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

ISBN 0-471-31999-6

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88

Section 5.6

2z 2z

u(z) = c1e + c2ze + (1/4)z + 1/4. Hence

22

y(x) = c^x + c2x lnx + (1/4)lnx + 1/4.

31. If x > 0, then |x| = x and |x|r1 = xr1 so we can choose c1 = k1. If x < 0, then |x| = -x and

ri ' ri ' ri-ri ------- ------ -1 = (-1) rik1,

|x| 1 = (-x) 1 = (-1) 1x 1 and we can choose c, = (-1) 1k1,

or k1 = (-1) r1c1 = (-1) r1c1. In both cases we have

c2 = k2.

Section 5.6, Page 271

2. If the D.E. is put in the standard form y" + p(x)y +

-1 2 q(x)y = 0, then p(x) = x and q(x) = 1 - 1/9x . Thus

x = 0 is a singular point. Since xp(x) ^ 1 and

x q(x) ^ -1/9 as x ^ 0 it follows that x = 0 is a

regular singular point. In determining a series solution

of the D.E. it is more convenient to leave the equation

2

in the form given rather than divide by the x , the

coefficient of y". If we substitute y = anxn+r, we have

n=0

^ ^ ^

X , x x n+r X / x n+r /2 1 x X n+r ~

> (n+r)(n+r-1)anx + > (n+r)anx + (x - )J. a^ = 0.

n=0 n=0 9 n=0

^ ^ ^

txt o_ o-i,.. o_ 2 X n+r X n+r+2 X n+r

Note that x > a^x = > a^x = > an-2x . Thus we

n=0 n=0 n=2

1 r 1 r+1

have [r(r-1) + r - -ja^x + [(r+1)r + (r+1) - ] a^x +

9 0 9 1

1 - n+r _ _

^ x = 0. From

{[(n+r)(n+r-1) + (n+r) - ]an + an-2}

n=2 9

2

the first term, the indicial equation is r - 1/9 = 0 with roots r1 = 1/3 and r2 = - 1/3. For either value of r it is necessary to take a1 = 0 in order that the

r+1

coefficient of x be zero. The recurrence relation is 2

[(n+r) - 1/9]an = -an_2. For r = 1/3 we have

1 2 1 2 2 (n + -) 2-(-) 2 (n + )n

3 3 3

, n = 2,3,4,...

CO

an-2

an =

Section 5.6

89

Thus one

Since a1 = 0 it follows from the recurrence relation that a3 = a5 = a7 = ... = 0. For the even coefficients it is convenient to let n = 2m, m = 1,2,3,... . Then 2 1

a2m - "a2m_2 ^ The flrst feW aĞ

given by

(-1)a0 (-1)a2 a0

a2 = , a4 = =

2 2 1 4 2 1 4 1 1 22(1 + )1 22(2 + )2 24(1 + )(2 + )2!

3 3 3 3

(-1)a4 (-1)a0

a6 = ------------- =--------------------------------------------------------------------, and the

6 2 1 6 1 1 1

22(3 + )3 26(1 + )(2 + )(3 + )3!

3 3 3 3

coefficent of x2m for m = 1, 2,... is

(-1) %

a2m 2m 1 1 1

2 m!(1 + )(2 + ) ... (m + -)

3 3 3

solution (on setting a0 = 1) is

/ - n m

, , 1/3,., ^ (-1) , xx 2m,

y3(x) = x ' [1 + X ( ) ].

1 11 12

m=i m! (1 + )(2 + )...(m + )

3 3 3

Since r2 = - 1/3 ^ r3 and r3 - r2 = 2/3 is not an integer,

we can calculate a second series solution corresponding

to r = - 1/3. The recurrence relation is

n(n-2/3)an = - an-2, which yields the desired solution

following the steps just outlined. Note that a3 = 0, as

in the first solution, and thus all the odd coefficients are zero.

4. Putting the D.E. in standard form y"+p(x)y'+q(x)y = 0, we see that p(x) = 1/x and q(x) = - 1/x. Thus x = 0 is a

2

singular point, and since xp(x) ^ 1 and xq(x) ^ 0, as x ^ 0, x = 0 is a regular singular point. Substituting

y = anxn+r in xy" + y' - y = 0 and shifting indices we

n=0

obtain

\ , -, \ / \ n+r V / -, \ n+r \ n+r

X an+3(r+n + 1)(r+n)x + X an+3(r+n + 1)x - X a^x = 0,

n=-1 n=-1 n=0

90

Section 5.6

11.

[r(r-1) + r]a0x 1 + r + [(r+n+1)

2 -, n+r

an+1 - an]x n=0

2

indicial equation is r 0 so r 0 is a double root. Thus we will obtain only one series of the form

y xr Y anxn. The recurrence relation is

n=0

2

(n+1) an+1 = an, n = 0,1,2,... . The coefficients are

/Ğ2 /'~i2 /-,2 /2 n 2

a1 = a0, a2 = a1/2 = a0/2 , a3 = a2/3 = a0/3 2 ,

*1 - &0, a-2 - a1/ ^ ^0' ^ a3 a2' J ~ CL0/

2 2 2 2 2

a4 = a3/4 = a0/4 -3-2 ,... and an = a0/(n!) . Thus one

solution (on setting a0 = 1) is y = ^Y xn/(n!)2.

n=0

If we make the change of variable t = x-1 and let y = u(t), then the Legendre equation transforms to 2

(t + 2t)u (t) + 2(t + 1)u (t) - a(a+1)u(t) = 0. Since x = 1 is a regular singular Point of the original equation, we know that t = 0 is a regular singular Point

of the transformed equation. Substituting u

n0

in the transformed equation and shifting indices, we obtain

Y, (n+r)(n+r-1)antn+r + 2 (n+r+1)(n+r)an+1tn+r

, n+r

ant

n=0 n=-1

+ 2 (n+r)antn+r + 2 (n+r+1)an+!tn+r

n+1

n=0 n=-1

a(a+1) antn+r = 0, or

n

n=0

[2r(r-1) + 2r]a0 tr1 + Y {2(n+r+1) 2an+1

n0

+ [(n+r)(n+r+1) - a(a+1)]an}tn+r 0.

2

The indicial equation is 2r 0 so r 0 is a double root. Thus there will be only one series solution of the

oo

oo

Section 5.6

91

14.

form y = antn+r. The recurrence relation is

n=0

2

2(n+1) an+1 = [a(a+1) - n(n+1)]an,n = 0,1,2,... . We have

2 9 9 9

a! = [a(a+1)]a0/2-12, a2 = [a(a+1)][a(a+1) - 1-2]a0/22-22-12,

3222

a3 = [a(a+1)][a(a+1) - 1-2][a(a+1) - 2-3]a0/23-32-22-12,..., and an = [a(a+1)][a(a+1)-1-2]..[a(a+1)-(n-1)n]a0/2n(n!) 2.

Reverting to the variable x it follows that one solution of the Legendre equation in powers of x-1 is ro

yi(x) = I [a(a+1)][a(a+1) - 1-2] ...

n=0

[a(a+1) - (n-1)n](x-1) n/2n(n!)2 where we have set a0 = 1,

which is equivalent to the answer in the text if a (-1) is taken out of each square bracket.

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