it is my little contribution to people to educate you abut Laplace transformation and its types.

1. Laplace Transformation:
 Let F(t) be a function of specified for Then the Laplace transform of denoted by is
defined by
 Where is complex parameter but we assume at present the parameter is real. Later it will
found useful to consider complex.
 The Laplace transform of is said to exist if the integral (Ⅰ) converges for some values of
 The Laplace transform of some elementary functions are given in the subsequent table.

2. 𝐹(𝑡) ℒ 𝐹 𝑡 = 𝑓 𝑠
1. 1 1
𝑠
, 𝑠 > 0
2. 𝑡 1
𝑠2
, 𝑠 > 0
3. 𝑡 𝑛, 𝑛 = 0, 1, 2, . . , 𝑛!
𝑠 𝑛+1
, 𝑠 > 0
4. 𝑒 𝑎𝑡
1
𝑠 − 𝑎
, 𝑠 > 𝑎
5. 𝑆𝑖𝑛(𝑎𝑡) 𝑎
𝑠2 + 𝑎2
, 𝑠 > 0
6. 𝑐𝑜𝑠(𝑎𝑡) 𝑠
𝑠2 + 𝑎2
, 𝑠 > 0
7. 𝑆𝑖𝑛ℎ(𝑎𝑡) 𝑎
𝑠2 − 𝑎2
, 𝑠 > |𝑎|
8. 𝑐𝑜𝑠ℎ(𝑎𝑡) 𝑠
𝑠2 − 𝑎2
, 𝑠 > |𝑎|

3. Some important Properties of Laplace transformation are,
1. Linearity property:
If "𝑐1" 𝑎𝑛𝑑 "𝑐2" are any while 𝐹1 𝑡 𝑎𝑛𝑑 𝐹2 𝑡 are functions with Laplace transforms 𝑓1 𝑠 𝑎𝑛𝑑 𝑓2 𝑠 respectively
then,
ℒ 𝑐1 𝐹1 𝑡 + 𝑐2 𝐹2 𝑡 = 𝑐1 ℒ 𝐹1 𝑡 + 𝑐2 ℒ{𝐹2 𝑡 }
𝑐2 𝑐1 𝐹1 𝑡 + 𝑐2 𝐹2 𝑡 = 𝑐1 𝑓1 𝑠 + 𝑐2 𝑓2 𝑠
2. Frist translation or shifting property
If ℒ{F(t)} = f(s) then
ℒ 𝑒 𝑎𝑡
F t = 𝑓(𝑠 − 𝑎)
3. Second translation or shifting property
If ℒ{F(t)} = f(s) and G t
F t − a t > a
0 t < a
, then
ℒ G t = 𝑒−𝑎𝑠 𝑓(𝑠)
4. Change of scale property.
If ℒ{F(t)} = f(s), then
ℒ F at =
1
𝑎
𝑓
𝑠
𝑎

4. 5. Laplace transformation of derivatives.
If ℒ{F(t)} = f(s), then
ℒ 𝐹′
t = s𝑓 𝑠 𝐹(0)
ℒ 𝐹′′ t = 𝑠2 𝑓 𝑠 − 𝑠 𝐹 0 − 𝐹′ 0
ℒ 𝐹(𝑛) t = 𝑠 𝑛 𝑓 𝑠 − 𝑠 𝑛−1 𝐹 0 − 𝑠 𝑛−2 𝐹′ 0 − ⋯ 𝑠𝐹 𝑛−2 0 − 𝐹 𝑛−1 0

5. APPLICATIONS OF LAPLACE TRANSFORMATION
1. APPLICATIONS TO MECHANICS
Preliminaries:
 Restoring force: The force which regain the original position of a body is called
restoring force.
 Hooke's law: According to Hooke’s law the restoring force on an extensible
string is directly proportional to the stretch or extension of the spring from the
equilibrium position.

6. APPLICATIONS TO MECHANICS:
Suppose a mass “𝑚” attached to a flexible spring fixed at equilibrium point “𝑂” is free to move
on a frictionless plan 𝑃𝑄 if due to some external force it covers some displacement 𝑋(𝑡) or
simply “X” from the equilibrium position “𝑂”. then according to newton 3rd law there will be a
force in opposite direction to the applied force called restoring force where k is a constant
depending on spring and is called spring constant. The magnitude of this force is equal to – 𝑘𝑥
follows from Hooke`s law.
Then by Newton`s 2nd law.
𝐹 = 𝑚𝑎
−𝑘𝑥 = 𝑚
𝑑2 𝑋
𝑑2 𝑡

7. 𝑚𝑋′′ + 𝑘𝑋 = 0 _______ (Ⅰ)
If in addition, there is a damping force proportional to the instantaneous speed of m, the
equation of motion is
𝑚𝑋′′
+ 𝛽𝑋′
+ 𝑘𝑋 = 0 _______ (Ⅱ)
Where, 𝛽 is constant called damping constant.
Further we modify the equation if there is time varying external force 𝑓(𝑡) also act on “𝑚”.
Then equation of motion becomes.
𝑚𝑋′′
+ 𝛽𝑋′
+ 𝑘𝑋 = 0 + 𝑓(𝑡)
𝑚𝑋′′
+ 𝛽𝑋′
+ 𝑘𝑋 = 𝑓(𝑡)________ (ⅡⅠ)
Then by applying Laplace transformation on (Ⅰ), (Ⅱ), (

8. 1. APPLICATIONS TO DIFFERENTIAL EQUATIONS.
The Laplace transform is useful in solving linear ordinary differential equations with
constant coefficients. Suppose we wish to solve the second order linear differential equation
𝑌′′
+ 𝛼𝑌′
+ 𝛽𝑌 = 𝐹(𝑡 )________(Ⅰ)
Where, 𝛼 𝑎𝑛𝑑 𝛽 are constants, subjected to the initial or boundary conditions.
𝑌 0 = 𝐴, 𝑌′ 0 = 𝐵________Ⅱ)
Where A and B are given constants,
Then we take the Laplace transform of both sides of (Ⅰ) and using (Ⅱ) we obtain algebraic
equation for determination of ℒ 𝑌 = 𝑦 𝑠 . Then required solution is obtained by finding the
inverse Laplace. The method is illustrated by mean of example as,

9. 1. Solve 𝑌′′
+ 𝑌 = 𝑡 , 𝑌 0 = 1, 𝑌′
0 = −2. Using Laplace transformations.
Solution:
we have,
𝑌′′ + 𝑌 = 𝑡
Taking Laplace on both sides we have,
ℒ 𝑌′′ + ℒ 𝑌 = ℒ 𝑡
𝑠2 𝑦 − sY 0 − 𝑌′ 0 + 𝑦 =
1
𝑠2
𝑠2 𝑦 − s 1 + 2 + 𝑦 =
1
𝑠2
𝑠2
𝑦 − s 1 + 𝑦 =
1
𝑠2 𝑠2 + 1
+
𝑠 − 2
𝑠2 + 1

10. ⇒ 𝑦 =
1
𝑠2
−
1
𝑠2 + 1
+
𝑠
𝑠2 + 1
−
2
𝑠2 + 1
⇒ 𝑦 =
1
𝑠2
+
𝑠
𝑠2 + 1
−
3
𝑠2 + 1
ℒ(𝑦) = ℒ{
1
𝑠2
+
𝑠
𝑠2 + 1
−
3
𝑠2 + 1
}
𝑌 = 𝑡 + cos 𝑡 − 3sin(𝑡)
Which is the required solution.