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

ISBN 0-471-31999-6

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25

?-1{H(s)} = (4/25)(5t - 4 + 4e-t/2cost - 3e-t/2sint),

which yields the desired solution. The graph of the forcing function is a ramp (f(t) = t) for 0 < t < p/2 and a constant (f(t) = p/2) for t > p/2. The solution will be a damped sinusoid oscillating about the "ramp" (20t-16)/25 for 0 < t < p/2 and oscillating about 2p/5 for t > p/2.

10. Note that g(t) = sint - up(t)sint = sint + up(t)sin(t-p). Proceeding as in Problem 8 we find

- ps 1

Y(s) = (1+e )------------------. The correct partial

22 (s +1)(s +s+5/4)

as+b cs+d

fraction expansion of the quotient is -------------- + ,

s2+1 s2+s+5/4

where

a+c = 0, a+b+d = 0, (5/4)a+b+c = 0 and (5/4)b+d = 1 by

equating coefficients. Solving for the constants yields the desired solution.

16b. Taking the Laplace transform of the D.E. we obtain

U(s2+s/4+1) = k(e 3s/2-e 5s/2)/s, since the I.C. are zero. Solving for U and using partial fractions yields

-3s/2 -5s/2 1 s +1/4

U(s) = k(e -e )(-). Thus, if

s s2 +s/4 +1

1 s +1/4

H(s) = (---------------), then

s s2+s/4+1

, , -t/8, 3\fl _ /7 . 3^7^, n

h(t) = 1 - e (cos 1+ sin 1) and

8 21 8

u(t) = ku3/2(t)h(t-3/2) - ku5/2(t)h(t-5/2).

16c. In all cases the plot will be zero for 0 < t < 3/2. For 3/2 < t < 5/2 the plot will be the system response

114

Section 6.4

(damped sinusoid) to a step input of magnitude k. For t > 5/2, the plot will be the system response to the

I.C. u(5-/2), u'(5_/2) with no forcing function. The graph shown is for k = 2.

Varying k will just affect the amplitude. Note that the amplitude never reaches 2, which would be the steady state response for the step input 2u3/2(t).

Note also that the solution and its derivative are continuous at t = 5/2.

19a. The graph on 0 ? t < 6p will depend on how large n is.

For instance, if n = 2 then

1 0 ? t < p, 2p ? t < 6p

f(t) =

For

-1 -p ? t < 2p

n > 6, f(t) =

1 0 ? t < P, 2p ? t < 3P, 4P ? t < 5P

-1 p ? t < 2 p, 3p ? t <4p, 5p t < 6 p

19b. Taking the Laplace transform of the D.E. and using the I.C.

have Y(s) =

1 [1 + 2 ? (-1) VPks],

/ 2 s(s +1)

-pks

since

?{upk(t)} =

e

Since

k=1

1

1s

s

2

-, we then obtain

s(s +1)

s +1

y(t) = 1 - cost + 2^^ (-1) kupk(t)[1 - cos(t-pk)], using line

k=1

13 in Table 6.2.1.

k

19d. Since cos(t-pk) = (-1) cost, the solution in part b can be written as

y(t) = 1 - cost + 2 ? (-1)

Upk(t) - 2

? (-1)

2k

cost

k=1

k=1

s

k

we

= 1 - cost - 2ncost + 2 (-1) kupk(t) which diverges for n?w.

k=1

Section 6.4

115

20. In this case

1

Y(s) = ----------------

2

s(s +.1S+1) fractions we have 1

H(s) = ----------------

2

[1 + 2 (-1) ke pks]. Using partial

n=1

s(s +.1s+1)

1 s+.1

s 2

s s +.1s+1 1 s+.05

s

.05

2 2 2 2 (s+.05) +b (s+.05) +b

, where

2

b = [1-(.05) ] = .9975. Now let

h(t) = ?-1{H(s)} = 1 - e .°5tcosbt - e .°5tsinbt. Hence,

b

y(t) = h(t) + 2 (-1) kupk(t) h(t-pk), and thus, for t > n the

n=1

solution will be approximated by

±1 - Ae .°5(t np) cos[b(t-np) + 8], and therefore converges as t.

20a. y(t) for n = 30

y(t) for n = 31

20b. From the graph of part a, A @ 12.5 and the frequency is 2p.

20c. From the graph

(or analytically) A = 10 and the frequency is 2 p.

116

Section 6.5

Section 6.5, Page 328

1. Proceeding as in Example 1, we take the Laplace transform of the D.E. and apply the I.C.:

(s2 + 2 s + 2 ) Y ( s ) = s + 2 + e-ps. Thus,

Y(s) = (s+2)/[(s + 1) 2 + 1] + e ps/[(s + 1) 2 + 1]. We write

22 the first term as (s + 1)/[(s + 1) + 1] + 1/[(s + 1) + 1].

Applying Theorem 6.3.1 and using Table 6.2.1, we obtain the solution, y = e-tcost + e-tsint - up(t)e (t p)sint.

Note that sin(t-p) = -sint.

3. Taking the Laplace transform and using the I.C. we have

-10s

2 1 - 5s e

(s + 3s+2)Y(s) = + e + ------------. Thus

2s

1/2 e-5s -10s 1/2 1/2 1

Y(s) = ---------- + + e (----------------1---------------) and hence

s2+3s+2 s2+3s+2 s s+2 s+1

y(t) = h(t) + u5(t)h(t-5) + u10(t)[ + ie-2(t-10)-e-(t-10)]

2 5 10 2 2

where h(t) = e t - e 2t.

5. The Laplace transform of the D.E. is

(s2+2s+3)Y(s) = ------- + e 3ps, so

s2+1

1 -3 ps 1

Y(s) = ------------------ + e [------]. Using partial

(s2+1)(s2+2s+3) s2+2s+3

fractions or a computer algebra system we obtain

1 . 1 1 -t r- 1

4 4 . 4 _ _ .

where h(t) = e tsi^V"2t.

y(t) = sint - cost + e cosy 2 t + ; u3p(t)h(t-3p),

7. Taking the Laplace transform of the D.E. yields

(s2+1)Y(s) - y'(0) = J0 e"st5 (t-2p)costdt. Since

S(t-2p) = 0 for t n 2p the integral on the right is equal

f -st 2 ps

to I e o(t-2p) costdt which equals e cos2p from

Eq.(16). Substituting for y'(0) and solving for Y(s)

-2 ps

1e

gives Y(s) = ---- + ------ and hence

22 s2 +1 s2 +1

[sint 0 ? t < 2p y(t) = sint + u2p (t)sin(t-2p) = <!

[ 2sint 2p ? t

10. See the solution for Problem 7.

Section 6.5

117

13a. From Eq. (22) y(t) will complete one cycle when \J15 (t-5)/4 = 2p or T = t - 5 = 8p/V 15 , which is consistent with the plot in Fig. 6.5.3. Since an impulse causes a discontinuity in the first derivative, we need

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