Higher Engineering Mathematics, Sixth Edition

The solution of differential equations using Laplace transforms 603

Whens= 0 ,− 120 =− 20 A, from which,A=6.

Whens= 5 , 180 = 45 B, from which,B=4.

Whens= 2 , 6 =− 6 D, from which,D=−1.

Equatings^3 terms gives: 0=A+B+C, from

which,C=−10.

Hence

1

3

L−^1

{

−s^2 + 65 s− 120

s(s− 5 )(s− 2 )^2

}

=

1

3

L−^1

{

6

s

+

4

s− 5

−

10

s− 2

−

1

(s− 2 )^2

}

=

1

3

[6+4e^5 x−10e^2 x−xe^2 x]

Thusy= 2 +

4

3

e^5 x−

10

3

e^2 x−

x

3

e^2 x

Problem 5. The current flowing in an electrical

circuit is given by the differential equation

Ri+L(di/dt)=E,whereE,LandRare

constants. Use Laplace transforms to solve the

equation for currentigiven that whent=0,

i=0.

Using the procedure:

(i) L{Ri}+L

{

L

di

dt

}

=L{E}

i.e. RL{i}+L[sL{i}−i( 0 )]=

E

s

(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 226 Further problemson solving

differential equations using Laplace

transforms

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=40e−^3 t and

(c)E=50sin5t.

⎡

⎢

⎢

⎣

(a)i= 2 ( 1 −e−^5 t)

(b)i= 10 (e−^3 t−e−^5 t)

(c)i=

5

2

(e−^5 t−cos5t+sin5t)

⎤

⎥

⎥

⎦