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
- 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)
⎤
⎥
⎥
⎦