Higher Engineering Mathematics

648 LAPLACE TRANSFORMS

(ii)i(0)=0, henceRL{i}+LsL{i}=

E

s

(iii) Rearranging gives:

(R+Ls)L{i}=

E

s

i.e. L{i}=

E

s(R+Ls)

(iv)i=L−^1

{

E

s(R+Ls)

}

E

s(R+Ls)

≡

A

s

+

B

R+Ls

≡

A(R+Ls)+Bs

s(R+Ls)

Hence E=A(R+Ls)+Bs

When s=0,E=AR,

from which, A=

E

R

When s=−

R

L

,E=B

(

−

R

L

)

from which, B=−

EL

R

HenceL−^1

{

E

s(R+Ls)

}

=L−^1

{

E/R

s

+

−EL/R

R+Ls

}

=L−^1

{

E

Rs

−

EL

R(R+Ls)

}

=L−^1

⎧

⎪⎨

⎪⎩

E

R

(

1

s

)

−

E

R

⎛

⎜

⎝

1

R

L

+s

⎞

⎟

⎠

⎫

⎪⎬

⎪⎭

=

E

R

L−^1

⎧

⎪⎪

⎨

⎪⎪

⎩

1

s

−

1

(

s+

R

L

)

⎫

⎪⎪

⎬

⎪⎪

⎭

Hencecurrenti=

E

R

(

1 −e−

Rt

L

)

Now try the following exercise.

Exercise 238 Further problems on solving

differential equations using Laplace trans-

forms

1. A first order differential equation involving
   currentiin a seriesR−Lcircuit is given by:
   di
   dt

+ 5 i=

E

2

andi=0 at timet=0.

Use Laplace transforms to solve for i

when (a)E=20 (b)E=40 e−^3 t and (c)

E=50 sin 5t.

⎡

⎢

⎢

⎣

(a)i=2(1−e−^5 t)

(b)i=10( e−^3 t−e−^5 t)

(c)i=

5

2

(e−^5 t−cos 5t+sin 5t)

⎤

⎥

⎥

⎦

In Problems 2 to 9, use Laplace transforms to

solve the given differential equations.

2. 9

d^2 y

dt^2

− 24

dy

dt

+ 16 y=0, given y(0)= 3

andy′(0)=3.

[

y=(3−t)e

4

3 t

]

3.

d^2 x

dt^2

+ 100 x=0, given x(0)= 2 and

x′(0)=0. [x=2 cos 10t]

4.

d^2 i

dt^2

+ 1000

di

dt

+ 250000 i=0, given

i(0)=0 andi′(0)=100. [i= 100 te−^500 t]

5.

d^2 x

dt^2

+ 6

dx

dt

+ 8 x=0, givenx(0)=4 and

x′(0)=8. [x=4(3e−^2 t−2e−^4 t)]

6.

d^2 y

dx^2

− 2

dy

dx

+y=3e^4 x,giveny(0)=−

2

3

andy′(0)= 4

1

(^3) [
y=(4x−1) ex+
1
3
e^4 x
]
7.
d^2 y
dx^2

• 16 y=10 cos 4x,giveny(0)=3 and
  y′(0)=4.
  [
  y=3 cos 4x+sin 4x+
  5
  4
  xsin 4x
  ]