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

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22 s (s +s+5/4)

= e-«s/)i/5 16/25 + (16/25)s-4/25 1

^ s2 s (s + 1/2) 2+1 ^

-ps/2

= (1-e 'lH(s), where we have used partial fractions and completed the square in the denominator of the last term. Since the numerator of the last term of H

16

can be written as —[(s+1/2) - 3/4], we see that

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"

(2 0t—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

, , —1/8, 3\[¿ _ ó[¿ . 3^/7 ^ ï

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 < 2 p

n > 6, f(t) =

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

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

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

have Y(s) =

1---- [1 + 2 X (-1) Vpks],

2

s(s +1) -pks

since

?{upk(t)} =

e

Since

k=1

1

1 s

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 X (-1)

Upk(t) - 2

X (-1)

2k

cost

k=1

k=1

s

k

we

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

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 .05tsinbt. 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 .05(t np) cos[b(t-np) + 8], and therefore converges as t——•.

20a. y(t) for n = 30

y(t) for n = 31

2 0b. From the graph of part a, A @ 12.5 and the frequency is 2p.

2 0c. From the graph

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

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].

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