Higher Engineering Mathematics, Sixth Edition

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

+ 2 x= 6

2

dy

dt

−

dx

dt

−y=− 1

given that att=0,x=8andy=3.

Using the above procedure:

(i) 3 L

{

dx

dt

}

− 5 L

{

dy

dt

}

+ 2 L{x}=L{ 6 } (1)

2 L

{

dy

dt

}

−L

{

dx

dt

}

−L{y}=L{− 1 } (2)

Equation (1) becomes:

3[sL{x}−x( 0 )]−5[sL{y}−y( 0 )]

+ 2 L{x}=

6

s

from equation (3), page 589, and Table 61.1,

page 584.

i.e. 3 sL{x}− 3 x( 0 )− 5 sL{y}

+ 5 y( 0 )+ 2 L{x}=

6

s

i.e.( 3 s+ 2 )L{x}− 3 x( 0 )− 5 sL{y}

+ 5 y( 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. 2 sL{y}− 2 y( 0 )−sL{x}

+x( 0 )−L{y}=−

1

s

i.e.( 2 s− 1 )L{y}− 2 y( 0 )−sL{x}

+x( 0 )=−

1

s

(2′)

(ii) x( 0 )=8andy( 0 )=3, hence equation (1′)
becomes
( 3 s+ 2 )L{x}− 3 ( 8 )− 5 sL{y}

+ 5 ( 3 )=

6

s

(1′′)

and equation (2′) becomes

( 2 s− 1 )L{y}− 2 ( 3 )−sL{x}

+ 8 =−

1

s

(2′′)

i.e.( 3 s+ 2 )L{x}− 5 sL{y}=

6

s

+9(1′′)

( 3 s+ 2 )L{x}− 5 sL{y}

=

6

s

+ 9 ( 1 ′′′)

−sL{x}+( 2 s− 1 )L{y}

=−

1

s

− 2 ( 2 ′′′)

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(A)

(iii) s×equation (1′′′)and( 3 s+ 2 )×equation (2′′′)

gives:

s( 3 s+ 2 )L{x}− 5 s^2 L{y}=s

(

6

s

+ 9

)

(3)

−s( 3 s+ 2 )L{x}+( 3 s+ 2 )( 2 s− 1 )L{y}

=( 3 s+ 2 )

(

−

1

s

− 2

)

(4)

i.e. s( 3 s+ 2 )L{x}− 5 s^2 L{y}= 6 + 9 s (3′)

−s( 3 s+ 2 )L{x}+( 6 s^2 +s− 2 )L{y}

=− 6 s−

2

s

−7(4′)

Adding equations (3′)and(4′)gives:

(s^2 +s− 2 )L{y}=− 1 + 3 s−

2

s

=

−s+ 3 s^2 − 2

s

from which, L{y}=

3 s^2 −s− 2

s(s^2 +s− 2 )

Using partial fractions

3 s^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 )