# Elementary differential equations 7th edition - Boyce W.E

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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