# Elementary differential equations 7th edition - Boyce W.E

ISBN 0-471-31999-6

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Now we return to Eq. (13) and set the coefficient of xr+n equal to zero. This gives the relation

[2(r + n)(r + n - 1) - (r + n) + 1]an + an-1 = 0 (16)

or

2(r + n)2 — 3(r + n) + 1

[(r + n) — 1][2(r + n) — 1]’

n > 1. (17)

For each root r1 and r2 of the indicial equation we use the recurrence relation (17) to determine a set of coefficients a17 a2, For r = r1 = 1, Eq. (17) becomes

Thus

a = n—1—, n > 1.

n (2n + 1)n -

a0

a = ——,

1 3 ? 1

a1

a2 =— "

5 ? 2 (3 ? 5)(1 ? 2)'

and

a3 = —

7 ? 3 (3 ? 5 ? 7)(1 ? 2 ? 3)

In general we have

(—1)n

[3 ? 5 ? 7???(2n + 1)]n!

a0, n > 1. (18)

an 1

an 1

a0

a2

a0

270

Chapter 5. Series Solutions ofSecond Order Linear Equations

Hence, if we omit the constant multiplier a0, one solution of Eq. (8) is

“ (—1)nxn

y1(x ) = x

1+

x > 0.

(19)

n=1 [3 ? 5 ? 7???(2n + 1)]n!

To determine the radius of convergence of the series in Eq. (19) we use the ratio test:

.n + 1

lim

n^?

an+1x

= lim ------------------------------= 0

n^? (2n + 3)(n + 1)

for all x. Thus the series converges for all x.

Corresponding to the second root r = r2 = 2, we proceed similarly. From Eq. (17) we have

Hence

an = — an-11

2n(n — 2)

n(2n — 1)

n > 1.

a, =--------.

1 11

and in general

2 ? 3 (1 ? 2)(1 ? 3)'

a, =------------

3 3-5

(1 ? 2 ? 3)(1 ? 3 ? 5)'

(—1)n

7a0.

n 1.

n![1 ? 3 ? 5 ? ? ? (2n — 1)]

Again omitting the constant multiplier a0, we obtain the second solution

y2 (x ) = x

_ x 1/2

1+

(—1)nxn

^ n![1 ? 3 ? 5 ? ? ? (2n — 1)]

x > 0.

(20)

(21)

As before, we can show that the series in Eq. (21) converges for all x. Since the leading terms in the series solutions y1 and y2 are x and x1/2, respectively, it follows that the solutions are linearly independent. Hence the general solution of Eq. (8) is

y = C1 y1(x) + C2y2(x), x > 0.

n

anx

a

a

0

a

0

The preceding example illustrates that if x = 0 is a regular singular point, then sometimes there are two solutions of the form (7) in the neighborhood of this point. Similarly, if there is a regular singular point at x = x0, then there may be two solutions of the form

y = (x - xo)r^2an(x - x0)n (22)

n=0

that are valid near x = x0. However, just as an Euler equation may not have two solutions of the form y = xr, so a more general equation with a regular singular point may not have two solutions of the form (7) or (22). In particular, we show in the next section that if the roots r1 and r2 of the indicial equation are equal, or differ by an

5.6 Series Solutions near a Regular Singular Point, Part I

271

PROBLEMS

integer, then the second solution normally has a more complicated structure. In all cases, though, it is possible to find at least one solution of the form (7) or (22); if rx and r2 differ by an integer, this solution corresponds to the larger value of r .If there is only one such solution, then the second solution involves a logarithmic term, just as for the Euler equation when the roots of the characteristic equation are equal. The method of reduction of order or some other procedure can be invoked to determine the second solution in such cases. This is discussed in Sections 5.7 and 5.8.

If the roots of the indicial equation are complex, then they cannot be equal or differ by an integer, so there are always two solutions of the form (7) or (22). Of course, these solutions are complex-valued functions of x. However, as for the Euler equation, it is possible to obtain real-valued solutions by taking the real and imaginary parts of the complex solutions.

Finally, we mention a practical point. If P, Q, and R are polynomials, it is often much better to work directly with Eq. (1) than with Eq. (3). This avoids the necessity of expressing xQ(x)/P (x) and x2 R(x)/P (x) as power series. For example, it is more convenient to consider the equation

x (1 + x) y" + 2y' + xy = 0

than to write it in the form

2x x2

x2 y + ^— y + ^~ y = 0,

1 + x 1 + x

which would entail expanding 2x/(1 + x) and x2/(1 + x) in power series.

In each of Problems 1 through 10 show that the given differential equation has a regular singular point at x = 0. Determine the indicial equation, the recurrence relation, and the roots of the indicial equation. Find the series solution (x > 0) corresponding to the larger root. If the roots are unequal and do not differ by an integer, find the series solution corresponding to the smaller

root also.

1. 2xy" + y' + xy = 0 2. x 2y" + xy' + (x2 — 1 )y = 0

3. xy" + y = 0 4. xy" + y' — y = 0

5. 3x2 y" + 2xy' + x2 y = 0 6. x2 y" + xy + (x — 2)y = 0

7. o II 1 77 I + Ay X 8. 2x2 y" + 3xy; + (2x2 — 1)y = 0

9. x y" — x (x + 3)y' + (x + 3)y = 0 10. x 2y" + (x2 + 1 )y = 0

11. The Legendre equation of order a is

(1 — x 2)y" — 2xy' + a(a + 1)y = 0.

The solution of this equation near the ordinary point x = 0 was discussed in Problems 22 and 23 of Section 5.3. In Example 5 of Section 5.4 it was shown that x = ±1 are regular singular points. Determine the indicial equation and its roots for the point x = 1. Find a series solution in powers of x — 1 for x — 1 > 0.

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