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

ISBN 0-471-31999-6

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to find the value of y' at t = 5 and t = 5 + T. From Eq.

(22) we have, for t > 5,

, -(t-5)/4 -1 1 ^15~

y = e / [----- sin------ (t-5) + —cos----------(t-5)]. Thus

2\j 15 4 24

. 1 . 1 -t/4

y (5) = — and y (5+T) = —e / . Since the original 22

impulse, 8(t-5), caused a discontinuity in y' of 1/2, we

-T/4

must choose the impulse at t = 5 + T to be -e , which

is equal and opposite to y' at 5 + T.

13b. Now consider 2y" + y' + 2y = 8 (t-5) + kd(t-5-T) with

y(0) = 0, y'(0) = 0. Using the results of Example 1 we

have

" ? ,,,e-(t-5)/^^— ^ 15

\f1s

+ ^u5+T(t)e-(t-5-T)/4si^^^(t-V 15 4

e (t 5)/4 [u5 (t)sin ^ 5 (t-5) +ku5+T (t)eT/4sin^ 5_

5 4 5+T 4

y(t) = < u5(t)e / sin------ (t-5)

e (t 5)/4[u5(t)+keT/4u5+T(t)]sin-----------------------—(t-5). If

\J~15

y(t) = 0 for t > 5 + T, then 1 + keT/4 = 0, or k = -e T/4, as found in part (a).

20 20

y _ -ks

kps 1 e

e so that Y(s) =

e kps so that Y(s) = ^

k=1 k=1 s +1

20

and hence y(t) = ukp (t)sin(t-kp)

k=1

= up (t)sin(t-p) + u2p (t)sin(t-2p) + ... + u10psin(t-10p). For 0 < t < p, y(t) = 0. For p < t < 2p, y(t) = sin(t-p) = -sint. For 2p < t < 3p,

y(t) = sin(t-p) + sin(t-2p) = -sint + sint = 0. Due to the periodicity of sint, the solution will exhibit this behavior in alternate intervals for 0 < t < 20p. After t = 20p the solution remains at zero.

21b. Taking the transform and using the I.C. we have

118

Section 6.6

15 15

e-(2k-1)p

(s2 + 1)Y(s) = e-(2k-1)p so that Y(s) = -------------

k=1 k=1 s +1

15

Thus y(t) = ^ u (2k-1)p (t)sin[t-(2k-1)p]

k=1

= sin(t-p) + sin(t-3p) ... + sin(t-29p)

= -sint - sint ... -sint = -15sint.

25b. Substituting for f(t) we have

y = JQe-(t-x * 8 (x-p)sin(t-x)dx. We know that the

integration variable is always less than t (the upper limit) and thus for t < p we have x < p and thus 8(x-p) = 0. Hence y = 0 for t < p. For t > p utilize Eq.(16).

Section 6.6, Page 335

1c. Using the format of Eqs.(2) and (3) we have f*(g*h) = J0 f(t-t)(g*h)(t)dt

• t ft

J0f(t-x)[J0g(x-h)h(h)dh]dx

tt

f(t-x)g(x-h)dx]h(h)(dh).

Jh

The last double integral is obtained from the previous line by interchanging the order of the h and x integrations. Making the change of variable w = x - h on the inside integral yields

f*(g*h) = [t [fth f(t-h-w)g(w)dw]h(h)dh

00

(f*g)(t-h)h(h)dh = (f*g)*h.

f0t

0

4. It is possible to determine f(t) explicitly by using

integration by parts and then find its transform F(s). However, it is much more convenient to apply Theorem

2

6.6.1. Let us define g(t) = t and h(t) = cos2t. Then,

f(t) = J0 g(t-x)h(x)dx. Using Table 6.2.1, we have

G(s) = ?{g(t)} = 2/s3 and H(s) = ?{h(t)} = s/(s2+4). Hence, by Theorem 6.6.1, ?{f(t)} = F(s) = G(s)H(s) =

2/s2(s2+4).

Section 6.6

119

As was done in Example 1 think of F(s) as the product of

-4 2 -1

s and (s +1) which, according to Table 6.2.1, are the

transforms of t3/6 and sint, respectively. Hence, by Theorem 6.6.1, the inverse transform of F(s) is

> t

3

f(t) = (1/6) JQ (t-t) 3sinxdx.

13. We take the Laplace transform of the D.E. and apply the 2 2 2

I.C.: (s + 2s + 2)Y(s) = a/(s + a). Solving for Y(s),

2 2 2 -1 we have Y(s) = [a/(s +a )][(s + 1) + 1] , where the second

factor has been written in a convenient way by completing the square. Thus Y(s) is seen to be the product of the transforms of sinat and e-tsint respectively. Hence,

according to Theorem 6.6.1, y = J0 e-(t-xsin(t-x)sinaxdx.

15. Proceeding as in Problem 13 we obtain Y(s) =

-i -s

1-e

2 2 s +s+5/4 s(s +s+5/4)

(s+1/2) - 1/2 1-e-s 1

(s+1/2) 2+1 s (s+1/2) 2+1

where the first term is obtained by completing the square in the denominator and the second term is written as the product of two terms whose inverse transforms are known, so that Theorem 6.6.1 can be used. Note that ?-1{(1-e-s)/s} = 1 - up(t). Also note that a different

form of the same solution would be obtained by writing

-ps a bs + c

the second term as (1-e )(— + -------------) and solving

s (s+1/2) 2+1

for a, b and c. In this case ?-1{1-e-s} = S(t) - S(t-p) from Section 6.5.

17. Taking the Laplace transform, using the I.C. and solving,

22

we have Y(s) = (s+3)/(s+1)(s+2) + s/(s+a)(s+1)(s+2).

As in Problem 15, there are several correct ways the

second term can be treated in order to use the

convolution integral. In order to obtain the desired

answer, write the second term as

s a b

( + -) and solve for a and b.

s2+a2 s + 1 s+2

20. To find F(s) you must recognize the integral that

120

Section 6.6

appears in the equation as a convolution integral. Taking the transform of both sides then yields

F(s)

F(s) + K(s)F(s) = F(s), or F(s) = ---------------.

1+K(s)

121

CHAPTER 7

Section 7.1, Page 344

2. As in Example 1, let x1 = u and x2 = u', then x1 = x2 and

x2 = u” = 3sint — .5x2 — 2x^

4. In this case let x1 = u, x2 = u', x3 = u”, and x4 = u”'.

5. Let x1 = u and x2 = u ; then x1 = x2 is the first of the

desired pair of equations. The second equation is

obtained by substituting u = x2, u = x2, and u = x1 in the given D.E. The I.C. become x1(0) = u0, x2(0) = u0.

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