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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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The standard form is y" + p(x)y' + q(x)y = 0, with
p(x) = 1/x and q(x) = 1. Thus x = 0 is a singular point;
2
and since xp(x) ^ 1 and xq(x) ^ 0 as x ^ 0, x = 0 is a
ro
regular singular point. Substituting y = I a^^ into
n=0
22
xy + xy + xy = 0 and shifting indices appropriately, we obtain
ro ro ro
X , x , .. x n+r X z X n+r X n+r ~
I (n+r)(n+r-1)anx + I (n+r)anx + I an-2x = 0,
n=0 n=0 n=2
or
[r(r-1)+r]a0xr + [(1+r)r+1+r]a1xr+1 ro
+1 [(n+r) 2an + an-2] xn+r = 0. The indicial equation
2
2n
n=2
2
is r = 0 so r = 0 is a double root. It is necessary to
r+1
take a1 = 0 in order that the coefficient of x be zero.
2
The recurrence relation in n an = -an-2, n = 2,3,... . Since a1 = 0 it follows that a3 = a5 = a7 = ... = 0. For
the even coefficients we let n = 2m, m = 1,2,... . Then
2 2 2- 2 2- 2- 2- 2
a2m = -a2m-2/2 m so a2 = -a0/2 -1 , a4 = a0/2 -2 -1 -2 ,... ,
and a2m = (-1)ma0/22m(m!) 2. Thus one solution of the Bessel
92
Section 5.6
equation of order zero is J0(x) = 1 + (-1) mx2m/22m(m!) 2
m=1
where we have set a0 = 1. Using the ratio test it can be shown that the series converges for all x. Also note that J0(x) 1 as x 0.
15. In order to determine the form of the integral for x near zero we must study the integrand for x small. Using the above series for J0, we have
1 1 1
x[J0(x)] 2 x[1 - x2/2 +...] 2 x[1 - x2 +...]
12
[1 + x + ...] for x small. Thus x
dx 1
= J0 (x) I [ + x + ...]dx
dx
y2(x) = J0(x) I----------------- = J0(x) I
x[J0(x)] 2 J x
2
x
= J0(x)[lnx + + ...], and it is clear that y2(x) 0 x 2
will contain a logarithmic term.
16a. Putting the D.E. in the standard form
y" + p(x)y' + q(x)y = 0 we see that p(x) = 1/x and 22
q(x) = (x -1)/x . Thus x = 0 is a singular point and
2
since xp(x) 1 and x q(x) -1 as x 0, x = 0 is a regular singular point. Substituting y = anxn+r into
n=0
22
xy + xy + (x -1)y = 0, shifting indices appropriately, and collecting coefficients of common powers of x we obtain [r(r-1) + r - 1]a^xr + [(1+r)r + 1 + r -1]a1xr+1
+ ^ {[(n+r)2 - 1]an + an-2}xn+r = 0.
n=2
2
The indicial equation is r -1 = 0 so the roots are r1 = 1
and r2 = -1. For either value of r it is necessary to
r+1
take a1 = 0 in order that the coefficient of x be zero.
The recurrence relation is [(n+r) - 1]an = -an-2,
n = 2,3,4... . For r = 1 we have an = -an-2/[n(n+2)],
n = 2,3,4,... . Since a1 = 0 it follows that a3 = a5 = a7 = ... = 0. Let n = 2m. Then a2m = -a2m-2/2 m(m+1), m =
co
Section 5.7
93
2 2
1,2,..., so a2 = -a0/2 -1-2, a4 = -a2/2 -1-2-3 =
a0/22 22 1-2-2-3,..., and a2m = (-1)m a0/22nm!(m+1)!. Thus one solution (set a0 = 1/2) of the Bessel equation of
order one is J2(x) = (x/2) (-1) nx2n/(n + 1)!n!22n. The
n=0
ratio test shows that the series converges for all x.
Also note that J2(x) ^ 0 as x ^ 0.
16b. For r = -1 the recurrence relation is 2
[(n-1) - 1]an = -an-2, n = 2,3,... . Substituting n = 2
2
into the relation yields [(2-1) - 1]a2 = 0 a2 = -a0.
Hence it is impossible to determine a2 and consequently impossible to find a series solution of the form
x-1 X bnxn-
n=0
Secton 5.7, Page 278
1. The D.E. has the form P(x)y" + Q(x)y' + R(x)y = 0 with
P(x) = x, Q(x) = 2x, and R(x) = 6ex. From this we find p(x) = Q(x)/P(x) = 2 and q(x) = R(x)/P(x) = 6ex/x and thus x = 0 is a singular point. Since xp(x) = 2x and
2x
x q(x) = 6xe are analytic at x = 0 we conclude that
x = 0 is a regular singular point. Next, we have
2
xp(x) ^ 0 = p0 and x q(x) ^ 0 = q0 as x ^ 0 and thus the
2
indicial equation is r(r-1) + 0-r + 0 = r - r = 0, which has the roots r1 = 1 and r2 = 0.
3. The equation has the form P(x)y" + Q(x)y' + R(x)y = 0
2
with P(x) = x(x-1), Q(x) = 6x and R(x) = 3. Since P(x),
Q(x), and R(x) are polynomials with no common factors and
P(0) = 0 and P(1) = 0, we conclude that x = 0 and x = 1
are singular points. The first point, x = 0, can be shown
to be a regular singular point using steps similar to
those to shown in Problem 1. For x = 1, we must put the
D.E. in a form similar to Eq.(1) for this case. To do
this, divide the D.E. by x and multiply by (x-1) to
23 obtain (x-1) y + 6x(x-1)y + (x-1)y = 0. Comparing this
x
to Eq.(1) we find that (x-1)p(x) = 6x and 2
(x-1) q(x) = 3(x-1)/x which are both analytic at
94
Section 5.7
x = 1 and hence x = 1 is a regular singular point. These last two exPressions aPProach P0 = 6 and q0 = 0
respectively as x ^ 1, and thus the indicial equation is
r(r-1) + 6r + 0 = r(r+5) = 0.
-(1+x) 2
9. For this D.E., p(x) = ------- and q(x) = and thus
x2(1-x) x(1-x)
x = 0, -1 are singular points. Since xp(x) is not
analytic at x = 0, x = 0 is not a regular singular point.
1+x 2 2(1-x)
Looking at (x-1)p(x) = -- and (x-1) q(x) = ---- we
x2 x
see that x = 1 is a regular singular point and that
p0 = 2 and q0 = 0.
sinx cosx
17a. We have p(x) = -- and q(x) = ---, so that x = 0 is
22
xx
a singular point. Note that xp(x) = (sinx)/x ^ 1 = p0
2
as x ^ 0 and x q(x) = -cosx ^ -1 = q0 as x ^ 0. In
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