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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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As in Problem 15, Section 10.2, we have
2cos(2n-1)x (-1) n+1sinnx
f(x) =-------+ > [-------- + -----------].
4 " n (2n-1) 2 n
n=1
n ^ 2cos(2k-1)x (—1) k+1sinkx
en(x) = f(x) +--------> [------------- +-------------- -------------].
4 f-' n (2k—1) 2 k
k=1
Using a computer algebra system, we find that for n = 5, 10 and 20 the maximum error occurs at x = —n in each case and is 1.6025, 1.5867 and 1.5787 respectively.
Note that the author's n values are 10, 20 and 40, since he has included the zero cosine coefficient terms and the sine terms are all zero at x = —n.
It's not possible in this case, due to Gibb's phenomenon, to satisfy |en(x)| < 0.01 for all x.
a0 = |1 (x-x3)dx = 0 and an = |1 (x-x3)cosnnxdx = 0 since
(x-x3) and (x-x3)cosnnx are odd functions. bn = l1(x—x3)sinnnxdx
x3 3x2 (n2n 2+6) (n2n 2+6) .
[ — cosnnx-----------sinnnx-----------xcosnnxH sinnnx] _1
22 33 44
nn n2n2 n3n3 n4n4
—12 12 ^ (—1) n
cosnn, so f(x) =----------> ---------sinnnx.
n3n3 n3 ^ n3
n=1
Section 10.3
209
12b.
14.
12 V1 (-1)
en(x) = f(x) + >--------3—sinknx. These errors will be
n k=1 k
much smaller than in the earlier problems due to the n3 factor in the denominator. Convergence is much more rapid in this case.
The solution to the corresponding homogeneous equation is found by methods presented in Section 3.4 and is y(t) = cicosrat + c2sinrat. For the nonhomogeneous terms we use the method of superposition and consider the sequence of equations y^ + ra2yn = bnsinnt for n = 1,2,3... . If ra > 0 is not equal to an integer, then the particular solution to this last equation has the form Yn = ancosnt + dnsinnt, as previously discussed in Section 3.6. Substituting this form for Yn into the equation and solving, we find an = 0 and dn = bn/(ra2-n2). Thus the formal general solution of the original nonhomogeneous D.E. is
y(t) = c1cosrat + c2sinrat + bn(sinnt)/(ra 2-n2), where
n=1
we have superimposed all the Yn terms to obtain the infinite sum. To evalute c1 and c2 we set t = 0 to obtain y(0) = c1 = 0 and
y'(0) = rac2 + nbn/(ra 2-n2) = 0 where we have formally
n=1
differentiated the infinite series term by term and evaluated it at t = 0. (Differentiation of a Fourier Series has not been justified yet and thus we can only consider this method a formal solution at this point).
Thus c2 = -(1/ra) nbn/(ra 2-n2), which when substituted
n=1
into the above series yields the desired solution.
If ra = m, a positive integer, then the particular solution of ym + m ym = bmsinrnt has the form Ym = t(amcosmt + dmsinmt) since sinmt is a solution of the related homogeneous D.E. Substituting Ym into the D.E. yields am = -bm/2m and dm = 0 and thus the general solution of the D.E. (when ra = m) is now y(t) = c1cosmt
+ c2sinmt - bmt(cosmt)/2m + bn(sinnt)/(m2-n2).
n=1,n^m
To evaluate c1 and c2 we set y(0) = 0 = c1 and
k
210
Section 10.3
y'(0) = c2m - bm/2m + bnn/(m2-n2) = 0. Thus
n=1,n?m
c2 = bm/2m2 - X bnn/m(m2-n2), which when substituted
n=1,n^m
into the equation for y(t) yields the desired solution.
15. In order to use the results of Problem 14, we must first find the Fourier series for the given f(t). Using Eqs.(2) and (3) with L = n, we find that
a0 = (1/rc)Jrcdx - (1/n)jn_2ndx = 0;
an = (1/n) l^cosnxdx - (1/n) 12ncosnxdx = 0; and
J0 Jn
bn = (1/n) j^sinnxdx - (1/n) j2nsinnxdx = 0 for n even and = 4/nn for n odd. Thus
f(t) = (4/n) sin(2n-1)t/(2n-1). Comparing this to the
n=1
forcing function of Problem 14 we see that bn of Problem 14 has the specific values b2n = 0 and b2n-i = (4/n)/(2n-1) in this example. Substituting these into the answer to Problem 14 yields the desired solution. Note that we have asumed ra is not a positive integer. Note also, that if the solution to Problem 14 is not available, the procedure for solving this problem would be exactly the same as shown in Problem 14.
16. From Problem 8, the Fourier series for f(t) is given by f(t) = 1/2 + (4/n2)
X,cos(2n-1)nt/(2n-1) 2 and thus we may
n=1
not use the form of the answer in Problem 14. The procedure outlined there, however, is applicable and will yield the desired solution.
18a. We will assume f(x) is continuous for this part. For the case where f(x) has jump discontinuities, a more detailed proof can be developed, as shown in part b. From Eq.(3)
1 |"l nnx
we have bn = — | f(x)sin dx. If we let u = f(x) and
L J-l L
nnx -L nnx
dv = sin dx, then du = f (x)dx and v = —cos---------------------------------
L nn L
Thus
Section 10.4
211
1 -L nnx iL L ^ nnx
bn = — [—f(x)cos-------------- |-L + —I f (x)cos dx]
L nn L nn J-l l
1 1 Tl . nnx
= -—[f(L)cosnn - f(-L)cos(-nn)] + —I f (x)cos dx
nn nn J-L L
1 |"l . nnx
= —I f (x)cos dx, since f(L) = f(-L) and
nn J-l l
1 fL , nnx
cos(-nn) = cosnn. Hence nbn = — f (x)cos dx, which
n J-L L
exists for all n since f' (x) is piecewise continuous.
Thus nbn is bounded as n ^ ^. Likewise, for an, we
1 fL . nnx
obtain nan = - — f (x)sin dx and hence nan is also
n J-l l
bounded as n ^ ^.
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