Mulugeta Andualem, Atinafu Asfaw

1. Application of Shehu transform To Mechanics, Newton’s Law of Cooling and Electric
circuit problems
Mulugeta Andualem1
, Atinafu Asfaw2
1
Department of Mathematics, Bonga University, Bonga, Ethiopia
Email: mulugetaandualem4@gmail.com
2
Department of Mathematics, Bonga University, Bonga, Ethiopia
Email: atinafu.asfaw2008@gmail.com
Abstract: In this study, we will discuss the problems occurred in first and second order ordinary
differential equations of constant coefficient based on law of newton’s law of cooling, Electric
circuit and damped and undamped motion using Shehu transform.
Keyword: Differential equation, Shehu transform, Newton’s law of cooling, Mechanics.
Introduction
In science and engineering Differential equations plays an important role. In order to solve the
differential equations, the integral transforms were extensively used. The importance of an
Integral Transforms is that they provide powerful operational methods for solving initial value
problems. Mostly we use Laplace Transform technique for solving differential equations.
Shehu Transform is the new integral transform, we apply this new transform technique for
solving the problems on law of damped and undamped motion, Newton law of cooling and
electric circuit.
Shehu Transform
Definition: The Shehu transform of the function 𝑣(𝑡) of exponential order is defined over the set
of functions,
𝐴 = { 𝑣(𝑡): ∃ 𝑁, 𝜂1, 𝜂2 > 0, |𝑣(𝑡)| < 𝑁𝑒
|𝑡|
𝜂 𝑖, 𝑖𝑓 𝑡 ∈ (−1)𝑖
× [0 ∞) }, by the following integral

2. 𝕊[𝑣(𝑡)] = 𝑉(𝑠, 𝑢) = ∫ 𝑒
(
−𝑠𝑡
𝑢
)
𝑣(𝑡)𝑑𝑡
∞
0
(1.1)
And the inverse Shehu transform is given by
𝕊−1[𝑉(𝑠, 𝑢)] = 𝑣(𝑡)for 𝑡 ≥ 0 (1.2)
Properties of the Shehu transform
1. Property 1. Linearity property of Shehu transform. Let the functions 𝛼𝑣(𝑡) and 𝛽𝑤(𝑡) be
in set 𝐴, then (𝛼𝑣(𝑡) + 𝛽𝑤(𝑡)) ∈ 𝐴, where 𝛼 and 𝛽 are nonzero arbitrary constants,
and 𝕊[𝛼𝑣(𝑡) + 𝛽𝑤(𝑡)] = 𝛼𝕊 [𝑣(𝑡)] + 𝛽𝕊 [𝑤(𝑡)]
Proof: Using the Definition (1.1) of Shehu transform, we get
𝕊 [𝛼𝑣(𝑡) + 𝛽𝑤(𝑡)] = ∫ 𝑒
(
−𝑠𝑡
𝑢
)
(𝛼𝑣(𝑡) + 𝛽𝑤(𝑡))𝑑𝑡 1.3
∞
0
= ∫ 𝑒
(
−𝑠𝑡
𝑢
)
𝛼𝑣(𝑡)𝑑𝑡
∞
0
+ ∫ 𝑒
(
−𝑠𝑡
𝑢
)
𝛽𝑤(𝑡)𝑑𝑡
∞
0
= 𝛼 ∫ 𝑒
(
−𝑠𝑡
𝑢
)
𝑣(𝑡)𝑑𝑡
∞
0
+ 𝛽 ∫ 𝑒
(
−𝑠𝑡
𝑢
)
𝑤(𝑡)𝑑𝑡
∞
0
= 𝛼𝕊[𝑣(𝑡)] + 𝛽𝕊[𝑤(𝑡)]
Property 2. Change of scale property of Shehu transform. Let the function 𝑣(𝛽𝑡) be in set 𝐴,
where 𝛽 is an arbitrary constant. Then
𝕊[𝛽𝑣(𝑡)] =
𝑢
𝛽
𝑉 (
𝑠
𝛽
, 𝑢)
Using the Definition 1.1 of Shehu transform, we deduce
𝕊[𝛽𝑣(𝑡)] = ∫ 𝑒(
−𝑠𝑡
𝑢
)
𝑣(𝛽𝑡)𝑑𝑡
∞
0
1.4

3. Substituting 𝑥 = 𝛽𝑡 ⇒ 𝑡 =
𝑥
𝛽
and
𝑑𝑡
𝑑𝑥
=
1
𝛽
⇒ 𝑑𝑡 =
𝑑𝑥
𝛽
in equation 1.4 yield
𝕊[𝛽𝑣(𝑡)] =
1
𝛽
∫ 𝑒
(
−𝑠𝑥
𝑢𝛽
)
𝑣(𝑥)𝑑𝑥
∞
0
=
1
𝛽
∫ 𝑒
(
−𝑠𝑡
𝑢𝛽
)
𝑣(𝑡)𝑑𝑡
∞
0
=
𝑢
𝛽
∫ 𝑒
(
−𝑠𝑡
𝛽
)
𝑣(𝑢𝑡)𝑑𝑡
∞
0
=
𝑢
𝛽
𝑉 (
𝑠
𝛽
, 𝑢)
Derivative of Shehu transform. If the function 𝑣(𝑛)
(𝑡) is the nth derivative of the function
𝑣(𝑡) ∈ 𝐴 with respect to 𝑡, then its Shehu transform is defined by
𝕊[𝑣(𝑛)
(𝑡)] =
𝑠 𝑛
𝑢 𝑛
𝑉(𝑠, 𝑢) − ∑ (
𝑠
𝑢
)
𝑛−(𝑘+1)
𝑣(𝑘)
(0)
𝑛−1
𝑘=0
1.5
When 𝑛 = 1, we obtain the following derivatives with respect to 𝑡.
𝕊[𝑣(1)
(𝑡)] = 𝕊[𝑣′(𝑡)] =
𝑠
𝑢
𝑉(𝑠, 𝑢) − 𝑣(0) 1.6
When 𝑛 = 1, we obtain the following derivatives with respect to 𝑡.
𝕊[𝑣(2)
(𝑡)] = 𝕊[𝑣′′(𝑡)] =
𝑠2
𝑢2
𝑉(𝑠, 𝑢) −
𝑠
𝑢
𝑣(0) − 𝑣′(0) 1.7
Assume that equation 1.5 true for 𝑛 = 𝑘. Now we want to show that for 𝑛 = 𝑘 + 1
𝕊[𝑣(𝑘+1)
(𝑡)] = 𝕊[(𝑣(𝑘)
(𝑡))′
] =
𝑠
𝑢
𝑆[𝑣(𝑘)
(𝑡)] − 𝑣(𝑘)
(0) using equation 1.6

4. =
𝑠
𝑢
[
𝑠 𝑘
𝑢 𝑘
𝕊[𝑣(𝑡)] − ∑ (
𝑠
𝑢
)
𝑘−(𝑖+1)
𝑣(𝑖)
(0)
𝑘−1
𝑖=0
] 𝑣(𝑘)
(0)
= (
𝑠
𝑢
)
𝑘+1
𝕊[𝑣(𝑡)] − ∑ (
𝑠
𝑢
)
𝑘−𝑖
𝑣(𝑖)
(0)
𝑘
𝑖=0
which implies that Eq (1.5) holds for n = k+1. By induction hypothesis the proof is complete
Property 3: Let the function 𝑣(𝑡) = 1 be in set 𝐴. Then its Shehu transform is given by
𝕊[1] =
𝑢
𝑠
Poof: Using equation 1.1
𝕊[1] = ∫ 𝑒
(
−𝑠𝑡
𝑢
)
∞
0
𝑑𝑡
= −
𝑢
𝑠
lim
𝜂→∞
[𝑒
(
−𝑠𝜂
𝑢
)
]
0
∞
=
𝑢
𝑠
Property 4: Let the function 𝑣(𝑡) = 𝑠𝑖𝑛(𝛼𝑡) be in set A. Then its Shehu transform is given by
𝕊[ 𝑠𝑖𝑛(𝛼𝑡)] =
𝛼𝑢2
𝑠2 + 𝛼2 𝑢2
Property 5: Let the function 𝑣(𝑡) = 𝑐𝑜𝑠(𝛼𝑡) be in set A. Then its Shehu transform is given by
𝕊[𝑐𝑜𝑠(𝛼𝑡] =
𝑢𝑠
𝑠2 + 𝛼2 𝑢2
Property 6: Let the function 𝑣(𝑡) = 𝑡𝑒𝑥𝑝(𝛼𝑡) and 𝑣(𝑡) = 𝑡𝑒𝑥𝑝(𝛼𝑡) be in set A. Then its
Shehu transform is given by
𝑢2
(𝑠−𝛼𝑢)2
and
𝑢
(𝑠−𝛼𝑢
respectively.

5. Shehu Transform for handling Newton’s law of cooling problem
The Newton's Law of Cooling model computes the temperature of an object of mass 𝑀 as
it is heated by a flame and cooled by the surrounding medium. The model assumes that
the temperature T within the object is uniform. This lumped system approximation is
valid if the rate of thermal energy transfer within the object is faster than the rate of
thermal energy transfer at the surface. Newtons law of cooling states that the rate at
which a warm body cools is approximately proportional to the difference between the
temperature of the warm object and the temperature of its environment. Newtons law of
cooling is generally limited to simple cases where the mode of energy transfer is
convection, from a solid surface to a surrounding fluid in motion, and where the
temperature difference is small, approximately less than 10º C (The Encyclopedia
Britannica, 1910).When the medium into which the hot body is placed varies beyond a
simple fluid, such as in the case of a gas, solid, or vacuum, etc., this becomes a residual
effect requiring further analysis (Whewell , 1866).
Newtons law of cooling based on the assumptions above stated mathematically as
𝑑𝑇
𝑑𝑡
= −𝑘(𝑇 − 𝑇𝑠)
Where, 𝑇 : Temperature of the cooling object at time t
𝑡 ∶ time in hours
𝑇𝑠:Temperature of surrounding medium (Ambient temperature)
𝑘: Constant of proportionality
Solution of Newton’s Equation
𝑑𝑇
𝑑𝑡
= −𝑘(𝑇 − 𝑇𝑠) 1.8

6. 𝑑𝑇
𝑑𝑡
+ 𝑘𝑇 = 𝑘𝑇𝑠 1.9
Taking Shehu transform both sides of Equation 1.9
𝕊 [
𝑑𝑇
𝑑𝑡
+ 𝑘𝑇 = 𝑘𝑇𝑠 ]
𝕊 [
𝑑𝑇
𝑑𝑡
] + 𝑘𝕊[𝑇] = 𝑘𝑇𝑠 𝕊[1]
𝑠
𝑢
𝑉(𝑠, 𝑢) − 𝑇(0) + 𝑘𝑉(𝑠, 𝑢) = 𝑘𝑇𝑠
𝑠
𝑢
𝑠
𝑢
𝑉(𝑠, 𝑢) + 𝑘𝑉(𝑠, 𝑢) = 𝑇(0) + 𝑘𝑇𝑠
𝑠
𝑢
𝑉(𝑠, 𝑢) =
𝑇(0) + 𝑘𝑇𝑠
𝑠
𝑢
𝑠
𝑢
+ 𝑘
=
𝑇(0)
𝑠
𝑢
+ 𝑘
+
𝑘𝑇𝑠
𝑠
𝑢
𝑠
𝑢
+ 𝑘
= 𝑇(0)
𝑢
𝑠 + 𝑢𝑘
+ 𝑇𝑠
𝑢
𝑠
− 𝑇𝑠
𝑢
𝑠 + 𝑢𝑘
Applying the inverse transform gives the solution:
𝑇(0) + 𝑒−𝑘𝑡
− 𝑇𝑠 𝑒−𝑘𝑡
= 𝑇(0) + (1 − 𝑇𝑠)𝑒−𝑘𝑡
Example: The rate of cooling of a body is proportional to the difference between the
temperature of the body and the surrounding air. If the air temperature is 20°C and the body
cools for 20 minutes from 140°C to 80°C, find when the temperature will be 35°C.
Solution: Given 𝑇𝑠(Temperature of surrounding medium)= 200
𝑐, initial temperature at time
𝑡 = 0 is 1400
𝑐 and after 𝑡 = 20 min is 800
𝑐.
Now, If T is the temperature of the body at time '𝑡' then from Newton's law of cooling

7. −𝑑𝑇
𝑑𝑡
∝ (𝑇 − 20) =
𝑑𝑇
𝑑𝑡
= −𝑘(𝑇 − 20)
The above equation can be rewritten as:
𝑑𝑇
𝑑𝑡
+ 𝑘𝑇 = 20𝑘
Take the Shehu Transform both sides:
𝕊 [
𝑑𝑇
𝑑𝑡
+ 𝑘𝑇 = 20𝑘]
⟹
𝑠
𝑢
𝑇(𝑢, 𝑠) − 𝑇(0) + 𝑘𝑇(𝑠, 𝑢) = 20𝑘
𝑢
𝑠
⟹ 𝑇(𝑠, 𝑢) [
𝑠 + 𝑢𝑘
𝑢
] = 20𝑘
𝑢
𝑠
+ 140
⟹ 𝑇(𝑠, 𝑢) = 20𝑘
𝑢2
𝑠(𝑠 + 𝑢𝑘)
+ 140
𝑢
𝑠 + 𝑢𝑘
⟹ 𝑇(𝑠, 𝑢) =
20𝑢
𝑠
+ 120
𝑢
𝑠 + 𝑢𝑘
Apply the invers Shehu transform both sides, then we get
⟹ 𝑇(𝑡) = 20 + 120𝑒−𝑘𝑡
Now, using the condition at time 𝑡 = 20, i.e 𝑇(20) = 80
⟹ 80 = 20 + 120𝑒−20𝑘
⟹
1
2
= 𝑒−20𝑘
𝑒−𝑘
= (
1
2
)
1
20
⟹ 𝑘 =
1
20
log(2) = 0.034657

8. Now substituting the value of 𝑘
𝑇(𝑡) = 20 + 120𝑒−0.034657𝑡
The above equation represent the temperature of the cooling body at any time 𝑡
≫ 𝑡 = [0: 10: 60];
>> 𝑇 = 20 + 120 ∗ 𝑒𝑥𝑝(−0.034657 ∗ 𝑡);
>> 𝑝𝑙𝑜𝑡(𝑡, 𝑇, ′ − ′);
>> 𝑥𝑙𝑎𝑏𝑒𝑙(′time′);
>> 𝑦𝑙𝑎𝑏𝑒𝑙(′𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒′);
>> 𝑙𝑒𝑔𝑒𝑛𝑑(′𝑇 = 20 + 120 ∗ 𝑒𝑥𝑝(−0.034657 ∗ 𝑡)′);
>> title('The graph represents graphical solution of the problem');

9. We are required to find 𝑡 when 𝑇 = 350
𝑐
35 = 20 + 120𝑒−𝑘𝑡
⟹ 15 = 120𝑒−𝑘𝑡
⟹ 𝑒−𝑘𝑡
=
1
8
⟹ (
1
2
)
𝑡
20
=
1
8
Now, taking both sides natural logarithmic, we obtained:
𝑡
20
ln (
1
2
) = ln (
1
8
)
⟹ 𝑡 = 20
ln (
1
8
)
ln (
1
2
)
min = 60 min
Application of Shehu transform on Damped Oscillations
Mechanical and electrical systems, and control systems in general, can exhibit oscillatory
behavior that after an initial disturbance slowly decays to zero. The process producing the decay
is a dissipative one that removes energy from the system, and it damping is called damping. To
see the prototype equation that exhibits this phenomenon we need only consider the following
very simple mechanical model. A mass M rests on a rough horizontal surface and is attached by
a spring of negligible mass to a fixed point. The mass–spring system is caused to oscillate along
the line of the spring by being displaced from its equilibrium position by a small amount and
then released. Figure-(a) below shows the system in its equilibrium configuration, and Figure-(b)
when the mass has been displaced through a distance 𝑥 from its rest position.

10. If 𝑡 is the time, the acceleration of the mass is
𝑑2 𝑥
𝑑𝑡2 , so the force acting due to the motion is 𝑀
𝑑2 𝑥
𝑑𝑡2
. The forces opposing the motion are the spring force, assumed to be proportional to the
displacement 𝑥 from the equilibrium position, and the frictional force, assumed to be
proportional to the velocity
𝑑𝑥
𝑑𝑡
of the mass 𝑀. If the spring constant of proportionality is 𝑝 and
the frictional constant of proportionality is 𝑘, the two opposing forces are 𝑘
𝑑𝑥
𝑑𝑡
due to friction and
𝑝𝑥 due to the spring. Equating the forces acting along the line of the spring and taking account of
the fact that the spring and frictional forces oppose the force due to the acceleration shows the
equation of motion to be the homogeneous second order linear equation.
𝑀
𝑑2
𝑥
𝑑𝑡2
= −𝑘
𝑑𝑥
𝑑𝑡
− 𝑝𝑥
⟹ 𝑀
𝑑2 𝑥
𝑑𝑡2
+ 𝑘
𝑑𝑥
𝑑𝑡
+ 𝑝𝑥 = 0
⟹
𝑑2 𝑥
𝑑𝑡2 + 𝑎
𝑑𝑥
𝑑𝑡
+ 𝑏𝑥 = 0
Where, 𝑎 =
𝑘
𝑀
> 0, 𝑏 =
𝑝
𝑀
> 0
If we apply Shehu transform we can have
⟹
𝑠2
𝑢2
𝑉(𝑠, 𝑢) −
𝑠
𝑢
𝑣(0) − 𝑣′(0) + 𝑎 [
𝑠
𝑢
𝑉(𝑠, 𝑢) − 𝑣(0)] + 𝑏𝑉(𝑠, 𝑢) = 0
⟹
𝑠2
𝑢2
𝑉(𝑠, 𝑢) + 𝑎
𝑠
𝑢
𝑉(𝑠, 𝑢) + 𝑏𝑉(𝑠, 𝑢) =
𝑠
𝑢
𝑣(0) + 𝑣′(0) + 𝑎𝑣(0)

11. ⟹ 𝑉(𝑠, 𝑢) [
𝑠2
𝑢2
+ 𝑎
𝑠
𝑢
+ 𝑏] = 𝑣(0) [
𝑠+𝑎𝑢
𝑢
] + 𝑣′(0)
⟹ 𝑉(𝑠, 𝑢) [
𝑠2+𝑎𝑠𝑢+𝑏𝑢2
𝑢2
] = 𝑣(0) [
𝑠+𝑎𝑢
𝑢
] + 𝑣′(0)
⟹ 𝑉(𝑠, 𝑢) =
𝑢2
𝑠2 + 𝑎𝑠𝑢 + 𝑏𝑢2
[𝑣(0)
𝑠 + 𝑎𝑢
𝑢
+ 𝑣′(0)]
=
𝑣′
(0)𝑢2
𝑠2 + 𝑎𝑠𝑢 + 𝑏𝑢2
+ 𝑣(0) [
𝑠𝑢 + 𝑎𝑢2
𝑠2 + 𝑎𝑠𝑢 + 𝑏𝑢2
]
=
𝑣′(0)
(
𝑠
𝑢
)
2
+ 𝑎
𝑠
𝑢
+ 𝑏
+
𝑣(0)
𝑠
𝑢
(
𝑠
𝑢
)
2
+ 𝑎
𝑠
𝑢
+ 𝑏
+
𝑎𝑣(0)
(
𝑠
𝑢
)
2
+ 𝑎
𝑠
𝑢
+ 𝑏
Now, if we see the roots of (
𝑠
𝑢
)
2
+ 𝑎
𝑠
𝑢
+ 𝑏
𝑟1, 𝑟2 =
−𝑎 ± √𝑎2 − 4𝑏
2
Case 1:
𝑎2
− 4𝑏 > 0 has two distinct real roots
𝑉(𝑠, 𝑢) =
𝑣′(0)
(
𝑠
𝑢
− 𝑟1) (
𝑠
𝑢
− 𝑟2)
+
𝑣(0)
𝑠
𝑢
(
𝑠
𝑢
− 𝑟1) (
𝑠
𝑢
− 𝑟2)
+
𝑎𝑣(0)
(
𝑠
𝑢
− 𝑟1) (
𝑠
𝑢
− 𝑟2)
When we apply partial decomposition, we can have
−𝑣′(0)
𝑟2 − 𝑟1
[
𝑢
𝑠 − 𝑟1 𝑢
] +
𝑣′(0)
𝑟2 − 𝑟1
[
𝑢
𝑠 − 𝑟2 𝑢
] −
𝑣(0)𝑟1
𝑟2 − 𝑟1
[
𝑢
𝑠 − 𝑟1 𝑢
]
+
𝑣(0)𝑟2
𝑟2 − 𝑟1
[
𝑢
𝑠 − 𝑟2 𝑢
] −
𝑎𝑣(0)
𝑟2 − 𝑟1
[
𝑢
𝑠 − 𝑟2 𝑢
] +
𝑎𝑣(0)
𝑟2 − 𝑟1
[
𝑢
𝑠 − 𝑟1 𝑢
]

12. 𝑢
𝑠 − 𝑟1 𝑢
[
−𝑣′(0)
𝑟2 − 𝑟1
−
𝑣(0)𝑟1
𝑟2 − 𝑟1
+
𝑎𝑣(0)
𝑟2 − 𝑟1
] +
𝑢
𝑠 − 𝑟2 𝑢
[
𝑣′(0)
𝑟2 − 𝑟1
−
𝑎𝑣(0)
𝑟2 − 𝑟1
+
𝑣(0)𝑟2
𝑟2 − 𝑟1
]
𝑐1 =
−𝑣′(0)
𝑟2 − 𝑟1
−
𝑣(0)𝑟1
𝑟2 − 𝑟1
+
𝑎𝑣(0)
𝑟2 − 𝑟1
and 𝑐2 =
𝑣′(0)
𝑟2 − 𝑟1
−
𝑎𝑣(0)
𝑟2 − 𝑟1
+
𝑣(0)𝑟2
𝑟2 − 𝑟1
⟹ 𝑉(𝑠, 𝑢) =
𝑐1 𝑢
𝑠 − 𝑟1 𝑢
+
𝑐2 𝑢
𝑠 − 𝑟2 𝑢
Applying the inverse transform gives the solution
𝑥(𝑡) = 𝑐1 𝑒 𝑟1 𝑡
+ 𝑐2 𝑒 𝑟2 𝑡
Case 2:
𝑎2
− 4𝑏 < 0 has complex roots
𝑟1, 𝑟2 =
−𝑎 ± 𝑖√4𝑏 − 𝑎2
2
From the above
𝑥(𝑡) = 𝑐1 𝑒 𝑟1 𝑡
+ 𝑐2 𝑒 𝑟2 𝑡
⟹ 𝑥(𝑡) = 𝑒
−𝑎𝑡
2 [𝑐1 cos(𝜔𝑡) + 𝑐2 𝑖 sin(𝜔𝑡)]
Case 3:
𝑎2
− 4𝑏 = 0 has one root 𝑟1, 𝑟2 =
−𝑎
2
𝑉(𝑠, 𝑢) =
𝑣′(0)
(
𝑠
𝑢
+
𝑎
2
)
2 +
𝑣(0)
𝑠
𝑢
(
𝑠
𝑢
+
𝑎
2
)
2 +
𝑎𝑣(0)
(
𝑠
𝑢
+
𝑎
2
)
2

13. ⟹ 𝑉(𝑠, 𝑢) =
𝑢2
𝑣′(0)
(𝑠 +
𝑎
2
𝑢)
2 +
𝑎𝑣(0)𝑢2
+ 𝑣(0)𝑢𝑠
(𝑠 +
𝑎
2
𝑢)
2
⟹ 𝑉(𝑠, 𝑢) =
𝑢2
𝑣′(0)
(𝑠 +
𝑎
2
𝑢)
2 +
𝑣(0)𝑢
(𝑠 +
𝑎
2
𝑢)
Applying the inverse transform gives the solution
𝑥(𝑡) = 𝑐1 𝑒
−𝑎𝑡
2 + 𝑡𝑐2 𝑒
−𝑎𝑡
2
Shehu Transform for handling Free Undamped Motion problem
We next consider the motion of a spring that is not subject to a frictional force then by using
Hooke’s Law, which says that if the spring is stretched (or compressed) x units from its natural
length, then it exerts a force that is proportional to x : where is k a positive constant (called the
spring constant). If we ignore any external resisting forces (due to air resistance or friction) then,
by Newton’s Second Law (force equals mass times acceleration/n), we have
𝑚
𝑑2
𝑥
𝑑𝑡2
= −𝑘(𝑥 + 𝑠) + 𝑚𝑔 = −𝑘𝑥 + 𝑚𝑔 − 𝑘𝑠 1.10
Since, 𝑚𝑔 − 𝑘𝑠 = 0 from newton second law Therefore equation 1.10 is equal to:
𝑚
𝑑2
𝑥
𝑑𝑡2
= −𝑘𝑥 1.11
DE of Free Undamped Motion: By dividing (1.11) by the mass m we obtain the second order
differential equation
𝑑2 𝑥
𝑑𝑡2
+ (
𝑘
𝑚
) 𝑥 = 0, this can be written as:
𝑑2
𝑥
𝑑𝑡2
+ 𝜔2
𝑥 = 0, where 𝜔2
=
𝑘
𝑚
1.12

14. Equation (1.12) is said to describe simple harmonic motion or free undamped motion.
In this section, we present Shehu Transform for handling free undamped motion problem given
by (1.12)
Taking Shehu transform both sides of Equation 1.12
𝕊 [
𝑑2
𝑥
𝑑𝑡2
+ 𝜔2
𝑥 = 0]
𝕊 [
𝑑2
𝑥
𝑑𝑡2
] + 𝜔2
𝕊[𝑥] = 0
⟹
𝑠2
𝑢2 𝑉(𝑠, 𝑢) −
𝑠
𝑢
𝑣(0) − 𝑣′(0) + 𝜔2
𝑉(𝑠, 𝑢) = 0
⟹ 𝑉(𝑠, 𝑢) [
𝑠2
𝑢2
+ 𝜔2
] =
𝑠
𝑢
𝑣(0) + 𝑣′(0)
⟹
𝑣(0)𝑠𝑢
𝑠2 + 𝜔2 𝑢2
+
𝑣′(0)𝑢2
𝑠2 + 𝜔2 𝑢2
=
𝑣(0)𝑠𝑢
𝑠2 + 𝜔2 𝑢2
+
𝑣′(0)
𝜔
𝜔𝑢2
𝑠2 + 𝜔2 𝑢2
Applying the inverse transform gives the solution
𝑣(0) cos(𝜔𝑡) +
𝑣′(0)
𝜔
𝑠𝑖𝑛(𝜔𝑡)
⟹ 𝑐1 cos(𝜔𝑡) + 𝑐2 sin(𝜔𝑡), where 𝑐1 = 𝑣(0), 𝑐2 =
𝑣′(0)
𝜔
Example: A spring with a mass of 2 kg has natural length 0.5 m. a force of 25.6 N is required to
maintain it stretched to length of 0.7 m. If the spring is stretched to a length of 0.7m and then
released with initial velocity 0, find the position of the mass at any time 𝑡.
Solution: From hook’s law the force required to stretched the spring is

15. 𝑘(0.2) = 25.6
So 𝑘 =
25.6
0.2
= 128. Using this value of the spring constant 𝑘, together with 𝑚 = 2
We have,
2
𝑑2
𝑥
𝑑𝑡2
+ 128𝑥 = 0
This also equal to
𝑑2
𝑥
𝑑𝑡2
+ 64𝑥 = 0
Now taking the Shehu Transform both sides
𝕊 [
𝑑2
𝑥
𝑑𝑡2
+ 64𝑥 = 0]
⟹
𝑠2
𝑢2
𝑋(𝑠, 𝑢) −
𝑠
𝑢
𝑥(0) − 𝑥′(0) + 64𝑋(𝑠, 𝑢) = 0
⟹
𝑠2
𝑢2
𝑋(𝑠, 𝑢) + 64𝑋(𝑠, 𝑢) =
𝑠
𝑢
𝑥(0) + 𝑥′(0)
⟹ 𝑋(𝑠, 𝑢) [
𝑠2
+ 64𝑢2
𝑢2
] =
𝑠
𝑢
𝑥(0) + 𝑥′(0)
⟹ 𝑋(𝑠, 𝑢) = 𝑥(0)
𝑢𝑠
𝑠2 + (8𝑢)2
+
1
8
𝑥′
(0)
8𝑢2
𝑠2 + (8𝑢)2
Now taking the inverse of Shehu Transform, we get
𝑥(𝑡) = 𝑥(0) cos(8𝑡) +
1
8
𝑥′
(0) sin(8𝑡)

16. We are given initial condition that 𝑥(0) = 0.2 and the initial velocity given as 𝑥′(0) =
0, therefore our solution is converted to:
𝑥(𝑡) =
1
5
cos(8𝑡)
Shehu Transform for handling an Electric circuit in series containing resistance and self-
inductance (RL series circuit) problem
In this section, we present Shehu Transform for handling RL series circuit. Consider a circuit
containing resistance R and inductance L in series with a constant voltage source E is given by
the differential equation.

17. 𝑅𝑖 + 𝐿
𝑑𝑖
𝑑𝑡
= 𝐸 Or
𝑑𝑖
𝑑𝑡
+ 𝑎𝑖 = 𝑏 𝑖(0) = 0 1.13
Where 𝑎 =
𝑅
𝐿
, 𝑏 =
𝐸
𝐿
Taking Shehu transform on both sides of (1.13), we have
𝕊 [
𝑑𝑖
𝑑𝑡
+ 𝑎𝑖 = 𝑏 ]
Now applying the properties of Shehu transform, we have
𝕊 [
𝑑𝑖
𝑑𝑡
] + 𝑎𝕊[𝑖(𝑡)] = 𝑏𝕊[1]
⟹
𝑠
𝑢
𝐼(𝑠, 𝑢) − 𝑖(0) + 𝑎𝐼(𝑠, 𝑢) =
𝑏𝑢
𝑠
⟹ 𝐼(𝑠, 𝑢) [
𝑠
𝑢
+ 𝑎] =
𝑏𝑢
𝑠
+ 0, since 𝑖(0) = 0
⟹ 𝐼(𝑠, 𝑢) =
𝑏𝑢2
𝑠(𝑠 + 𝑎𝑢)
This also can be rewritten as:
⟹ 𝐼(𝑠, 𝑢) =
𝑏𝑢
𝑎𝑠
−
𝑏𝑢
𝑎𝑠 + 𝑎2 𝑢
=
𝑏
𝑎
[
𝑢
𝑠
−
𝑢
𝑠 − (−𝑎𝑢)
] 1.14
Operating inverse Shehu transform on both sides of (1.14)

18. ⟹ 𝑖(𝑡) =
𝑏
𝑎
𝕊−1
[
𝑢
𝑠
] −
𝑏
𝑎
𝕊−1
[
𝑢
𝑠 − (−𝑎𝑢)
]
⟹ 𝑖(𝑡) =
𝑏
𝑎
−
𝑏
𝑎
𝑒−𝑎𝑡
1.15
Since, 𝑎 =
𝑅
𝐿
, 𝑏 =
𝐸
𝐿
⟹
𝑏
𝑎
=
𝐸
𝑅
. Therefore equation 1.15 is equal to:
⟹ 𝑖(𝑡) =
𝐸
𝑅
−
𝐸
𝑅
𝑒
(−
𝑅
𝐿
)𝑡
=
𝐸
𝑅
(1 − 𝑒
(−
𝑅
𝐿
)𝑡
)
This equation gives the required amount of current in the circuit at any time 𝑡 as shown in figure
below
Conclusion
In this paper, the Shehu Transform Method was proposed for solving first and second order
ordinary differential equation with initial conditions occurred in Engineering and Physics
problems. We successfully found an exact solution in all the examples. It may be concluded that
Shehu transform is very powerful and efficient in finding analytical solutions for wide classes of
linear ordinary differential equations.

19. Reference
[1] Aggarwal, S., Gupta, A.R., Singh, D.P., Asthana, N. and Kumar, N., Application of Laplace
transform for solving population growth and decay problems, International Journal of Latest
Technology in Engineering, Management & Applied Science, 7(9), 141-145, 2018.
[2] Lokenath Debnath and Bhatta, D., Integral transforms and their applications, Second edition,
Chapman & Hall/CRC, 2006.
[3] Chauhan, R. and Aggarwal, S., Solution of linear partial integro-differential equations using
Mahgoub transform, Periodic Research, 7(1), 28-31, 2018.
[4] Aggarwal, S., Sharma, N., Chauhan, R., Gupta, A.R. and Khandelwal, A., A new application
of Mahgoub transform for solving linear ordinary differential equations with variable
coefficients, Journal of Computer and Mathematical Sciences, 9(6), 520-525, 2018.
[5] Zill, D.G., Advanced engineering mathematics, Jones & Bartlett, 2016.
[6] Sadikali Latif Shaikh “Introducing a new Integral Transform Sadik Transform”, American
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Transform for solving Volterra integral equation of first kind”, International Journal of Research
in Advent Technology, vol-6.No.8 Aug 2018 ISSN: 2321-9637.
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