# Elementary differential equations 7th edition - Boyce W.E

ISBN 0-471-31999-6

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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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