The solution of simultaneous differential equations using Laplace transforms 607
Problem 2. Solve the following pair of
simultaneous differential equations
3
dx
dt
− 5
dy
dt
+ 2x = 6
2
dy
dt
−
dx
dt
− y = −1
given that at t = 0, x = 8 and y = 3.
Using the above procedure:
(i) 3L
dx
dt
− 5L
dy
dt
+ 2L{x} = L{6}
(1)
2L
dy
dt
− L
dx
dt
− L{y} = L{−1}
(2)
Equation (1) becomes:
3[sL{x} − x(0)] − 5[sL{y} − y(0)]
+ 2L{x} =
6
s
from equation (3), page 589, and Table 61.1,
page 584.
i.e. 3sL{x} − 3x(0) − 5sL{y}
+ 5y(0) + 2L{x} =
6
s
i.e. (3s + 2)L{x} − 3x(0) − 5sL{y}
+ 5y(0) =
6
s
(1
)
Equation (2) becomes:
2[sL{y} − y(0)] − [sL{x } − x(0)]
− L{y} = −
1
s
from equation (3), page 589, and Table 61.1,
page 584,
i.e. 2sL{y} − 2y(0) − sL{x}
+ x(0) − L{y} = −
1
s
i.e. (2s − 1)L{y} − 2y(0) − sL{x}
+ x(0) = −
1
s
(2
)
(ii) x(0) = 8 and y(0) = 3, hence equation (1 )
becomes
(3s + 2)L{x} − 3(8) − 5sL{y}
+ 5(3) =
6
s
(1 )
and equation (2
) becomes
(2s − 1)L{y} − 2(3) − sL{x }
+ 8 = −
1
s
(2 )
i.e. (3s + 2)L{x} − 5sL{y} =
6
s
+ 9
( 1 )
(3s + 2)L{x} − 5sL{y}
=
6
s
+ 9
(1 )
− sL{x} + (2s − 1)L{y}
= −
1
s
− 2
(2 )
⎫
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭
(A)
(iii) s × equation (1 ) and (3s + 2) × equation (2 )
gives:
s(3s + 2)L{x} − 5s
2
L{y} = s
6
s
+ 9
(3)
−s(3s + 2)L{x} + (3s + 2)(2s − 1)L{y}
= (3s + 2)
−
1
s
− 2
(4)
i.e. s(3s + 2)L{x} − 5s
2
L{y} = 6 + 9s (3 )
−s(3s + 2)L{x} + (6s
2
+ s − 2)L{y}
= −6s −
2
s
− 7
(4 )
Adding equations (3
) and (4
) gives:
(s
2
+ s − 2)L{y} = −1 + 3s −
2
s
=
−s + 3s 2 − 2
s
from which, L{y} =
3s 2 − s − 2
s(s 2 + s − 2)
Using partial fractions
3s 2 − s − 2
s(s 2 + s − 2)
≡
A
s
+
B
(s + 2)
+
C
(s − 1)
=
A(s + 2)(s − 1) + Bs(s − 1) + Cs(s + 2)
s(s + 2)(s − 1)