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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
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Since n(x) converges to zero we have c^6 n(x) = 0.
n=1 n=1
Multiplying and integrating as suggested yields
JX[/ c 6 (x)]r(x)6 (x)dx = 0 or
0 nTn m
n=1
r(x)6n(x)6m(x)dx = 0. The integral that multiplies c^
n=1
is just 5 [Eq.(22) of Section 11.2]. Thus the infinite sum
nm
becomes c and the last equation yields c = 0.
mm
A twice differentiable function v satisfying the boundary conditions can be found by assuming that v = ax + b.
Thus v(0) = b = 1 and v(1) = a + 1 while v'(1) = a.
Hence 2a + 1 = -2 or a = -3/2 and v(x) = 1 - 3x/2.
Section 11.3
245
19.
22.
Assuming y = u + v we have (u+v)" + 2(u+v) = u" + 2u + 2(1-3x/2) = 2 - 4x or u" + 2u = -x, u(0) = 0, u(1) + u'(1) = 0 which is the same as Example 1 of the text.
From Eq.(30) we assume u(x,t) = bn(t)6n(x), where the
n=1
6n are the eigenfunctions of the related eigenvalue problem y" + Xy = 0, y(0) = 0, y'(1) = 0 and the bn(t) are given by Eq.(42). From Problem 2, we have 6 = V~2sin[(2n-1) nx/2] and X = (2n-1) 2n 2/4. To
nn
evaluate Eq.(42) we need to calculate
a = 11sin(nx/2) V"2sin[(2n-1)nx/2dx [Eq.(41) with n0
r(x) = 1 and f(x) = sin(nx/2)], which is zero except for n = 1 in which case a1 = /2, and
Yn = J1(-x) ^"2sin[(2n-1)nx/2]dx [Eq.(35) with
F(x,t) = -x]. This integral is the negative of the cn in Problem 2 and thus
Y n = -4^2 (-1) n+1/(2n-1) 2n2 = -cn, n = 1,2,... . Setting
Y = -c in Eq.(42) we then have
nn
b = ^e-*2t/4 - c fVn2(t-s)/4ds
0
1 2 ^0
\/~o 4ci
V 2 e~n 2t/4 - _________-e_n 2(t-s)/4
2
\fo 4c 4c
^e‘n 2t/4 - — + —1 e-n 2t/4 and
2
ft -X (t-s) -X (t-s)
similarly b = -c e n ds = -(c /X )e n n 0 n n
0
= -(c /X ) (1-e n ), where X = (2n-1)2n2/4,
n n n
n = 2,3,... . Substituting these values for bn along
with 6n = V"2sin[(2n-1)nx/2] into the series for u(x,t) yields the solution to the given problem.
In this case a = 0 for all n and Y is given by
= i1 " "
n
1
Yn = J1e t(1-x) V~2sin[(2n-1)nx/2]dx
= e-tl 1(1-x) V~2sin[(2n-1)nx/2]dx. This last integral 0
can be written as the sum of two integrals, each of which has been evaluated in either Problem 6 or 7 of Section
t
0
t
246
Section 11.3
11.2. Letting c denote the value obtained, we then have
n
*t -1 (t-s)
' n n n nj0
-t ft -1 (t-s)
Y = c e t and thus b = c e n e sds =
n n n ni0
-1 tft (1 -1)s -1 t
c e n e n ds = [c /(1 -1)](e t-e n ), where n Jo n n
1n = (2n-1) 2n 2/4. Substituting these values into Eq.(30) yields the desired solution.
24. Using the approach of Problem 23 we find that v(x) satisfies v" = 2, v(0) = 1, v(1) = 0. Thus v(x) = x2 + c1x + c2 and the B.C. yield v(0) = c2 = 1 and
v(1) = 1 + c1 + 1 = 0 or c1 = -2. Hence
v(x) = x2 - 2x + 1 and w(x,t) = u(x,t) - v(x) where, from
Problem 23, we have w = w , w(0,t) = 0, w(1,t) = 0 and
t xx
w(x,0) = x2 - 2x + 2 - v(x) = 1. This last problem can be solved by methods of this section or by methods of Chapter 10. Using the approach of this section we have
w(x,t) = > b (t)6 (x) where 6 (x) = V2 sinnnx
n n n
n=1
[which are the normalized eigenfunctions of the associated eigenvalue problem y" + 1y = 0, y(0) = 0, y(1) = 0] and the bn are given by Eq.(42). Since the
P.D.E. for w(x,t) is homogeneous Eq.(42) reduces to
b = a e n (1 = n2n2 from the above eigenvalue problem),
n n n
where
1
a =
n
111^V~2sinnnxdx = \p2 [1-(-1)n]/nn. Thus 0
^ / n
u(x,t) = x2-2x + 1 + [--(--)—] e-n2n2^V"2sinnnx,
^ nn
n=1
which simplifies to the desired solution.
28a. Since yc = ci+c2x, we assume that
Y(x) = u1(x) + xu2 (x). Then Y' = u2 since we require
u1 + xu^ = 0. Differentiating again yields Y" = u^ and
thus u"2 = -f(x) by substitution into the D.E. Hence
u (x) = - xf(s)ds, u' = xf(x), and u (x) = xsf(s)ds.
2 J0 1 1 J0
Therefore Y = xsf(s)ds - ^xf(s)ds = - x(x-s)f(s)ds and J0 J0 J0
6(x) = c + c x - Ix(x-s)f(s)ds.
1 2 0
247
28b. From part a we have y(0) = ci = 0. Thus
y(x) = c2x - x(x-s)f(s)ds and hence
Jo
y(1) = c2 - (1-s)f(s)ds = 0, which yields the desired
Jo
value of c2.
28c. From parts a and b we have
^(x) = xj1(1-s)f(s)ds - Jx(x-s)f(s)ds
I xx(1-s)f(s)ds + I 1x(1-s)f(s)ds - I Jo Jx Jc
Jx(x-xs-x+s)f(s)ds + Jlx(1-s)f(s)ds Jxs(1-x)f(s)ds + J1x(1-s)f(s)ds.
28d. We have 6(x) = I1G(x,s)f(s)ds
^(x) = I1
Jo
= JxG(x,s)f(s)ds + |lG(x,s)f(s)ds = Jxs(1-x)ds + J1x(1-s)ds, which is the same as found in part c.
30b. In this case y1(x) = sinx and y2(x) = sin(1-x) [assume
y (x) = c cosx + c sinx, let x = 1, solve for c in terms 2 1 2 2
of c1 using y(1) = 0 and then let c1 = sin1]. Using
these functions for y1 and y2 we find W(y1,y2) = -sin1
and thus G(x,s) = -sins sin(1-x)/(-sin1), since p(x) = 1, for 0 < s < x. Interchanging the x and s verifies G(x,s) for x < s < 1.
30c. Since W(y1,y2)(x) = y1(x)y^(x) - y2(x)y1(x) we find that [p(x)W(y1,y2)(x)]' = p'(x)[y1(x)y'2(x) - y2(x)y1(x)]
+ p(x)[y1(x)y'2(x) + y1(x)y'2(x) - y'2(x)y1(x) - y2(x)y'1(x)]
= y1[py,2]' - y2[py1]' = y1[q(x)y2] - y2[q(x)y1] = 0.
30d. Let c = p(x)W(y1,y2)(x). If 0 < s < x, then
G(x,s) = -y1(s)y2(x)/c. Since the first argument in
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