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Elementary differential equations 7th edition - Boyce W.E

Boyce W.E Elementary differential equations 7th edition - Wiley publishing , 2001. - 1310 p.
ISBN 0-471-31999-6
Download (direct link): elementarydifferentialequat2001.pdf
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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 —
n—0
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
n—0
+ [(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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