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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.
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Multiplying and integrating as suggested yields
J1[/ c ô (x)]r(x^ (x)dx = 0 or 0 ëøà nTn m
n=1
I cnt1 r(x)ô.(x^m(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^ n(x), where the
n=1
ôn are the eigeefuectioes 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 ô = 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) "\/~2sin[(2n-1)TCx/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)TCx/2]dx [Eq.(35) with
F(x,t) = -x]. This integral is the negative of the cn in Problem 2 and thus
Yn = -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
\/~2 4ci
V 2 e~n 2t/4 - ________1e_n 2(t-s)/4
2
\Ã2 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 ô n = "\/"2sin[(2n-1)TCx/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 = |1 " "
n
1
Yn = Je b(1-x) V~2sin[(2n-1)nx/2]dx
= e_ti 1(1-x) '\/~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 -X (t-s)
' n n n nJo
-t ft -X (t-s)
Y = c e t and thus b = c e n e sds =
n n n nJo
-X tft (X -1)s -X t
c e n e n ds = [c /(X -1)](e t-e n ), where n Jo n n
X. = (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 ô (x) = V2 sinnnx
n n n
n=1
[which are the normalized eigenfunctions of the associated eigenvalue problem y" + Xy = 0, y(0) = 0, y(1) = 0] and the b. are given by Eq.(42). Since the
P.D.E. for w(x,t) is homogeneous Eq.(42) reduces to
b = a e n (X = n2n2 from the above eigenvalue problem),
n n n
where
1
à =
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
ô(x) = c + c x - Ix(x-s)f(s)ds.
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