608 Higher Engineering Mathematics
i.e. 3 s^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= 3 C, hence C= 0
When s=−2, 12= 6 B, hence B= 2
Thus L{y}=
3 s^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−^2 t
Returning to equations (A) to determineL{x}and
hencex:
( 2 s− 1 )×equation (1′′′)and5s×( 2 ′′′)gives:
( 2 s− 1 )( 3 s+ 2 )L{x}− 5 s( 2 s− 1 )L{y}
=( 2 s− 1 )
(
6
s
+ 9
)
(5)
and −s( 5 s)L{x}+ 5 s( 2 s− 1 )L{y}
= 5 s
(
−
1
s
− 2
)
(6)
i.e.( 6 s^2 +s− 2 )L{x}− 5 s( 2 s− 1 )L{y}
= 12 + 18 s−
6
s
−9( 5 ′)
and − 5 s^2 L{x}+ 5 s( 2 s− 1 )L{y}
=− 5 − 10 s (6′)
Adding equations (5′)and(6′)gives:
(s^2 +s− 2 )L{x}=− 2 + 8 s−
6
s
=
− 2 s+ 8 s^2 − 6
s
from which, L{x}=
8 s^2 − 2 s− 6
s(s^2 +s− 2 )
=
8 s^2 − 2 s− 6
s(s+ 2 )(s− 1 )
Using partial fractions
8 s^2 − 2 s− 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. 8 s^2 − 2 s− 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= 3 C, hence C= 0
When s=−2, 30= 6 B, hence B= 5
Thus L{x}=
8 s^2 − 2 s− 6
s(s+ 2 )(s− 1 )
=
3
s
+
5
(s+ 2 )
Hence x=L−^1
{
3
s
+
5
s+ 2
}
= 3 +5e−^2 t
Therefore the solutions of the given simultaneous dif-
ferential equations are
y= 1 +2e−^2 t and x= 3 +5e−^2 t
(These solutions may be checked by substituting the
expressions forxandyinto 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 att=0,x=2,y=−1,
dx
dt
= 0
and
dy
dt
=0.