1.3 Solutions
67
1.3.9 Laplace Transforms
1.72
dN A (t)
dt
= −λ A N A (t)
( 1 )
dN B (t)
dt
= −λ B N B (t) + λ A N A (t)
( 2 )
Applying Laplace transform to (1)
sL(N A ) − N A (0) = −λ A L(N A )
or L(N A ) =
N
0
A
s + λ A
=
N
0
A
s − (−λ A )
(3)
∴ N A = N
0
A exp(–λ A t)
( 4 )
Applying the Laplace transform to (2)
sL(N B ) − N B (0) = −λ B L(N B ) + λ A L(N A )
( 5 )
Using (3) in (4) and putting N 2 (0) = 0
L(N B )(s + λ B ) =
λ A N
0
A
s + λ A
or L(N B ) =
λ A N
0
A
(s + λ A )(s + λ B )
=
λ A N
0
A
λ B − λ A
1
s + λ A
−
1
s + λ B
=
λ A N
0
A
λ B − λ A
1
s − (−λ A )
−
1
s − (−λ B )
∴ N B =
λ A N
0
A
λ B − λ A
e
−λAt
− e
−λB t
1.73
dN A
dt
= −λ A N A
(1)
dN B
dt
= −λ B N B + λ A N A
(2)
dN C
dt
= +λ B N B
(3)
Applying the Laplace transform to (3)
sL{N C } − N C (0) = λ B L{N B } =
λ B λ A N
0
A
(s + λ A )(s + λ B )
Given N c (0) = 0
L{N c } =
λ A λ B N
0
A
s(s + λ A )(s + λ B )
=
λ A λ B N
0
A
(λ B − λ A )s
1
s + λ A
−
1
s + λ B