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Elementary differential equations 7th edition - Boyce W.E

Boyce W.E Elementary differential equations 7th edition - Wiley publishing , 2001. - 1310 p.
ISBN 0-471-31999-6
Download (direct link): elementarydifferentialequat2001.pdf
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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 $(x) = I1G(x,s)f(s)ds
Jo
y1(s)y2 (x)f(s) f1 y1(x)y2(s)f(s)
ds - I --------------------ds
c Jx c
4>(x) = I Jo
= fxyi(s)y2(x)f(s) ds I Jo C Jx
(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)
Jo
-fly1(x)y2(s)f(s)ds. From this we obtain
-c(p^')' = (py2)' Ixyi (s)f(s)ds + py2yif(x)
Jo
+ (pyi),|l y2(s)f(s)ds - py1y2f(x). Dividing by c and adding q(x)^(x) we get
(-p^')'+q^ = (py2 ) fxy1(s)f(s)ds - fxy1(s)f(s)ds
c Jo c Jo
(pyl)' fi qyi fi
I y2(s)f(s)ds - I
Jx c Jx
c
(py,2)/-qy2 fx , , , , (py!),-qyi fi
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 y1 and y2 satisfy L[y] = o. Using ^(x) and (x) as found above, the B.C. are both satisfied since y1(x) satisfies one B.C. and y2(x) satisfies the other B.C.
33. In general y(x) = cicosx + c2sinx. For y'(o) = o we must choose c2 = o and thus y1(x) = cosx. For y(1) = o we have cicos1 + c2sin1 = o, which yields c2 = -ci(cos1)/sin1 and thus y2(x) = c1cosx - c1(cos1)sinx/sin1
= c1(sin1cosx - cos1sinx)/sin1 = sin(1-x) [by setting c1 = sin1].
Furthermore, W(y1,y2) = -cos1 and thus
G(x,s) =
and hence
coss sin(1-x) cos1 cosx sin(1-s) cos1
o < s < x
x < s < 1
6(x) = Ix[cosssin(1-x)f(s)/cos1]ds Jo
+ I1[cosxsin(1-s)f(s)/cos1]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(YX x) + c2Y0(YX x). The B.C. at x = 0 requires
that c2 = 0, and the B.C. at x = 1 requires
c^YX J'0(YX ) = 0. For X = 0 we have 60(x) = J0(0) = 1
and if X is the nth positive root of j' (v X ) = 0 then
n 0 '
6 (x) = J (a/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.
I1
For n ^ 0, set y = J0(^X>) in the D.E. and integrate
from 0 to 1 to obtain - Xxj' )'dx = X XxJ (aIX x)dx.
J0 0 n0 0 V n
Integrating the left side of this equation yields
1
(xj' )'dx = xj' (aIX x) = j' (a/X ) - 0 = 0 since the X
0 0 V n 0yn n
0
are eigenvalues from part a. Thus J1xJ0(^Xn x)dx = 0.
For other n and m, we let L[y] = -(xy')'. Then
L[J (a/X~x)] = X xJ (a[X"x) and 0 V n n 0 V n
L[J (a/X x)] = X xJ (a/X x). Multiply the first equation 0Vm m0Vm
by Jh/X x), the second by J (a/X x), subtract the
0ym 0 v n
second from the first, and integrate from 0 to 1 to obtain
J1{J h/X"x)L[J (a/X~x)] - J (a/X~x)L[J (a/X~x)]}dx =
J0 0 V m 0 V n 0 V n 0 V m
(X - X ) 11xJ (a/X x)J (a/X x)dx. Again the left side
n mj0^Vn 0 V m
is zero after each term is integrated by parts once, as was done above. If X ^ X , the result follows with
nm
6 (x) = J (a/X"x).
T n 0 V n
2c. Since X = 0 is an eigenvalue we assume that y = b + > b J (y X x). Since
0 n 0 V n
n=1
-[xJ^ (yX x)]' = X xJ (yX x), n = 0,1,..., we find that 0 V n n 0 V n
250
Section 11.4
4a.
4b.
-(xy')' = x> 1 b J (i/1 x) [note that 1 = 0 and b are
n n 0 v n 0 0
n=1
missing on the right]. Now assume
f(x)/x = c0 + cnJ0 (^ 1n x). Multiplying both sides by
n=1
xJ0(^ 1m x), integrating from 0 to 1 and using the
orthogonality relations of part b, we find
c = 11f(x)J (i 11 x)dx/11xJ2 (i/1 x)dx, n = 0,1,2,... .
n J0 0 V n J0 M n
[Note that c = 2 f(x)dx since the denominator can be 00
integrated.] Substituting the series for y and f(x)/x into the D.E., using the above result for - (xy')', and simplifying we find that
(Ub + c ) + "V [c - b (1 - M)]Jn h Fk~ x) = 0. Thus
r 0 0 n n n r 0 V n
n=1
b = -c /u and b = c /(1 -u), n = 1,2,..., where 1/1 00r n n n r Vn
are obtained from j' (i/1 ) = 0.
0 v n
Let L[y] = -[(1-x2)y']'. Then L[6 ] = 16 and
T n nT n
L[6 ] = 16 . Multiply the first equation by 6 , the
Tm mTm Tm
second by 6 , subtract the second from the first, and
n
integrate from 0 to 1 to obtain
11(6 L[6 ] - 6 L[6 ])dx = (1 - 1 ) 116 6 dx. The integral
j0 m T n TnTm n m J0 rm
on the left side can be shown to be 0 by integrating each term once by parts. Since 1 * 1 if m * n, the result
nm
follows. Note that the result may also be written as TP (x)P (x)dx = 0, m * n.
j0 2m-1 2n-1
First let f(x) = cn6n(x), multiply both sides by 6m(x),
n=1
and integrate term by term from x = 0 to x = 1. The orthogonality condition yields
c = f(x)6 (x)dx/ Pb2(x)dx, n = 1,2,... where it is n Jo n Jo n
understood that 6 (x) = P (x). Now assume
Tn 2n-1
y = > b 6 (x). As in Problem 2 and in the text
nn
nn
n=1
251
-[(1-x2)y'] ' = bn6n since the 6n are eigenfunctions.
n=1
Thus, substitution of the series for y and f into the D.E. and simplification yields V [b (X -u) - c ]6 (x) = 0. Hence
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