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

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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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