Higher Engineering Mathematics, Sixth Edition

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:

1. 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. 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

]

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=3cost+cos(
  √
  3 t)and
  y=3cost−cos(
  √
  3 t)
  ]