Higher Engineering Mathematics

Laplace transforms

68

The solution of simultaneous

differential equations using

Laplace transforms

68.1 Introduction

It is sometimes necessary to solve simultaneous dif-

ferential equations. An example occurs when two

electrical circuits are coupled magnetically where

the equations relating the two currentsi 1 andi 2 are

typically:

L 1

di 1

dt

+M

di 2

dt

+R 1 i 1 =E 1

L 2

di 2

dt

+M

di 1

dt

+R 2 i 2 = 0

whereL represents inductance,R resistance, M

mutual inductance andE 1 the p.d. applied to one

of the circuits.

68.2 Procedure to solve simultaneous

differential equations using

Laplace transforms

(i) Take the Laplace transform of both sides of each

simultaneous equation by applying the formu-

lae for the Laplace transforms of derivatives (i.e.

equations (3) and (4) of Chapter 65, page 634)

and using a list of standard Laplace transforms,

as in Table 64.1, page 628 and Table 65.1,

page 632.

(ii) Put in the initial conditions, i.e.x(0),y(0),x′(0),

y′(0).

(iii) Solve the simultaneous equations forL{y}and

L{x}by the normal algebraic method.

(iv) Determineyandxby using, where necessary,

partial fractions, and taking the inverse of

each term.

68.3 Worked problems on solving

simultaneous differential

equations by using Laplace

transforms

Problem 1. Solve the following pair of simul-

taneous differential equations

dy

dt

+x= 1

dx

dt

−y+4et= 0

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

Using the above procedure:

(i)L

{

dy

dt

}

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

L

{

dx

dt

}

−L{y}+ 4 L{et}=0(2)

Equation (1) becomes:

[sL{y}−y(0)]+L{x}=

1

s

(1′)

from equation (3), page 634 and Table 64.1,

page 628.

Equation (2) becomes:

[sL{x}−x(0)]−L{y}=−

4

s− 1

(2′)

(ii)x(0)=0 andy(0)=0 hence

Equation (1′) becomes:

sL{y}+L{x}=

1

s

(1′′)