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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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Section 10.3
209
12b.
14.
12 V (-1)
en(x) = f(x) + > ---3sinknx. These errors will be
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 = bmsinmt 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 - 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/n)Jndx - (1/n)jn_2ndx = 0;
an = (1/n) l^cosnxdx - (1/) 12ncosnxdx = 0; and
J0 Jk
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 2-1 = (4/n)/(2n-1) in this example. Substituting these into the answer to Problem 14 yields the desired solution. Note that we have asumed 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 + (4/2)
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 = I f(x)sin--------dx. If we let u = f(x) and
L -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
1 1 , 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
1 . nnx
cos(-nn) = cosnn. Hence nbn = f (x)cos------------------dx, which
J-L L
exists for all n since f' (x) is piecewise continuous.
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