Higher Engineering Mathematics

654 LAPLACE TRANSFORMS

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}= 2 s^3 +s=s(2s^2 +1)

from which, L{x}=

s(2s^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 239 Further problems on solving

simultaneous differential equations using

Laplace transforms

Solve the following pairs of simultaneous dif-

ferential equations:

1. 2

dx

dt

+

dy

dt

=5et

dy

dt

− 3

dx

dt

= 5

given that whent=0,x=0 andy= 0

[x=et−t−1 andy= 2 t− 3 + 3 et]

2. 2

dy

dt

−y+x+

dx

dt

−5 sint= 0

3

dy

dt

+x−y+ 2

dx

dt

−et= 0

given that att=0,x=0 andy= 0

[

x=5 cost+5 sint−e^2 t−et−3 and

y=e^2 t+ 2 et− 3 −5 sint

]

3.

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=3 cost+cos (

√

3 t) and

y=3 cost−cos (

√

3 t)

]