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Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
Download (direct link): elementarydifferentialequations2001.pdf
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-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
^ ^ ^
n+r n+r 2 1 n+r
X (n+rHn+r-Da^ + X (n+r)a^ + (x - ) X a^ = 0.
n=0 n=0 9 n=0
^ ^ ^
2 n+r n+r+2 n+r
Note that x X a^x = X a^x = X an-2x . Thus we
n=0 n=0 n=2
1 r 1 r+1
have [r(r-1) + r - ]a^x + [(r+1)r + (r+1) - ]a1x +
9 0 9 1
1 n+r
n , x = 0. From
X {[(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,...
oo
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
21
a2m 'a2-2/2 m(m + 3. The ?lrst few =oeffl = lents are
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) ma0
a2m 2m 1 1 1
2 m!(1 + )(2 + ) ... (m + -)
3 3 3
solution (on setting a0 = 1) is
to
m
, , 1/3r-, ^ (-1) , x4 2m,
y,(x) = x ' [1 + X -------------------------------------- ( ) ].
1 11 12
m=i m! (1 + )(2 + )...(m + )
3 3 3
Since r2 = - 1/3 r1 and r1 - 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 a1 = 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
to
y = anxn+r in xy" + y' - y = 0 and shifting indices we
n=0
obtain
to to to
\ , -, \ / \ n+r V / -1 \ n+r V n+r
X an+1(r+n + 1)(r+n)x + X an+1(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 : I anxn. The recurrence relation is
n=0
2
(n+1) an+1 = an, n = 0,1,2,... . The coefficients are
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