Higher Engineering Mathematics

652 LAPLACE TRANSFORMS

Equation (1) becomes:

3[sL{x}−x(0)]−5[sL{y}−y(0)]

+ 2 L{x}=

6

s

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

page 628.

i.e. 3 sL{x}− 3 x(0)− 5 sL{y}

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

6

s

i.e. (3s+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 634, and Table 64.1,

page 628,

i.e. 2 sL{y}− 2 y(0)−sL{x}

+x(0)−L{y}=−

1

s

i.e. (2s−1)L{y}− 2 y(0)−sL{x}

+x(0)=−

1

s

(2′)

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

+5(3)=

6

s

(1′′)

and equation (2′) becomes

(2s−1)L{y}−2(3)−sL{x}

+ 8 =−

1

s

(2′′)

i.e. (3s+2)L{x}− 5 sL{y}=

6

s

+9(1′′)

(3s+2)L{x}− 5 sL{y}

=

6

s

+9( 1 ′′′)

−sL{x}+(2s−1)L{y}

=−

1

s

−2( 2 ′′′)

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(A)

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

gives:

s(3s+2)L{x}− 5 s^2 L{y}=s

(

6

s

+ 9

)

(3)

−s(3s+2)L{x}+(3s+2)(2s−1)L{y}

=(3s+2)

(

−

1

s

− 2

)

(4)

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

−s(3s+2)L{x}+(6s^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)

i.e. 3 s^2 −s− 2 =A(s+2)(s−1)

+Bs(s−1)+Cs(s+2)

When s=0,− 2 =− 2 A, hence A= 1

When s=1, 0= 3 C, hence C= 0

When s=−2, 12= 6 B, hence B= 2

Thus L{y}=

3 s^2 −s− 2

s(s^2 +s−2)

=

1

s

+

2

(s+2)

(iv) Hence y=L−^1

{

1

s

+

2

s+ 2

}

= 1 +2e−^2 t