# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

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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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