solution in any interval not containing the origin is y = c1 |x |r1 + c2|x |r2. Show that the general solution can also be written as
5.6 Series Solutions near a Regular Singular Point, Part I
267
y = kjxr1 + k2 |x |r2.
Hint: Show by a proper choice of constants that the expressions are identical for x > 0, and by a different choice of constants that they are identical for x < 0.
5.6 Series Solutions near a Regular Singular Point, Part I
We now consider the question of solving the general second order linear equation
P (x) y" + Q(x) y'+ R (x) y = 0 (1)
in the neighborhood of a regular singular point x = x0. For convenience we assume that x0 = 0. If x0 = 0, the equation can be transformed into one for which the regular singular point is at the origin by letting x — x0 equal t.
The fact that x = 0 is a regular singular point of Eq. (l)meansthat xQ(x)/P (x) = xp(x) and x2R(x)/P(x) = x2q(x) have finite limits as x ^ 0, and are analytic at x = 0. Thus they have convergent power series expansions of the form
CO CO
xp(x) = ^2 pnxn, x 2q (x) = ^2 qnxn, (2)
n=0 n=0
on some interval |x | < p about the origin, where p > 0. To make the quantities xp(x)
and x2q (x) appear in Eq. (1), it is convenient to divide Eq. (1) by P (x) and then to
multiply by x 2, obtaining
x2 y" + x [xp(x)] y' + [x 2q (x )]y = 0, (3)
or
x V + x (p0 + p1 x + ¦¦¦ + pnxn + ¦ ¦ ¦) y
+ (q0 + q1 x + ¦¦¦ + qnxn + ¦¦¦) y = 0. (4)
If all of the coefficients pn and qn are zero except possibly
r xQ(x) x2 R(x)
p = lim----------- and qn = lim-----------------------------------, (5)
x^0 P (x) x^0 P (x) W
then Eq. (4) reduces to the Euler equation
x 2y" + ?0xy + q0 y = 0> (6)
which was discussed in the preceding section. In general, of course, some of the pn and qn, n > 1, are not zero. However, the essential character of solutions of Eq. (4) is identical to that of solutions of the Euler equation (6). The presence of the terms p1 x + ¦¦¦ + pnxn + ¦¦¦ and q1 x + ¦¦¦ + qnxn + ¦¦¦ merely complicates the calculations.
We restrict our discussion primarily to the interval x > 0. The interval x < 0 can be treated, just as for the Euler equation, by making the change of variable x = — and then solving the resulting equation for f > 0.
268
Chapter 5. Series Solutions of Second Order Linear Equations
EXAMPLE
1
Since the coefficients in Eq. (4) are “Euler coefficients” times power series, it is natural to seek solutions in the form of “Euler solutions” times power series. Thus we assume that
CO CO
y = xr (a0 + a1x +----------+ anxn +-----) = xr ^ anxn = ^ anxr+n, (7)
n=0 n=0
where a0 = 0. In other words, r is the exponent of the first term in the series and a0 is its coefficient. As part of the solution we have to determine:
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