Page 146 - Advanced engineering mathematics

126 CHAPTER 4 Series Solutions

1 1

8
4
= a 0 1 − x + x + ···
12 672
1 1

5
9
+ a 1 x − x + x + ··· .
20 1440
This is the general solution, since a 0 and a 1 are arbitrary constants. Because a 0 = y(0) and
a 1 = y (0), a unique solution is determined by specifying these two constants.

SECTION 4.1 PROBLEMS

In each of Problems 1 through 10, find the recurrence rela- 5. y − xy + y = 3

tion and use it to generate the first five terms of a power 6. y + xy + xy = 0

series solution about 0.
2
7. y − x y + 2y = x

1. y − xy = 1 − x
8. y + xy = cos(x)

3
2. y − x y = 4

9. y + (1 − x)y + 2y = 1 − x 2

2
3. y + (1 − x )y = x

10. y + xy = 1 − e x

4. y + 2y + xy = 0

4.2 Frobenius Solutions
We will focus on the differential equation
P(x)y + Q(x)y + R(x)y = F(x). (4.7)

If P(x) = 0 on some interval, then we can divide by P(x) to obtain the standard form

y + p(x)y + q(x)y = f (x). (4.8)

If P(x 0 ) = 0, we call x 0 a singular point of equation (4.7). This singular point regular if
Q(x) R(x)
(x − x 0 ) and (x − x 0 ) 2
P(x) P(x)
are analytic at x 0 . A singular point that is not regular is an irregular singular point.
EXAMPLE 4.3
2
3
2
x (x − 2) y + 5(x + 2)(x − 2)y + 3x y = 0

has singular points at 0 and 2. Now
Q(x) 5x(x + 2)(x − 2) 5 x + 2
(x − 0) = =
P(x) x (x − 2) 2 x 2 x − 2
3
is not analytic (or even defined) at 0, so 0 is an irregular singular point. But
Q(x) 5(x + 2)
(x − 2) =
P(x) x 3

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October 14, 2010 14:17 THM/NEIL Page-126 27410_04_ch04_p121-136