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 )=0andy′( 0 )= 0
hence s^2 L{x}− 2 s−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}= 2 s (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 ) 2 s (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 ) 2 s−s
i.e. s^4 L{x}= 2 s^3 +s=s( 2 s^2 + 1 )
from which, L{x}=
s( 2 s^2 + 1 )
s^4
=
2 s^2 + 1
s^3
=
2 s^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 deter-
miney:
1 ×equation (3) and (s^2 − 1 )×equation (4) gives:
(s^2 − 1 )L{x}−L{y}= 2 s (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}
= 2 s+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 differ-
ential equations:
- 2
dx
dt
+
dy
dt
=5et
dy
dt
− 3
dx
dt
= 5
given that whent=0,x=0andy= 0.
[x=et−t−1andy= 2 t− 3 + 3 et]
- 2
dy
dt
−y+x+
dx
dt
−5sint= 0
3
dy
dt
+x−y+ 2
dx
dt
−et= 0
given that att=0,x=0andy= 0.
[
x=5cost+5sint−e^2 t−et−3and
y=e^2 t+ 2 et− 3 −5sint
]
- d
(^2) x
dt^2
- 2 x=y
d^2 y
dt^2 - 2 y=x
given that att=0,x=4,y=2,
dx
dt
= 0
and
dy
dt
= 0.
[
x=3cost+cos(
√
3 t)and
y=3cost−cos(
√
3 t)
]