606 Higher Engineering Mathematics
L
dx
dt
− L{y} + 4L{e
t
} = 0
( 2 )
Equation (1) becomes:
[sL{y} − y(0)] + L{x} =
1
s
(1 )
from equation (3), page 589 and Table 61.1,
page 584.
Equation (2) becomes:
[sL{x} − x(0)] − L{y} = −
4
s − 1
(2 )
(ii) x(0) = 0 and y(0) = 0 hence
Equation (1
) becomes:
sL{y} + L{x} =
1
s
(1 )
and equation (2 ) becomes:
sL{x} − L{y} = −
4
s − 1
or −L{y} + sL{x} = −
4
s − 1
(2 )
(iii) 1 × equation (1 ) and s × equation (2 ) gives:
sL{y} + L{x} =
1
s
(3)
−sL{y} + s
2
L{x} = −
4s
s − 1
(4)
Adding equations (3) and (4) gives:
(s
2
+ 1)L{x} =
1
s
−
4s
s − 1
=
(s − 1) − s(4s)
s(s − 1)
=
−4s 2 + s − 1
s(s − 1)
from which, L{x} =
−4s 2 + s − 1
s(s − 1)(s 2 + 1)
(5)
Using partial fractions
−4s 2 + s − 1
s(s − 1)(s 2 + 1)
≡
A
s
+
B
(s − 1)
+
Cs + D
(s 2 + 1)
=
A(s − 1)(s 2 + 1) + Bs(s 2 + 1)
+ (Cs + D)s(s − 1)
s(s − 1)(s 2 + 1)
Hence
−4s
2
+ s − 1 = A(s − 1)(s
2
+ 1) + Bs(s
2
+ 1)
+ (Cs + D)s(s − 1)
When s = 0, −1 =−A hence A = 1
When s = 1, −4 = 2B hence B =−2
Equating s 3 coefficients:
0 = A + B + C hence C = 1
(since A = 1 and B =−2)
Equating s 2 coefficients:
−4 = −A + D − C hence D =−2
(since A = 1 and C = 1)
Thus L{x} =
−4s 2 + s − 1
s(s − 1)(s 2 + 1)
=
1
s
−
2
(s − 1)
+
s − 2
(s 2 + 1)
(iv) Hence
x = L
−1
1
s
−
2
(s − 1)
+
s − 2
(s 2 + 1)
= L
−1
1
s
−
2
(s − 1)
+
s
(s 2 + 1)
−
2
(s 2 + 1)
i.e. x = 1 −2e t + cos t − 2 sin t,
from Table 63.1, page 593
From the second equation given in the question,
dx
dt
− y + 4 e
t
= 0
from which,
y =
dx
dt
+ 4 e
t
=
d
dt
(1 − 2 e
t
+ cos t − 2 sin t ) + 4 e
t
= −2 e
t
− sin t − 2 cost + 4 e
t
i.e. y = 2e
t
− sin t − 2 cos t
[Alternatively, to determine y, return to
equations (1 ) and (2 )]