Page 310 - Schaum's Outline of Differential Equations

CHAP. 29] SOME CLASSICAL DIFFERENTIAL EQUATIONS 293

If n = 3, then we must show that the equation H 4(x) = 2xH 3(x) - 6H 2(x) is satisfied by the appropriate Hermite
polynomials. Direct substitution gives

We see that the right-side does indeed equal the left side, hence, the recurrence relation is verified.

29.6. Legendre polynomials satisfy the recurrence formula

Use this formula to find P 5 (x).
Letting n = 4 and solving for P$(x), we have P 5(x) = (9xP 4(x) - 4P 3 (x)). Substituting for P 3(x) and for P 4(x),
3
5
we have P 5(x) = (63x - 70x + I5x).
1
29.7. Chebyshev polynomials, T n (x), can also be obtained by using the formula T n(x) = cos(« cos (x)). Verify
2
this formula for T 2(x) = 2x - 1.
1
1
2
2
2
Letting n = 2, we have cos(2cos" (jc)). Let a= cos" ^). Then cos(2a) = cos (a) - sin (a) = cos (a)
2
1
2
2
1
- (1 - cos (a)) = 2 cos (a) -1. But if a = cos" ^), then x = cos(a). Hence, cos(2 cos" ^)) = 2x -1 = T 2(x).
2
29.8. The differential equation (1 - x )y" + Axy' + By = 0 closely resembles both the Chebyshev and Legendre
equations, where A and B are constants. A theorem of differential equations states that this differential
equation has two finite polynomial solutions, one of degree m, the other of degree «, if and only if
A = m + n — 1 and B = —mn, where m and n are nonnegative integers and n + m is odd.
2
For example, the equation (1 - x )y" + 4xy' -6y = 0 has polynomial solutions of degree 2 and 3:
y = 1 + 3x 2 and y = x H (these are obtained by using the series techniques discussed in Chapter 27).
We note here that A = 4 = n + m— 1 and B = —6 = —mn necessarily imply that m = 2, n = 3
(or conversely). Hence our theorem is verified for this equation.
Determine whether the three following differential equations have two polynomial solutions:
2
2
2
a) (1 - x )y" + 6xy' - Uy = 0; b) (1 - Jt )/' + xy' + 8y = 0; c) (1 - x )y" - xy' + 3y = 0.
a) Here A = 6 = n + m- 1, B = -mn = -12 implies m = 3, n = 4; hence we have two finite polynomial solutions, one
of degree 3, the other of degree 4.
b) Here A = 1 and B = 8; this implies m = 2, n = -4; therefore, we do not have two such solutions. (We will have
one polynomial solution, of degree 2.)
c) Since A = -l, B = 3 implies m = -^j3, w = -V3, we do not have two polynomial solutions to the differential
equation.

Supplementary Problems

29.9. Verify H 2(x) and H 3(x) are orthogonal with respect to the weight function e~ x on the interval (—°°, °°).

29.10. Find H 5(x) by using the recurrence formula H n + i(x) = 2xH n(x) - 2nH n _ l(x).

29.11. The Rodrigues formula for the Legendre polynomials is given by