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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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24. Under the standard assumptions, the Lapace transform of
the left side of the D.E. is s2Y(s) - sy(0) - y'(0) + 4Y(s). To transform the right side we must revert to the definition of the Laplace trasnform to determine f“ -st
|0 e f(t)dt. Since f(t) is piecewise continuous we are
able to calculate ?{f(t)} by
108 Section 6. 2
/•• fP pM
e-stf(t)dt = e-st dt + lim (e-st)(0)dt
J0 J0 m —
fK
J0 e-stdt = (1 - e-Ps)/s.
Hence, the Laplace transform Y(s) of the solution is given by Y(s) = s/(s2+4) + (1 - e-Ps)/s(s2+4).
27b. The Taylor series for f about t = 0 is
f(t) = ^^(-1) n t2n/(2n+1)!, which is obtained from
n=0
part(a) by dividing each term of the sine series by t.
Also, f is continuous for t > 0 since lim (sint)/t = 1.
t —— 0 +
Assuming that we can compute the Laplace transform of f term by term, we obtain ?{f(t)} = ?{^(-1) nt2n/(2n+1)!}
[(-1) n/(2n+1)! L{t2n}
[(-1) n(2n)!/(2n + 1)!]s-(2n+1)
[(-1) n/(2n + 1)]s (2n+1, which converges for s > 1.
The Taylor series for arctan x is given by
^^(-1) n x2n+1/(2n+1), for |x| < 1. Comparing ?{f(t)} with
n=0
the Taylor series for arctanx, we conclude that ?{f(t)} = arctan(1/s), s > 1.
30. Setting n = 2 in Problem 28b, we have
?{t2sinbt} = — [b/(s2+b2)] = — [-2bs/(s2+b2) 2] =
ds2 ds
-2b/(s2+b2) 2 + 8bs2/(s2+b2) 3 = 2b(3s2-b2)/(s2+b2) 3.
32. Using the result of Problem 28a. we have
?{teat} = --d (s-a) -1 = (s-a) -2 ds
?{t2eat} = -—(s-a) -2 = 2(s-a) "3ds
?{t3eat} = --^2(s-a) -3 = 3!(s-a) "4. Continuing in this ds
Section 6.3
109
fashion, or using induction, we obtain the desired result.
36a. Taking the Laplace transform of the D.E. we obtain
?{y/,} - ?{ty} = ?{y"} + ?{-ty}
= s2Y(s) - sy(0) - y'(0) + Y'(s) = 0.
Hence, Y satisfies Y' + s2^ = s.
P(rk)
3 8a. From Eq(i) we have Ak = lim (s-rk) ------, since Q has
s^rk Q(rk)
s-rk P(rk)
distinct zeros. Thus Ak = P(rk) lim ------------------------------ = -, by
k k s^rK Q(rk) Q'(rk)
L'Hopital's Rule.
38b. Since ? 1<j s-r [ = erkt, the result follows.
Section 6.3, Page 314
2. From the definition of uc(t) we have:
g(t) = (t-3)u2(t) - (t-2)u3(t)
0 - 0 = 0, 0?t<2
(t-3) - 0 = t-3, 2?t<3.
(t-3) - (t-2) = -1, 3 ?t
‘ «Ut
I 1 J t
-1 j /
As indicated in the discussion following Example 1, the unit step function can be used to translate a given function f, with domain t>0, a distance c to the right by the multiplication uc(t)f(t-c).
Hence the required graph of y = u3(t)f(t-3) for f(t) = sint is shown.
In order to use Theorem 6.3.1 we must write f(t) in terms
22 of uc(t). Since t - 2t + 2 = (t-1) + 1 (by completing
the square), we can thus write f(t) = u1(t)g(t-1), where g(t) = t +1. Now applying Theorem 6.3.1 we have ?{f(t)} = ?{u1(t)g(t-1)} = e-s ?{g(t)} = e-s(2/s3 + 1/s).
110
Section 6.3
14.
21.
22.
27.
Use partial fractions to write
-2s -1 -1
F(s) = e [(s-1) - (s+2) ] /3. For ease in calculations
let us define G(s) = (s-1)-1 and H(s) = (s+2)-1. Then F(s) = [e- 2s G(s) - e-2s H(s)]/3. Using the fact that ?{eat} = (s-a)-1 and applying Theorem 6.3.1, we have F(s) = [e-2s ?{et} - e-2s ?{e-2t}]/3. Thus F(s) = [?{u2(t)e(t-2)} - ?{u2(t)e"2(t_2)}]/3. Using the
linearity of the Laplace transform, we have ?{f(t)} = ?{u2(t)[et-2 - e-2(t-2)]/3}. Hence,
f(t) = [u2(t)(et-2 - e"2(t"2)]/3. An alternate method is to complete the square in the denominator:
-2s
e
F(s) = -----------------. This gives
(s+1/2) 2- 9/4
3 2
to be the same as that found above.
f(t) = (2/3)u2(t)e (t2)/2 sinh—(t-2), which can be shown
22
By completing the square in the denominator of F we can write F(s) = (2s+1)/[(2s+1) + 4]. This has the form
G(2s+1) where G(u) = u/(u +4). We must find ?-1{G(2s+1)}. Applying the results of Problem 19(c), we
-1r -³ 1 -t/2 2t
have ? {F(s)} = —e / cos(—).
22
If the approach of Problem 21 is used we find
f(t) = (1/3)e2t/3 sinh(t/3), which is equivalent to the given answer using the definition of sinht.
Assuming that term-by-term integration of the infinite series is permissible and recalling that ?{uc(t)} = e-cs/s
for s > 0, we have ?{f(t)} = (1/s) + (-1)k ?{uk(t)}
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