Higher Engineering Mathematics

PROPERTIES OF LAPLACE TRANSFORMS 633

K

=

4(s−3)

s^2 − 6 s+ 9 + 25

=

4(s−3)

s^2 − 6 s+ 34

Problem 2. Determine (a)L{e−^2 tsin 3t}

(b)L{3eθcosh 4θ}.

(a) From (ii) of Table 65.1,

L{e−^2 tsin 3t}=

3

(s−(−2))^2 + 32

=

3

(s+2)^2 + 9

=

3

s^2 + 4 s+ 4 + 9

=

3

s^2 + 4 s+ 13

(b) From (v) of Table 65.1,

L{3eθcosh 4θ}= 3 L{eθcosh 4θ}=

3(s−1)

(s−1)^2 − 42

=

3(s−1)

s^2 − 2 s+ 1 − 16

=

3 (s− 1 )

s^2 − 2 s− 15

Problem 3. Determine the Laplace transforms

of (a) 5e−^3 tsinh 2t(b) 2e^3 t(4 cos 2t−5 sin 2t).

(a) From (iv) of Table 65.1,

L{5e−^3 tsinh 2t}= 5 L{e−^3 tsinh 2t}

= 5

(

2

(s−(−3))^2 − 22

)

=

10

(s+3)^2 − 22

=

10

s^2 + 6 s+ 9 − 4

=

10

s^2 + 6 s+ 5

(b) L{2e^3 t(4 cos 2t−5 sin 2t)}

= 8 L{e^3 tcos 2t}− 10 L{e^3 tsin 2t}

=

8(s−3)

(s−3)^2 + 22

−

10(2)

(s−3)^2 + 22

from (iii) and (ii) of Table 65.1

=

8(s−3)−10(2)

(s−3)^2 + 22

=

8 s− 44

s^2 − 6 s+ 13

Problem 4. Show that

L

{

3e−

1

2 xsin^2 x

}

=

48

(2s+1)(4s^2 + 4 s+17)

.

Since cos 2x= 1 −2 sin^2 x, sin^2 x=

1

2

(1−cos 2x).

Hence,

L

{

3e−

1

2 xsin^2 x

}

=L

{

3e−

1

2 x

1

2

(1−cos 2x)

}

=

3

2

L

{

e−

1

2 x

}

−

3

2

L

{

e−

1

2 xcos 2x

}

=

3

2

⎛

⎜

⎜

⎝

1

s−

(

−

1

2

)

⎞

⎟

⎟

⎠−

3

2

⎛

⎜

⎜

⎜

⎝

(

s−

(

−

1

2

))

(

s−

(

−

1

2

)) 2

+ 22

⎞

⎟

⎟

⎟

⎠

from (iii) of Table 64.1 (page 628) and (iii)

of Table 65.1 above,

=

3

2

(

s+

1

2

)−

3

(

s+

1

2

)

2

[(

s+

1

2

) 2

+ 22

]

=

3

2 s+ 1

−

6 s+ 3

4

(

s^2 +s+

1

4

+ 4

)

=

3

2 s+ 1

−

6 s+ 3

4 s^2 + 4 s+ 17

=

3(4s^2 + 4 s+17)−(6s+3)(2s+1)

(2s+1)(4s^2 + 4 s+17)

=

12 s^2 + 12 s+ 51 − 12 s^2 − 6 s− 6 s− 3

(2s+1)(4s^2 + 4 s+17)

=

48

(2s+1)(4s^2 + 4 s+17)