Higher Engineering Mathematics, Sixth Edition

Properties of Laplace transforms 591

2. Use the Laplace transform of the first deriva-
   tive to derive the transforms:
   (a)L{eat}=

1

s−a

(b)L{ 3 t^2 }=

6

s^3

3. Derive the Laplace transform of the second
   derivative from the definition of a Laplace
   transform. Hence derive the transform
   L{sinat}=

a

s^2 +a^2

4. Use the Laplace transform of the second
   derivative to derive the transforms:
   (a)L{sinhat}=

a

s^2 −a^2

(b)L{coshat}=

s

s^2 −a^2

62.4 The initial and final value

theorems

There are several Laplace transform theorems used to
simplify and interpret the solution of certain problems.
Two such theorems are the initialvalue theorem and the
final value theorem.

(a) The initialvalue theorem states:

limit

t→ 0

[f(t)]=limit

s→∞

[sL{f(t)}]

For example, iff(t)=3e^4 tthen

L{3e^4 t}=

3

s− 4

from (iii) of Table 61.1, page 584.

By the initial value theorem,

limit

t→ 0

[3e^4 t]=limit

s→∞

[

s

(

3

s− 4

)]

i.e. 3e^0 =∞

(

3

∞− 4

)

i.e. 3 = 3 , which illustrates the theorem.

Problem 7. Verify the initial value theorem for

the voltage function( 5 +2cos3t)volts, and state its

initial value.

Let f(t)= 5 +2cos3t

L{f(t)}=L{ 5 +2cos3t}=

5

s

+

2 s

s^2 + 9

from (ii) and (v) of Table 61.1, page 584.

By the initial value theorem,

limit

t→ 0

[f(t)]=limit

s→∞

[sL{f(t)}]

i.e. limit

t→ 0

[5+2cos3t]=limit

s→∞

[

s

(

5

s

+

2 s

s^2 + 9

)]

=limit

s→∞

[

5 +

2 s^2

s^2 + 9

]

i.e. 5 + 2 ( 1 )= 5 +

2 ∞^2

∞^2 + 9

= 5 + 2

i.e. 7 = 7 , which verifies the theorem in this case.

The initial value of the voltage is thus7V.

Problem 8. Verify the initial value theorem for

the function( 2 t− 3 )^2 and state its initial value.

Let f(t) =( 2 t− 3 )^2 = 4 t^2 − 12 t+ 9

Let L{f(t)}=L( 4 t^2 − 12 t+ 9 )

= 4

(

2

s^3

)

−

12

s^2

+

9

s

from (vii), (vi) and (ii) of Table 61.1, page 584.

By the initial value theorem,

limit

t→ 0

[( 2 t− 3 )^2 ]=limit

s→∞

[

s

(

8

s^3

−

12

s^2

+

9

s

)]

=limit

s→∞

[

8

s^2

−

12

s

+ 9

]

i.e. ( 0 − 3 )^2 =

8

∞^2

−

12

∞

+ 9

i.e. 9 = 9 , which verifies the theorem in this case.

The initial value of the given function is thus 9.

(b) The final value theorem states:

limit

t→∞

[f(t)]=limit

s→ 0

[sL{f(t)}]

For example, iff(t)=3e−^4 tthen:

limit

t→∞

[3e−^4 t]=limit

s→ 0

[

s

(

3

s+ 4

)]