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

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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+x

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 —3ps 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 Ï 2p the integral on the right is equal

¥ -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)

.. -2ps

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

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

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

y/Ti

+ ^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_ àÄ¯ 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

\[¿¿

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

20 20

Ó _ -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) = X 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 = j0e-(t-t * 8 (x-p)sin(t-T)dx. We know that the

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

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