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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.
Download (direct link): elementarydifferentialequations2001.pdf
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1 sin t + cos(t nj4), t > nj4,
find L{ f (t)}. The graph of y = f (t) is shown in Figure 6.3.5. Note that f (t) = sin t + g(t), where
g(t) = ( t < nj4,
B( ) \ cos(t nj4), t > nj4.
Thus
g(t) = uXj4(t) cos(t nj4)
and
L{ f (t)} = L{sin t} + L{unj4(t) cos(t nj4)}
= L{sin t} + e~nsj4L{cos t}.
Introducing the transforms of sin t and cos t, we obtain
1 s _i_ sensj4
L{ f(t)} =-2-------+ ensj4-2--------= -+-------------.
s2 + 1 s2 + 1 s2 + 1
You should compare this method with the calculation of L{ f (t)} directly from the definition.
6.3 Step Functions
313
Find the inverse transform of
F(s)=
1 - e
-2s
From the linearity of the inverse transform we have
f(t) = L-{F(s)} = L-1 (-U - C~A
= t - u2(t)(t - 2)-The function f may also be written as
t,
f (t) =
2,
0 < t < 2, t> 2.
2s
The following theorem contains another very useful property of Laplace transforms that is somewhat analogous to that given in Theorem 6.3.1.
Theorem 6.3.2 If F(s) = L{ f (t)} exists for s > a > 0, and if c is a constant, then
L{ s > a + c. (5)
e
t
(f
)
=
(F
s
1
)c
Conversely, if f (t) = L 1{F(s)}, then
e c)}. (6)
t
(f
)
=
L
1
(F
s
1
According to Theorem 6.3.2, multiplication of f (t) by ect results in a translation of the transform F(s) a distance c in the positive s direction, and conversely. The proof of this theorem requires merely the evaluation of L{ect f (t)}. Thus
r TO
L{ect f(t)} = e~stect f(t) dt = /
Jo Jo
= F( s - c),
e
-(s-c)t
f(t) dt
2
s
2
s
s
314
Chapter 6. The Laplace Transform
EXAMPLE
4
PROBLEMS
which is Eq. (5). The restriction s > a + c follows from the observation that, according to hypothesis (ii) of Theorem 6.1.2, | f(t)l < Keat; hence |ect f(t)l < Ke(a+c)t. Equation (6) follows by taking the inverse transform of Eq. (5), and the proof is complete.
The principal application of Theorem 6.3.2 is in the evaluation of certain inverse transforms, as illustrated by Example 4.
Find the inverse transform of
G(s) =
1
s2 4s 5
By completing the square in the denominator we can write
1
G (s) =
= F(s - 2),
Previous << 1 .. 169 170 171 172 173 174 < 175 > 176 177 178 179 180 181 .. 609 >> Next