Higher Engineering Mathematics

THE SOLUTION OF SIMULTANEOUS DIFFERENTIAL EQUATIONS USING LAPLACE TRANSFORMS 653

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Returning to equations (A) to determineL{x}and
hencex:

(2s−1)×equation (1′′′) and 5s×(2′′′) gives:

(2s−1)(3s+2)L{x}− 5 s(2s−1)L{y}

=(2s−1)

(

6

s

+ 9

)

(5)

and −s(5s)L{x}+ 5 s(2s−1)L{y}

= 5 s

(

−

1

s

− 2

)

(6)

i.e. (6s^2 +s−2)L{x}− 5 s(2s−1)L{y}

= 12 + 18 s−

6

s

−9( 5 ′)

and − 5 s^2 L{x}+ 5 s(2s−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

differential 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 simul-

taneous differential equations

d^2 x

dt^2

−x=y

d^2 y

dt^2

+y=−x

given that at t=0, x=2, y=−1,

dx

dt

= 0

and

dy

dt

=0.

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 635

(ii)x(0)=2,y(0)=−1,x′(0)=0 andy′(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)2s (5)

L{x}+(s^2 +1)L{y}=−s (6)