590 Higher Engineering Mathematics
(b) Second derivative
Let the second derivative of f (t ) be f (t ), then from
equation (1),
L{ f
(t )} =
∞
0
e
−st f
(t ) dt
Integrating by parts gives:
∞
0
e
−st f
(t ) dt =
e
−st f
(t )
∞
0
+ s
∞
0
e
−st f
(t ) dt
= [0 − f
(0)] + sL{ f
(t )}
assuming e −st f (t ) → 0 as t → ∞, and f (0) is the
value of f
(t ) at t = 0. Hence
{ f (t )} = − f (0) + s[s( f (t )) − f (0)], from equation (3),
i.e.
L{ f (t)}
= s 2 L{ f (t)} −sf (0)− f (0)
or L
d 2 y
dx 2
= s
2
L{ y} −sy(0) − y
(0)
⎫
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭
(4)
where y (0) is the value of
dy
dx
at x = 0.
Equations (3) and (4) are important and are used in
the solution of differential equations (see Chapter 64)
and simultaneous differential equations (Chapter 65).
Problem 5. Use the Laplace transform of the first
derivative to derive:
(a) L{k} =
k
s
(b) L{2t } =
2
s 2
(c) L{e −at } =
1
s + a
From equation (3), L{ f
(t )} =sL{ f (t )} − f (0).
(a) Let f (t ) = k, then f (t ) = 0 and f (0) = k.
Substituting into equation (3) gives:
L{0} = sL{k} − k
i.e.
k = sL{k}
Hence L{k} =
k
s
(b) Let f (t ) = 2t then f (t ) = 2 and f (0) = 0.
Substituting into equation (3) gives:
L{2} =sL{2t } − 0
i.e.
2
s
= sL{2t }
Hence L{2t}=
2
s 2
(c) Let f (t ) = e −at then f (t ) =−ae −at and f (0) = 1.
Substituting into equation (3) gives:
L{−ae
−at
} = sL{e
−at
} − 1
−aL{e
−at
} = sL{e
−at
} − 1
1 = sL{e
−at
} + aL{e
−at
}
1 = (s + a)L{e
−at
}
Hence L{e
−at
} =
1
s + a
Problem 6. Use the Laplace transform of the
second derivative to derive
L{cos at } =
s
s 2 + a 2
From equation (4),
L{ f
(t )} = s
2
L{ f (t )} − s f (0) − f
(0)
Let f (t ) = cos at , then f (t ) = −a sin at and
f (t ) = −a 2 cosat , f (0) = 1 and f (0) = 0
Substituting into equation (4) gives:
L{−a 2 cos at } = s 2 {cos at } − s(1) − 0
i.e.
−a 2 L{cos at } = s 2 L{cos at } − s
Hence
s = (s 2 + a 2 )L{cos at }
from which, L{cos at } =
s
s 2 + a 2
Now try the following exercise
Exercise 221 Further problems on the
Laplace transforms of derivatives
1. Derive the Laplace transform of the first
derivative from the definition of a Laplace
transform. Hence derive the transform
L{1} =
1
s
(b) Second derivative
Let the second derivative of f (t ) be f (t ), then from
equation (1),
L{ f
(t )} =
∞
0
e
−st f
(t ) dt
Integrating by parts gives:
∞
0
e
−st f
(t ) dt =
e
−st f
(t )
∞
0
+ s
∞
0
e
−st f
(t ) dt
= [0 − f
(0)] + sL{ f
(t )}
assuming e −st f (t ) → 0 as t → ∞, and f (0) is the
value of f
(t ) at t = 0. Hence
{ f (t )} = − f (0) + s[s( f (t )) − f (0)], from equation (3),
i.e.
L{ f (t)}
= s 2 L{ f (t)} −sf (0)− f (0)
or L
d 2 y
dx 2
= s
2
L{ y} −sy(0) − y
(0)
⎫
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭
(4)
where y (0) is the value of
dy
dx
at x = 0.
Equations (3) and (4) are important and are used in
the solution of differential equations (see Chapter 64)
and simultaneous differential equations (Chapter 65).
Problem 5. Use the Laplace transform of the first
derivative to derive:
(a) L{k} =
k
s
(b) L{2t } =
2
s 2
(c) L{e −at } =
1
s + a
From equation (3), L{ f
(t )} =sL{ f (t )} − f (0).
(a) Let f (t ) = k, then f (t ) = 0 and f (0) = k.
Substituting into equation (3) gives:
L{0} = sL{k} − k
i.e.
k = sL{k}
Hence L{k} =
k
s
(b) Let f (t ) = 2t then f (t ) = 2 and f (0) = 0.
Substituting into equation (3) gives:
L{2} =sL{2t } − 0
i.e.
2
s
= sL{2t }
Hence L{2t}=
2
s 2
(c) Let f (t ) = e −at then f (t ) =−ae −at and f (0) = 1.
Substituting into equation (3) gives:
L{−ae
−at
} = sL{e
−at
} − 1
−aL{e
−at
} = sL{e
−at
} − 1
1 = sL{e
−at
} + aL{e
−at
}
1 = (s + a)L{e
−at
}
Hence L{e
−at
} =
1
s + a
Problem 6. Use the Laplace transform of the
second derivative to derive
L{cos at } =
s
s 2 + a 2
From equation (4),
L{ f
(t )} = s
2
L{ f (t )} − s f (0) − f
(0)
Let f (t ) = cos at , then f (t ) = −a sin at and
f (t ) = −a 2 cosat , f (0) = 1 and f (0) = 0
Substituting into equation (4) gives:
L{−a 2 cos at } = s 2 {cos at } − s(1) − 0
i.e.
−a 2 L{cos at } = s 2 L{cos at } − s
Hence
s = (s 2 + a 2 )L{cos at }
from which, L{cos at } =
s
s 2 + a 2
Now try the following exercise
Exercise 221 Further problems on the
Laplace transforms of derivatives
1. Derive the Laplace transform of the first
derivative from the definition of a Laplace
transform. Hence derive the transform
L{1} =
1
s