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

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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 ³ _i_ se—nsj4

L{ f(t)} =-2-------+ e—nsj4-2--------= -+-------------.

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 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 ï Ñ

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),

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