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

Boyce W.E Elementary differential equations 7th edition - Wiley publishing , 2001. - 1310 p.
ISBN 0-471-31999-6
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- st
all t. To determine ?{cosat}
J0
cosatdt we
must integrate by parts twice to get
J0
e st cosatdt
. r , -1 -st . -2 -st .x iM
lim [(-s e cos at + as e sinat)|0
m ^ •
pM
- (a /s ) I e 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 ! 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+lb)}-(1/2i) ?{e(a-lb)}. 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.
f 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 l
converges and hence I (t + 1) dt also converges.
1
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) = JQ e stf(t)dt
= lim - — e"stf(t)|M + — f e-stf'(t)dt. Now use an m ^ • s s J0
argument similar to that given to establish Theorem 6.—.2.
27a. Make a transformation of variables with x = st and dx = sdt. Then use the definition of T(P+—) from Problem 26.
27b. From part a, ?{tn} = J~e xxndx = J“e xxn 1dx
sn+iJo sn+iJo
n! ,• _x
sn+1Jo
:J0’
e dx, using integration by
parts successively. Evaluation of the last integral yields the desired answer.
27c. From part a, ?{t 1/2} = —— J~e xx 1/2dx. Let x = y2, then
yfs Jo
2dy = x-1/2dx and thus ?{t-1/2} = , J“e"ydy.
A/a "0
— /2
27d. Use the definition of ?{t / } and integrate by parts once to get ?{t1/2} = (—/2s) ?{t-—/2}. The result follows from part c.
Section 6.2, Page 307
Problems — through —0 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.—.
4 2!
2. We have ---------- = 2--------- and thus the inverse Laplace
/ i\3 / -.-.2 + —
(s-— (s-—)
2 t
transform is 2t e , using line ——.
3s 3s 9/5 6/5
4. We have ---------- = = + using partial
s2-s-6 (s-3)(s+2) s-3 s+2
3t _2t
fractions. Thus (9/5)e + (6/5)e is the inverse
transform, from line 2.
2s+— 2s+— 2(s-—) 3
7. We have ---------- = = + , where
s2-2s+2 (s-—)2+— (s-—)2+— (s-—)2+—
106
Section 6.2
we first used the concept of completing the square (in the denominator) and then added and subtracted appropriately to put the numerator in the desired form. Lines 9 and 10 may now be used to find the desired result.
In each of the Problems 11 through 23 it is assumed that the
I.V.P. has a solution y = f(t) which, with its first two derivatives, satisfies the conditions of the Corollary to Theorem 6.2.1.
II. Take the Laplace transform of the D.E., using Eq.(1) and Eq.(2), to get
s2Y(s) - sy(0) - y'(0) - [sY(s) - y(0)] - 6Y(s) = 0.
Using the I.C. and solving for Y(s) we obtain s-2
Y(s) = . Following the pattern of Eq.(12) we have
s2-s-6
s-2 a b a(s-3)+b(s+2)
= +-- = -----. Equating like
s2-s-6 s+2 s-3 (s+2)(s-3)
powers in the numerators we find a+b = 1 and -3a + 2b = -2. Thus a = 4/5 and b = 1/5 and 4+5 1/5
Y(s) = ----- + , which yields the desired solution
s+2 s-3
using Table 6.2.1.
14. Taking the Laplace transform we have s2Y(s) - sy(0) -y'(0) -4[sY(s)-y(0)] + 4Y(s) = 0. Using the I.C. and solving for
s-3
Y(s) we find Y(s) = -----------. Since the denominator is a
2
s -4s+4
s-3
perfect square, the partial fraction form is
2
s -4s+4
a b
+ . Solving for a and b, as shown in examples of
(s-2)2 s-2
this section or in Problem 11, we find a = -1 and b = 1.
1 1
Thus Y(s) = ------- - -------, from which we find
s-2 (s-2)2
y(t) = e2t - te2t (lines 2 and 11 in Table 6.2.1).
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