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

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1 2 0

247

2 8b. From part a we have y(0) = c1 = 0. Thus

y(x) = c2x - x(x-s)f(s)ds and hence

J0

y(1) = c2 - (1-s)f(s)ds = 0, which yields the desired

J0

value of c2.

28c. From parts a and b we have

ô^) = xj1(1-s)f(s)ds - Jx(x-s)f(s)ds

[ xx(1-s)f(s)ds + f 1x(1-s)f(s)ds - [

0x0

Jx(x-xs-x+s)f(s)ds + J1x(1-s)f(s)ds Jxs(1-x)f(s)ds + J1x(1-s)f(s)ds.

28d. We have ô^) = 11G(x,s)f(s)ds

ÔÛ = f1

0

= JxG(x,s)f(s)ds + f1G(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)y'1(x)]

+ p(x)[y,1(x)y,2(x) + y1(x)y'2(x) - y,2(x)y,1(x) - y2(x)y1(x)]

= y^py',,]' - y2[py,1]' = 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

G(s,x) is less than the second argument, the bottom expression of formula (iv) must be used to determine G(s,x). Thus, G(s,x) = -y1(s)y2(x)/c. A similar argument

holds if x < s < 1.

248

Section 11.3

30e. We have ô^) = 11G(x,s)f(s)ds

0

y1(s)y2 (x)f(s) f1 y1(x)y2(s)f(s)

ds - I -------------------ds

c Jx c

ô (x) = I

0

= Ãxyi(s)y2(x)f(s) ds I

0 c x

(where c = p(x)W(y1fy2) and thus, by Leibnitz's rule,

cô/ (x) = -yi (x)y2 (x)f(x) - I xyi (s)y2 (x)f(s)ds + yi(x)y2(x)f(x)

0

-|ly1(x)y2(s)f(s)ds. From this we obtain

-c(pô/)/ = (py2)'Ixyi (s)f(s)ds + py2yif(x)

0

+ (pyi),|i y2(s)f(s)ds - pyiy2f(x). Dividing by c and adding q(x^(x) we get

(^ô^'+äô = (py2 ) [xy1 (s)f(s)ds - [xy1(s)f(s)ds

c Jo c Jo

(pyi)' fi qyi fi

I y2(s)f(s)ds - --------I

Jx c Jx

y2(s)f(s)ds + f(x)

³ fx , , *, , n (pyi) -qyi fi *, ,

- yi(s)f(s)ds + ----------------- y2(s)f(s)ds + f(x)

Jo c Jx

= f(x),

since yi and y2 satisfy L[y] = 0. Using ô^) and ô'(x) as found above, the B.C. are both satisfied since yi(x) satisfies one B.C. and y2(x) satisfies the other B.C.

33. In general y(x) = cicosx + c2sinx. For y'(0) = 0 we must choose c2 = 0 and thus yi(x) = cosx. For y(i) = 0 we have cicosi + c2sini = 0, which yields c2 = -c^cosD/sini and thus y2(x) = cicosx - ci(cosi)sinx/sini

= ci(sinicosx - cosisinx)/sini = sin(i-x) [by setting ci = sini].

Furthermore, W(yi,y2) = -cosi and thus

G(x,s) =

and hence

coss sin(i-x) cosi cosx sin(i-s) cosi

0 < s < x

x < s < i

ô(x) = Ix[cosssin(i-x)f(s)/cosi]ds

0

+ ^^[cosx sin(i-s)f(s)/cosi]ds is the solution of the given B.V.P.

+

c

249

Section 11.4, Page 661

2a. The D.E. is the same as Eq.(6) and thus, from Eq.(9), the general solution of the D.E. is

y = cj0(VX x) + c2Y0(V"X x). The B.C. at x = 0 requires

that c2 = 0, and the B.C. at x = 1 requires

c^V"X J'0(VX) = 0. For X = 0 we have ô0(x) = J0(0) = 1

and if X is the nth positive root of j' (v X ) = 0 then

n0

ô (x) = J (ä/X x). Note that for X = 0 the D.E. becomes

T n 0 v n

(xy')' = 0, which has the general solution y = c^nx + c2.

To satisfy the bounded conditions at x = 0 we must choose c1 = 0, thus obtaining the same solution as above.

2b.

I01

For n Ô 0, set y = J0(^/X. x) in the D.E. and integrate from 0 to 1 to obtain -l^xj' )'dx = X ^xJ (äIX x)dx.

J0 0 nfo 0 V n

Integrating the left side of this equation yields

1

(xj' )'dx = xj' (äIX x) = j' (ä/X ) - 0 = 0 since the X

0 0 V n 0yn n

0

are eigenvalues from part a. Thus J1xJ0(^X. x)dx = 0.

For other n and m, we let L[y] = - (xy')'. Then

L[J (ä/X"x)] = X xj (ä/X”x) and

0 V n n 0 V n

L[J (ä/X x)] = X xj (ä/X x). Multiply the first equation 0Vm m0Vm

by JA\ X x), the second by J (ä/X x), subtract the 0Vm 0 V n

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