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

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
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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
j0 f(t-t)[j0g(t-h)h(h)dh]dt
tt
f(t-t)g(t-h)dt]h(h)(dh).
j0t[j
Jh
The last double integral is obtained from the previous line by interchanging the order of the h and t integrations. Making the change of variable w = t - h on the inside integral yields
f*(g*h) = jt [jth f(t-h-w)g(w)dw]h(h)dh
00
(f*g)(t-h)h(h)dh = (f*g)*h.
jt
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-t)h(t)dt. 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-tsin(t-x)sinaxdx.
15. Proceeding as in Problem 13 we obtain Y(s) =
-s
1-e
22 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.
8. Follow the steps outlined in Problem 7. Solve the first
3 1 '
D.E. for x2 to obtain x2 = —x1 - — x1. Substitute this
// /
into the second D.E. to obtain x”1 - x1 - 2x1 = 0, which
2t -t
has the solution x1 = c1e + c2e . Differentiating this and substituting into the above equation for x2 yields x2
1 2t -t
= -c,e + 2c2e . The I.C. then give
2 1 2
11 ci + c2 = 3 and — c1 + 2c2 = —, which yield
11 2 11 2t 2 -t
c = —, c2 = -—. Thus x1 = —e - — e and
1 3 2 3 1 3 3
11 2t 4 -t
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