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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 6.1, Page 298
2.
The graph of f(t) is shown. Since the function is continuous on each interval, but has a jump discontinuity at t = 1, f(t) is piecewise continuous.
Note that lim t^1+
(t-1)
5b. Since t is continuous for 0 ? t ? A for any positive A and since t2 ? eat for any a > 0 and for t sufficiently large, it follows from Theorem 6.1.2 that ?{t2} exists
λΜ
{t2} = I e stt2dt = lim I J0 m ^  J0
for s > 0. ?
st 2
e t dt
-t -st M 2
lim [--------e L + 
0
m ^  s
2 1 =  lim [- te
s m ^  s
2 1 _ =  lim - e 2m ^  s
/M
J0 e sttdt]
1 J
sJC
stM 1 I -stn.n
0 + - J0e dt]
st| M |0
2
ss That f(t) = cosat satisfies the hypotheses of Theorem 6.1.2 can be verified by recalling that |cosat| ? 1 for
st
all t. To determine ?{cosat}
I0
cosatdt we
must integrate by parts twice to get
I0
e st cosatdt
. γ , -1 -st . -2 -st .X iM
lim [(-s e cos at + as e sinat)|0
m ^ 
pM
- (a /s ) I0e s cos at dt]. Evaluating the first two terms, letting M ^ , and adding the third term to both
1/s, s > 0.
sides, we obtain [1 + a2/s2]J0
e-st cos at dt
22
Division by [1 + a /s ] and simplification yields the desired solution.
s
9. From the definition for coshbt we have
?{eatcoshbt} = ?{1[e(a+b)t + e(a-b)tj}. Using the linearity
2
104
Section 6.1
13.
16.
19.
21.
property of ?, Eq.(5), the right side becomes
?{e(a+b)} + ?{e(a-b)} which can be evaluated using the 22
result of Example 5 and thus
r at r 1/2 1/2
?{e coshbt} = ------------ +
s-(a+b) s-(a-b)
s-a
(s-a)2-b2
for s-a >
We write sinat = (eiat - e iat)/2i, then the linearity of the Laplace transform operator allows us to write
?{eatsinbt} = (1/2i) ?{e(a+ib) }-(1/2i) ?{e(a-ib)}. Each of these two terms can be evaluated by using the result of Example 5, where we now have to require s to be greater than the real part of the complex numbers a ± ib in order for the integrals to converge. Complex algebra then gives the desired result. An alternate method of evaluation would be to use integration on the integral appearing in the definition of ?{eatsinbt}, but that method requires integration by parts twice.
As in Problem 13,
?{tsinat} = (1/2i) ?{teiat} - (1/2i) ?{te-iat}. Using the result of Problem 15 we obtain
?{tsinat} = (1/2i)[(s-b) -2 - (s+b) -2] where b = ia and s > 0. Hence ?{tsinat} = 2as/(s2+a2)2, s > 0.
Use the approach shown in Problem 16 with the result of Problem 18, for n = 2. A computer algebra system may also be used.
P A
I 2 -1
The integral I (t + 1) dt can be evaluated in terms of
the arctan function and then Eq. (3) can be used. To illustrate Theorem 6.1.1, however, consider that
11 f -2
----- <  for t > 1 and, from Example 3, t dt
t2+1 t2 j1
f 2 ³
converges and hence I (t + 1) dt also converges.
₯1
I 2 -1
I0 (t + 1) dt is finite and hence does not affect the
f  2 -1 convergence of I (t + 1) dt at infinity.
Section 6.2
105
25. If we let u = f and dv = e stdt then F(s) = J0 e stf(t)dt
= lim -  e"stf(t)|M +  Γ e"stf'(t)dt. Now use an m ^  s s J0
argument similar to that given to establish Theorem 6.1.2.
27a. Make a transformation of variables with x = st and dx = sdt. Then use the definition of Γ(Π+1) from Problem 26.
27b. From part a, ?{tn} = ----------fe"xxndx = ------fe"xxn-1dx
sn+1 0 sn+1 0
n! , _x
sn+1 0
e dx, using integration by
parts successively. Evaluation of the last integral yields the desired answer.
2 7c. From part a, ?{t 1/2} = ,_ fe xx 1/2dx. Let x = y2, then
yfs Jo
2dy = x-1/2dx and thus ?{t-1/2} = ,_ fe"ydy.
λ/ΰ "0
1/2
2 7d. Use the definition of ?{t / } and integrate by parts once to get ?{t1/2} = (1/2s) ?{t-1/2}. The result follows from part c.
Section 6.2, Page 307
Problems 1 through 10 are solved by using partial fractions and algebra to manipulate the given function into a form matching one of the functions appearing in the middle column of Table 6.2.1.
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