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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
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00
1 ^m^ndx = Xnl ^mdx and thus (Xn - Xm) l ^m^ndx = 0. If
o o o
Xn ^ Xm the desired result follows.
24b. The general solution of the D.E. is
y = c^inpx + c2cospx + c3sinhpx + c4coshpx where X = p4.
The B.C. require that c2 + c4 = 0, -c2 + c4 = 0, c1sinpL + c2cospL + c3sinhpL + c4coshpL = 0, and c1cospL - c2sinpL + c3coshpL + c4sinhpL = 0. The first two
equations yield c2 = c4 = 0, and the last two have nontrivial solutions if and only if sinpLcoshpL - cospLsinhpL = 0. In this case the third equation yields c3 = -c^inpL/sinhpL and thus the desired eigenfunctions are obtained. The quantity pL can be approximated by finding the intersetion of f(x) = tanx and g(x) = tanhx, where x = pL. The first intersection is at x = 3.9266, which gives X1 = 237.72/L4 and the second intersection is at x = 7.0686, which gives X2 = 2,496.5/L4.
Section 11.2, Page 639
1. We have y(x) = cicosV"X x + c2sin^/~X x and thus y(0) = 0
yields c1 = 0 and y'(1) = 0 yields \fX c2cosV~X = 0. X = 0 gives y(x) = 0, so is not an eigenvalue. Otherwise X = (2n-1)2n2/4 and the eigenfuctions are sin[(2n-1)nx/2], n = 1,2,... . Thus, by Eq.(20), we must choose kn so that
J1{k sin[(2n-1)nx/2]}2dx = 1, since the weight function
0n
r(x) = 1 (by comparing the D.E. to Eq.(1)). Evaluating the integral yields ^/2 = 1 and thus k^ = and the desired
normalized eigenfuctions are obtained.
3. Note here that ^0(x) = 1 satisifes Eq.(20) and hence it
5. From Problem 17 of Section 11.1 we have exsinnnx,
n = 1,2,... as the eigenfunctions and thus k must be chosen
so that l1r(x)k2e2xsin2nnxdx = 1. To determine r(x), we must
0n
write the D.E. in the form of Eq.(1). That is, we multiply
240
Section 11.2
the D.E. by r(x) to obtain ry" - 2ry' + ry + Xry = 0. Now choose r so that (ry')' = ry"-2ry', which yields r' = -2r or
-2x
r(x) = e . Thus the above integral becomes
J*1 2 2 /— /— x
k sin nnxdx = 1 and k = v 2 . Hence 6 (x) = y 2 e sinnnx
0n n n
are the normalized eigenfunctions.
7. Using Eq.(34) with r(x) = 1, we find that the
coefficients of the series (32) are determined by
a„ - f,6 „> = V2 JO' xsin[(2n-1)nx/2]dx
= (4 j/"2/(2n-1) 2n 2)sin(2n-1)n/2. Thus Eq.(32) yields
4\p2 ^ (-1)n-1 i—
f(x) = ----- / ----------- V 2 sin[(2n-1)nx/2], 0 < x < 1,
n2 ^ (2n-1)2
n=1
which agrees with the expansion using the approach developed in Problem 39 of Section 10.4.
10. In this case 6 (x) = (y2 /a ) cosy/X x, where
T n # n V n
an = (1 + sin2^)1/2. Thus Eq.(34) yields a = (V"2/a ) 1co^/X xdx = \[2 sirn/X /a i/X .
n # nJ0 yn^Vn
14. In this case L[y] = y" + y' + 2y is not of the form shown in Eq.(3) and thus the B.V.P. is not self adjoint.
17. In this case L[y] = [(1+x2)y']' + y and thus the D.E. has
the form shown in Eq.(3). However, the B.C. are not separated and thus we must determine by integration whether Eq.(8) is satisfied. Therefore, for u and v satisfying the B.C., integration by parts yields the following:
(L[u],v) = 11{[(1+x2)u']'+u}vdx = vu'(1+x2) -11{(1+x2)v'u'+uv}dx J0 0 *0
= vu' (1+x2)
1
-uv' (1+x2)
0
+ 11{[(1+x2)v']'+v}udx
0
= (u,L[v])
since the integrated terms add to zero with the given B.C. Thus the B.V.P. is self-adjoint.
21a. Substituting 6 for y in the D.E., multiplying both sides by 6, and integrating form 0 to 1 yields
X|1r6 2dx = 11{-[p(x)6']'6 + q(x)6 2}dx. Integrating the 00
first term on the right side once by parts, we obtain
1
0
Section 11.2
241
Xlrto 2dx = -p(1) 6'(1) 6 (1) + p(0)6'(0)6(0) + f1(p6' 2+q62)dx.
00
If a2 ^ 0, b2 ^ 0, then 6'(1) = -b16(1)/b2 and
6'(0) = -a16(0)/a2 and the result follows. If a2 = 0,
then 6(0) = 0 and the boundary term at 0 will be missing.
A similar result is obtained if b2 = 0.
21b. From the text, p(x) > 0 and r(x) > 0 for 0 < x < 1 (following Eq.(4)). If q(x) > 0 and if b1/b2 and -a1/a2 are non-negative, then all terms in the final equation of part a are non-negative and thus X must be non-negative. Note that X could be zero if q(x) = 0, b1 = 0, a1 = 0 and 6(x) = 1.
21c. If either q(x) ^ 0 or a1 ^ 0 or b1 ^ 0, there is at least one
positive term on the right and thus X must be positive.
23a. Using 6(x) = U(x) + iV(x) in Eq.(4) we have
L[6] = L[U(x) + iV(x)] = Xr(x)[U(x)+iV(x)]. Using the linearity of L and the fact that X and r(x) are real we have L[U(x)] + iL[V(x)] = Xr(x)U(x) + iXr(x)V(x).
Equating the real and imaginary parts shows that both U and V satisfy Eq.(1). The B.C. Eq.(2) are also satisfied
by both U and V, using the same arguments, and thus both
U and V are eigenfunctions.
23b. By Theorem 11.2.3 each eigenvalue X has only one linearly independent eigenfuction. By part a we have U and V being eigenfunctions corresponding to X and thus U and V must be linearly dependent.
23c. From part b we have V(x) = cU(x) and thus 6(x) = U(x) + icuU(x) = (1+ic)U(x).
24. This is an Euler equation, so for y = xr we have
r2 - (X+1)r + X = 0 or (r-1)(r-X) = 0. If X = 1, the general solution to the D.E. is y = c1x + c2xlnx. The B.C. require
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