Higher Engineering Mathematics, Sixth Edition

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.