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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.
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-2 2 nnx 8x nnx 16 nnx
= [-- (2-x )cos------ - sin------ - -cos---]
nn 2 n2n2 2 n3n3 2
4 16
=  (1+cosnn) + ---------- (1-cosnn) and thus
nn n3n3
4n2n 2(1+cosnn) + 16(1-cosnn) , nnx
4n n (1+cosnn) + 16(1-cosnn) 1
(x) = V (------------------n----------------)sin-
n3n 3
n=1
2
28b.For the cosine series(even extension) we have
2 |*1 j 1
a0 =  xdx = 
2 0 2
Section 10.4
215
2 f1 nnx
 xcos-----------dx
2 J0 2
2x nnx
[ sin---------- +
nn 2
4 nnx
cos-------]
2 2 n2n 2
2 nn
 sin +
nn 2 n2n2
4 nn
cos  2
4
22 n2n 2
so
ro
(x)  4 + ^
n=1
4cos(nn/2) + 2nnsin(nn/2) - 4 nnx
------------------------------------cos----
n2n2 2
For the sine series (odd extension) we have
- 2f1  I
20
2 Γ1 nnx -2x nnx
 I xsin--------dx  [-------cos------ +
2 ^0 2 nn 2 n2n2
4 nnx
sin------]
2 nn
----cos------ +
nn 2 n2n2
4 nn
sin------
2
so
h(x)
4sin(nn/2)  2nncos(nn/2) nnx
--------------------------------sin-----
n2n2 2
28c.
28d. The maximum error does not approach zero in either case, due to Gibb's phenomenon. Note that the coefficients in both series behave like 1/n as n ^ ro since there is an n in the numerator.
31. We have JLf(x)dx = J
f(x)dx + I Lf(x)dx. Now, if we let
-L J0
y in the first integral on the right, then
f(x)dx
f(x)dx
= I0 = -I
0f(-y)(-dy) = JLf(-y)dy
0
Lf(y)dy + JLf(x)dx
0
= -I³
f(y)dy. Thus
0.
a
n
32. To prove property 2 let f1 and f2 be odd functions and let f(x) = f1(x) ± f2(x). Then f(-x) = f1(-x) ± f2(-x) =
-f1(x) ± [-f2(x)] = -f1(x) + f2(x) = -f(x), so f(x) is
odd. Now let g(x)= f1(x)f2(x), then g(-x) = f1(-x)f2(-x) = [-f1(x)][-f2(x)] = f1(x)f2(x) = g(x)
and thus g(x) is even. Finally, let h(x) = f1(x)/f2(x) and hence h(-x) = f1(-x)/f2(-x) = [-f1(x)]/[-f2(x)] =
f1(x)/f2(x) = h(x), which says h(x) is also even.
Property 3 is proven in a similar manner.
216
Section 10.4
34. Since F(x) = xf(t)dt we have
0
F(-x) = f(t)dt = -I f(-s)ds by letting t = -s. If f
00
is an even function, f(-s) = f(s), we then have
F(-x) = - xf(s)ds = -F(x) from the original definition of 0
F. Thus F(x) is an odd function. The argument is similar if f is odd.
35. Set x = L/2 in Eq.(6) of Section 10.3. Since we know f
is continuous at L/2, we may conclude, by the Fourier
theorem, that the series will converge to
f(L/2) = L at this point. Thus we have
L = L/2 + (2L/n) Σ (-1) n+1/(2n-1), since
n=1
n+1
sin[(2n-1)n/2] = (-1) n+1. Dividing by L and simplifying
-V-, . 1 ^ -V-,
(-1)n+1 V³ (-1)n
4 ' (2n-1) ' 2n+1
n=1 n=0
37a. Multiplying both sides of the equation by f(x) and integrating from 0 to L gives
L[f(x)]2dx = I L[f(x) Σ bnsin(nnx/L)]dx
JoL[f(x)] 2dx = JQL[f(x) Σ1
n=1
= bnJLf(x)sin(nnx/L)dx = (L/2) όΟ, by Eq.(8).
n=1 n=1
This result is identical to that of Problem 17 of Section
10.3 if we set an = 0, n = 0,1,2,... , since
1 L 2 2 L 2
 [f(x)]2dx =  [f(x)]2dx . In a similar manner, it
L -L L 0
can be shown that
(2/L)|L[f(x) ] 2dx = a2/2 + Σ a°.
n=1
37b. Since f(x) = x and bn = 2L(-1) n+1/nn (from Eq.(9)), we
n
obtain
(2/L)JL[f(x)]2dx = (2/L)JLx2dx = 2L2/3 = Σ^
όΟ =
n=1
Σ4L2/n2n2 = 4L2/n2 Σ(1/n2) or ο2/6 = Σ(1/n2).
n=1 n=1 n=1
Section 10.5
217
We assume that the extensions of f and f' are piecewise continuous on [-2L,2L]. Since f is an odd periodic function of fundamental period 4L it follows from properties 2 and 3 that f(x)cos(nnx/2L) is odd and f(x)sin(nnx/2L) is even. Thus the Fourier coefficients of f are given by Eqs.(8) with L replaced by 2L; that is an = 0, n = 0,1,2,... and
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