The solution of simultaneous differential equations using Laplace transforms 609
Using the procedure:
(i) [s
2
L{x} − sx(0) − x
(0)] − L{x} = L{y}
(1)
[s
2
L{y} − sy(0) − y
(0)] + L{y} = −L{x} (2)
from equation (4), page 590
(ii) x(0) = 2, y(0) = −1, x (0) = 0 and y (0) = 0
hence s
2
L{x} − 2s − L{x} = L{y}
(1
)
s
2
L{y} + s + L{y} = −L{x}
(2 )
(iii) Rearranging gives:
(s
2
− 1)L{x} − L{y} = 2s
(3)
L{x} + (s
2
+ 1)L{y} = −s
(4)
Equation (3) ×(s 2 + 1) and equation (4) ×1
gives:
(s
2
+ 1)(s
2
− 1)L{x} − (s
2
+ 1)L{y}
= (s
2
+ 1)2s
(5)
L{x} + (s
2
+ 1)L{y} = −s
(6)
Adding equations (5) and (6) gives:
[(s
2
+ 1)(s
2
− 1) + 1]L{x} = (s
2
+ 1)2s − s
i.e. s
4
L{x} = 2s
3
+ s = s(2s
2
+ 1)
from which, L{x} =
s(2s 2 + 1)
s 4
=
2s 2 + 1
s 3
=
2s 2
s 3 +
1
s 3 =
2
s
+
1
s 3
(iv) Hence x = L
−1
2
s
+
1
s 3
i.e.
x = 2 +
1
2
t
2
Returning to equations (3) and (4) to determine y:
1 × equation (3) and (s 2 − 1) × equation (4) gives:
(s
2
− 1)L{x} − L{y} = 2s
(7)
(s
2
− 1)L{x} + (s
2
− 1)(s
2
+ 1)L{y}
= −s(s
2
− 1)
(8)
Equation (7) −equation (8) gives:
[−1 − (s
2
− 1)(s
2
+ 1)]L{y}
= 2s + s(s
2
− 1)
i.e.
−s
4
L{y} = s
3
+ s
and
L{y} =
s 3 + s
−s 4 = −
1
s
−
1
s 3
from which,
y = L
−1
−
1
s
−
1
s 3
i.e.
y = −1 −
1
2
t
2
Now try the following exercise
Exercise 227 Further problems on solving
simultaneous differential equations using
Laplace transforms
Solve the following pairs of simultaneous differential equations:
1. 2
dx
dt
+
dy
dt
= 5 e
t
dy
dt
− 3
dx
dt
= 5
given that when t = 0, x = 0 and y = 0.
[x = e t − t − 1 and y = 2t − 3 + 3e t ]
2. 2
dy
dt
− y + x +
dx
dt
− 5 sin t = 0
3
dy
dt
+ x − y + 2
dx
dt
− e
t
= 0
given that at t = 0, x = 0 and y = 0.
x = 5 cos t + 5 sin t − e
2t
− e
t
− 3 and
y = e 2t + 2e t − 3 − 5 sin t
3. d
2 x
dt 2 + 2x = y
d 2 y
dt 2 + 2y = x
given that at t = 0, x = 4, y = 2,
dx
dt
= 0
and
dy
dt
= 0.
x = 3 cos t + cos(
√
3 t ) and
y = 3 cos t − cos(
√
3 t )