608 Higher Engineering Mathematics
i.e. 3s
2
− s − 2 = A(s + 2)(s − 1)
+ Bs(s − 1) + Cs(s + 2)
When s = 0, −2 =−2 A, hence A =1
When s = 1, 0 = 3C, hence C = 0
When s =−2, 12 =6B, hence B =2
Thus L{y} =
3s 2 − s − 2
s(s 2 + s − 2)
=
1
s
+
2
(s + 2)
(iv) Hence y = L
−1
1
s
+
2
s + 2
= 1 +2e
−2t
Returning to equations (A) to determine L{x} and
hence x:
(2s − 1) × equation (1 ) and 5s × (2 ) gives:
(2s − 1)(3s + 2)L{x} − 5s(2s − 1)L{y}
= (2s − 1)
6
s
+ 9
(5)
and −s(5s)L{x} + 5s(2s − 1)L{y}
= 5s
−
1
s
− 2
(6)
i.e. (6s
2
+ s − 2)L{x} − 5s(2s − 1)L{y}
= 12 + 18s −
6
s
− 9
( 5 )
and − 5s
2
L{x} + 5s(2s − 1)L{y}
= −5 − 10s
(6 )
Adding equations (5
) and (6
) gives:
(s
2
+ s − 2)L{x} = −2 + 8s −
6
s
=
−2s + 8s 2 − 6
s
from which, L{x} =
8s 2 − 2s − 6
s(s 2 + s − 2)
=
8s
2
− 2s − 6
s(s + 2)(s − 1)
Using partial fractions
8s 2 − 2s − 6
s(s + 2)(s − 1)
≡
A
s
+
B
(s + 2)
+
C
(s − 1)
=
A(s + 2)(s − 1) + Bs(s − 1) + Cs(s + 2)
s(s + 2)(s − 1)
i.e. 8s
2
− 2s − 6 = A(s + 2)(s − 1)
+ Bs(s − 1) + Cs(s + 2)
When s = 0, −6 =−2 A, hence A = 3
When s = 1, 0 =3C, hence C = 0
When s =−2, 30 = 6B, hence B = 5
Thus L{x} =
8s 2 − 2s − 6
s(s + 2)(s − 1)
=
3
s
+
5
(s + 2)
Hence x = L
−1
3
s
+
5
s + 2
= 3 + 5e
−2t
Therefore the solutions of the given simultaneous differential equations are
y = 1 +2e
−2t and x = 3 +5e
−2t
(These solutions may be checked by substituting the
expressions for x and y into the original equations.)
Problem 3. Solve the following pair of
simultaneous differential equations
d
2 x
dt 2 − x = y
d
2 y
dt 2 + y = −x
given that at t = 0, x = 2, y =−1,
dx
dt
= 0
and
dy
dt
= 0.