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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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2s-4 2s-4 2(s-1) 2
15. Note that Y(s) = ------------=----------=------------------ .
s2-2s-2 (s-1) 2-3 (s-1) 2-3 (s-1) 2-3
Three formulas in Table 6.2.1 are now needed: F(s-c) in
line 14 in conjunction with the ones for coshat and
sinhat, lines 7 and 8.
Section 6.2
107
17. The Laplace transform of the D.E. is
s4Y(s) - s3y(0) - s2y'(0) - sy"(0) - y"'(0) - 4[s3Y(s)-s2y(0)
-sy'(0) - y"(0)] + 6[s2Y(s) - sy(0) - y'(0)] - 4[sY(s) - y(0)] +Y(s) = 0. Using the I.C. and solving for Y(s) we find 2
s2 - 4s + 7
Y(s) = ---------------. The correct partial fraction
432
s4-4s3+6s2-4s+1
a b c d
form for this is
(s-1) 4 (s-1) 3 (s-1) 2 s 1
Setting this equal to Y(s) above and equating the
22 numerators we have s -4s+7 = a + b(s-1) + c(s-1) +
d(s-1)3. Solving for a,b,c, and d and use of Table 6.2.1
yields the desired solution.
20. The Laplace transform of the D.E. is
s2Y(s) - sy(0) - y'(0) + w 2Y(s) = s/(s2+4). Applying the
2 2 2
I.C. and solving for Y(s) we get Y(s) = s/[(s +4)(s +w )] 22
+ s/(s +w ). Decomposing the first term by partial
fractions we have
s s s
Y(s) =-------------------------------------+ --------
(w2-4)(s2+4) (w2-4)(s2+w2) s2+w2
2
2 -1 (w -5)s s
= (w2-4) 1[----------- + ].
2 2 2
s2 +w 2 s2 +4
Then, using Table 6.1.2, we have
2 -1 2 y = (w -4) [(w -5)coswt + cos2t].
2
22. Solving for Y(s) we find Y(s) = 1/[(s-1) + 1] +
2
1/(s+1)[(s-1) + 1]. Using partial fractions on the second term we obtain Y(s) = 1/[(s-1) 2 + 1] + {1/(s + 1) - (s-3)/[(s-1) 2 + 1]}/5 = (1/5){(s + 1) -1-(s-1)[(s-1) 2 + 1] -1 + 7[(s-1) 2 + 1] -1}. Hence, y = (1/5)(e-t - etcost + 7etsint).
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
Jo J m ^ J*
i*p
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} = --d (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' + s^Y = 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
tUI 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
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