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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.
Download (direct link): elementarydifferentialequations2001.pdf
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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
^) = 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 ^) = 11G(x,s)f(s)ds
= 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 ^) = 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
(x) = I
0
= xyi(s)y2(x)f(s) ds I
0 c x
(where c = p(x)W(y1fy2) and thus, by Leibnitz's rule,
c/ (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(p/)/ = (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
(^^'+ = (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)
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 ^) and '(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
(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 0(x) = J0(0) = 1
and if X is the nth positive root of j' (v X ) = 0 then
n0
(x) = J (/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 0, set y = J0(^/X. x) in the D.E. and integrate from 0 to 1 to obtain -l^xj' )'dx = X ^xJ (IX x)dx.
J0 0 nfo 0 V n
Integrating the left side of this equation yields
1
(xj' )'dx = xj' (IX x) = j' (/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 (/X"x)] = X xj (/Xx) and
0 V n n 0 V n
L[J (/X x)] = X xj (/X x). Multiply the first equation 0Vm m0Vm
by JA\ X x), the second by J (/X x), subtract the 0Vm 0 V n
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