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MATHEMATICAL PHYSICS UNIT – 6
LAPLACE TRANSFORM AND APPLICATION
DR. RAJESH MATHPAL
ACADEMIC CONSULTANT
SCHOOL OF SCIENCES
UTTARAKHAND OPEN UNIVERSITY
TEENPANI, HALDWANI
UTTRAKHAND
MOB:9758417736,7983713112
Email: [email protected]
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Structure of Unit
1. Introduction
2. Objectives
3. Laplace Transform
4. Linearity of the Laplace Transform
5. Change of Scale Property
6. First Shifting Theorem
7. Second Shifting Theorem (Heaviside’s Shifting Theorem):
8. Laplace Transform of the Derivative of 𝑓 𝑡
9. Laplace Transform of the Derivative of Order N
10. Laplace Transform of the Integral of 𝑓 𝑡
11. Laplace Transform of Some Important Functions
12. Laplace Transform of 1
𝑡𝑓(𝑡)
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13. Laplace Transform Of 𝑡. 𝑓(𝑡)
14. Unit Step Function
15. Laplace Transform of Unit Step Function
16. Periodic Functions:
17. Some Important Formulae of Laplace Transform
18. Inverse Laplace Transform
19. Some Important Formulae of Inverse Laplace Transform
20. Multiplication by s
21. Division by s (Multiplication By 1
𝑠)
22. First Shifting Property
23. Second Shifting Property
24. Inverse Laplace Transforms of Derivatives:
25. Inverse Laplace Transform of Integrals
26. Inverse Laplace Transform by Partial Fraction Method
27. Solution of Differential Equations by Laplace Transforms
28. Self Assessment Question
29. References
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1. INTRODUCTION
The Laplace transform is named for the French mathematician
Laplace,who studied this transform in 1782. Laplace transforms is
an integraltransform. It helps in solving the differential
equations with boundaryvalues without finding the general solution
and values of the arbitraryconstants. The method of Laplace
transforms is a system that relies onalgebra (rather than
calculus-based methods) to solve linear differentialequations.
While it might seem to be a somewhat cumbersomemethod at times, it
is a very powerful tool that enables us to readilydeal with linear
differential equations with discontinuous forcingfunctions.
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2. OBJECTIVES
After studying this chapter we will learn about how Laplace
transformsis useful for solving differential equations with
boundary values withoutfinding the general solution. With the use
of different properties ofLaplace transform and Inverse Laplace
transform one can solve manyimportant problem of physics with very
simple way. Thus we will learnfrom this unit to use the Laplace
transform for solving the differentialequations.
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3. LAPLACE TRANSFORM
Definition: The Laplace transform of a function f(t) is defined
as follows
F (s) = e−st∞
0
𝑓(𝑡)𝑑𝑡
For all positive values of t and integral should exits. The
Laplace transform is denoted by
𝑳 𝒇(𝒕) = 𝑭 (𝒔) = 𝒆−𝒔𝒕∞
𝟎
𝒇(𝒕)𝒅𝒕
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4. LINEARITY OF THE LAPLACE TRANSFORM
The Laplace transform is a linear operation; that is, for any
functions 𝑓(𝑡) and 𝑔(𝑡)whose
transforms exist and any constants 𝑎 and 𝑏 the transform of
𝑎𝑓(𝑡) + 𝑏𝑔(𝑡) exists, and
𝑳 𝒂𝒇(𝒕) + 𝒃𝒈(𝒕) = 𝒂𝑳 𝒇(𝒕) + 𝒃[𝒈(𝒕)]
Proof:
𝐿 𝑎𝑓(𝑡) + 𝑏𝑔(𝑡) = 𝑒−𝑠𝑡∞
0
[𝑎𝑓(𝑡) + 𝑏𝑔(𝑡)]𝑑𝑡
As we know that integration is a linear operation. So we can use
the linearity property of
integration in above equation
𝐿 𝑎𝑓(𝑡) + 𝑏𝑔(𝑡) = 𝑎 𝑒−𝑠𝑡∞
0
𝑓(𝑡)𝑑𝑡 + 𝑏 𝑒−𝑠𝑡∞
0
𝑔(𝑡)𝑑𝑡
𝐿 𝑎𝑓(𝑡) + 𝑏𝑔(𝑡) = 𝑎𝐿[ 𝑓(𝑡)] + 𝑏𝐿[𝑔(𝑡)]
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5. CHANGE OF SCALE PROPERTY
If the Laplace transform of 𝑓 𝑡 is 𝐹 𝑠 then
𝑳 𝒇 𝒂𝒕 =𝟏
𝒂𝑭
𝒔
𝒂Proof: From the definition of Laplace transform
𝐿 𝑓 𝑎𝑡 = 0
∞
𝑒−𝑠𝑡 𝑓 𝑎𝑡 𝑑𝑡
𝑝𝑢𝑡 𝑎𝑡 = 𝑟 ⟹ 𝑑𝑡 =𝑑𝑟
𝑎𝑎𝑛𝑑 𝑎𝑙𝑠𝑜 𝑡 =
𝑟
𝑎
⇒ 𝐿 𝑓 𝑎𝑡 = 0
∞
𝑒−𝑠𝑟𝑎 𝑓 𝑟
𝑑𝑟
𝑎=
1
𝑎 0
∞
𝑒−𝑆𝑟𝑓 𝑟 𝑑𝑟 𝑤ℎ𝑒𝑟𝑒 𝑆 =𝑠
𝑎
=1
𝑎𝐹 𝑆 =
1
𝑎𝐹
𝑠
𝑎
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6. FIRST SHIFTING THEOREM
If 𝐹 𝑠 has the Laplace transform of 𝑓 𝑡 then𝑳 𝒆𝒂𝒕𝒇 𝒕 = 𝑭(𝒔 −
𝒂)
Proof: Using the definition of Laplace transform
𝐹 𝑠 − 𝑎 = 0
∞
𝑒− 𝑠−𝑎 𝑡𝑓 𝑡 𝑑𝑡
= 0
∞
𝑒−𝑠𝑡+𝑎𝑡𝑓 𝑡 𝑑𝑡 = 0
∞
𝑒−𝑠𝑡𝑒𝑎𝑡𝑓 𝑡 𝑑𝑡
= 0∞
𝑒−𝑠𝑡 𝑒𝑎𝑡𝑓 𝑡 𝑑𝑡 = 𝐿 𝑒𝑎𝑡𝑓 𝑡
Alternative Method:
𝐿 𝑒𝑎𝑡𝑓 𝑡 = 0
∞
𝑒𝑎𝑡𝑒−𝑠𝑡𝑓 𝑡 𝑑𝑡 = 0
∞
𝑒−𝑠𝑡+𝑎𝑡𝑓 𝑡 𝑑𝑡
= 0
∞
𝑒− 𝑠−𝑎 𝑡𝑓 𝑡 𝑑𝑡 = 0
∞
𝑒−𝑢𝑡𝑓 𝑡 𝑑𝑡
Using 𝑠 − 𝑎 = 𝑢
= 𝐹 𝑢 = 𝐹(𝑠 − 𝑎)
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7. SECOND SHIFTING THEOREM(HEAVISIDE’S SHIFTING THEOREM)
If 𝐿 𝑓 𝑡 = 𝐹(𝑠) and 𝑔 𝑡 =𝑓 𝑡 − 𝑎 , 𝑓𝑜𝑟 𝑡 > 𝑎0, 𝑓𝑜𝑟 0 < 𝑡
< 𝑎
Then 𝐿 𝑔 𝑡 = 𝑒−𝑎𝑠𝐹(𝑠)
Proof: As per the definition of Laplace transform
𝐿 𝑔 𝑡 = 0
∞
𝑒−𝑠𝑡 𝑔 𝑡 𝑑𝑡
𝐿 𝑔 𝑡 = 0
𝑎
𝑒−𝑠𝑡 𝑔 𝑡 𝑑𝑡 + 𝑎
∞
𝑒−𝑠𝑡𝑔 𝑡 𝑑𝑡
Using the given condition 𝑔 𝑡 = 0 𝑓𝑜𝑟 0 < 𝑡 < 𝑎 𝑎𝑛𝑑 𝑔 𝑡 =
𝑓 𝑡 − 𝑎 𝑓𝑜𝑟 𝑡 > 𝑎
𝐿 𝑔 𝑡 =0 + 𝑎
∞
𝑒−𝑠𝑡𝑓 𝑡 − 𝑎 𝑑𝑡
𝑁𝑜𝑤 𝑢𝑠𝑖𝑛𝑔 𝑡 − 𝑎 = 𝑟 ⇒ 𝑑𝑡 = 𝑑𝑟 𝑎𝑛𝑑 𝑡 = (𝑟 + 𝑎) we get
𝐿 𝑔 𝑡 = 0
∞
𝑒−𝑠(𝑟+𝑎)𝑓 𝑟 𝑑𝑟 = 𝑒−𝑠𝑎 0
∞
𝑒−𝑠𝑟𝑓 𝑟 𝑑𝑟 = 𝑒−𝑠𝑎𝐹(𝑠)
Hence ⇒ 𝐿 𝑔 𝑡 = 𝑒−𝑎𝑠𝐹 𝑠
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8. LAPLACE TRANSFORM OF THE DERIVATIVE OF 𝒇 𝒕
If 𝐿 𝑓 𝑡 = 𝐹(𝑠) and 𝑓′(𝑡) is the derivative of 𝑓′(𝑡) then𝑳 𝒇′(𝒕)
= 𝒔𝑳 𝒇(𝒕) − 𝒇(𝟎)
Proof: As we know
𝐿 𝑓′(𝑡) = 𝑎
∞𝑒−𝑠𝑡𝑓′ 𝑡 𝑑𝑡
Solving above equation using integration by parts we get
𝐿 𝑓′(𝑡) = 𝑒−𝑠𝑡𝑓 𝑡 0∞ −
0
∞
(−𝑠𝑒−𝑠𝑡)𝑓 𝑡 𝑑𝑡
As we know that𝑒−∞ = 0 𝑎𝑛𝑑 𝑒0 = 1 ⇒ 𝑒−𝑠𝑡𝑓 𝑡 = 0 𝑤ℎ𝑒𝑛 𝑡 = ∞ 𝑎𝑛𝑑
𝑒−𝑠𝑡𝑓 𝑡 = 𝑓 0 𝑤ℎ𝑒𝑛 𝑡 = 0
⇒ 𝐿 𝑓′ (𝑡) = −𝑓(0) + 𝑠 0∞
𝑒−𝑠𝑡𝑓 𝑡 𝑑𝑡 = −𝑓(0) + 𝑠𝐿 𝑓(𝑡)⇒ 𝑳 𝒇′ (𝒕) = 𝒔𝑳 𝒇 𝒕 − 𝒇 𝟎
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9. LAPLACE TRANSFORM OF THE DERIVATIVE OF ORDER n
𝐿 𝑓𝑛(𝑡) = 𝑠𝑛𝐿 𝑓 𝑡 − 𝑠𝑛−1𝑓 0 − 𝑠𝑛−2𝑓′ 0 − 𝑠𝑛−3𝑓′′ 0 − ⋯−
𝑓𝑛−1(0)
Proof: As we know that the Laplace transform of derivative is
given by 𝐿 𝑓′(𝑡) = 𝑠𝐿 𝑓(𝑡) − 𝑓(0)……………… . 1
Using this equation we can find the Laplace transform of 𝑓′′(𝑡)
𝐿 𝑓′′(𝑡) = 𝑠𝐿 𝑓′(𝑡) − 𝑓′(0)
Using equation 1 we get𝐿 𝑓′′(𝑡) = 𝑠{𝑠𝐿 𝑓 𝑡 − 𝑓(0)} − 𝑓′(0)
𝐿 𝑓′′(𝑡) = 𝑠2𝐿 𝑓(𝑡) − 𝑠𝑓(0) − 𝑓′(0)………… .2
Similarly we can find the value of 𝐿 𝑓′′′(𝑡) by using equation 1
&2𝐿 𝑓′′′(𝑡) = 𝑠3𝐿 𝑓(𝑡) − 𝑠2𝑓(0) − 𝑠𝑓′(0) − 𝑓′′(0)………… .3
Similarly, using above method we get𝐿 𝑓𝑛(𝑡) = 𝑠𝑛𝐿 𝑓 𝑡 − 𝑠𝑛−1𝑓 0
− 𝑠𝑛−2𝑓′ 0 − 𝑠𝑛−3𝑓′′ 0 − ⋯− 𝑓𝑛−1(0)
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10. LAPLACE TRANSFORM OF THE INTEGRAL OF 𝒇 𝒕
If 𝐿 𝑓 𝑡 = 𝐹(𝑠) and 𝑓′(𝑡) is the derivative of 𝑓′(𝑡) then
𝑳 𝟎
𝒕
𝒇 𝒕 𝒅𝒕 =𝟏
𝒔𝑭(𝒔)
Proof: Let 𝑔(𝑡) = 0𝑡𝑓 𝑡 𝑎𝑛𝑑 𝑔 0 = 0 then 𝑔′(𝑡) = 𝑓(𝑡)
As we know that 𝐿 𝑔′ 𝑡 = 𝑠𝐿 𝑔 𝑡 − 𝑔(0)⇒ 𝐿 𝑔′ 𝑡 = 𝑠𝐿 𝑔 𝑡 𝑎𝑠 𝑔 0 =
0
⇒ 𝐿 𝑔 𝑡 =1
𝑠𝐿 𝑔′ 𝑡
Using the value of 𝑔(𝑡) = 0𝑡𝑓 𝑡 𝑎𝑛𝑑 𝑔′ 𝑡 = 𝑓(𝑡) we will get
⇒ 𝐿 0
𝑡
𝑓 𝑡 𝑑𝑡 =1
𝑠𝐿 𝑓 𝑡
⇒ 𝑳 𝟎
𝒕
𝒇 𝒕 𝒅𝒕 =𝟏
𝒔𝑭 𝒔
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11.LAPLACE TRANSFORM OF SOME IMPORTANT FUNCTIONS
1. 𝑳(𝟏) =𝟏
𝑺
Proof: From the definition of Laplace transform, the Laplace
transform of L(1) can be written as
𝐿(1) = 0
∞
1. 𝑒−𝑠𝑡 𝑑𝑡 =𝑒−𝑠𝑡
−𝑠0
∞
= −1
𝑠0 − 1 =
1
𝑠
𝐴𝑠 𝑒−∞ = 0 𝑎𝑛𝑑 𝑒0 = 1
2. 𝑳(𝒆𝒂𝒕) =𝟏
𝒔−𝒂𝑤ℎ𝑒𝑟𝑒 𝑠 > 𝑎
Proof: As per the definition of Laplace transform
𝐿(𝑒𝑎𝑡) = 0
∞
𝑒𝑎𝑡 . 𝑒−𝑠𝑡 𝑑𝑡 = 0
∞
𝑒 𝑎−𝑠 𝑡 𝑑𝑡
= 0
∞
𝑒−(𝑠−𝑎)𝑡 𝑑𝑡 =𝑒− 𝑠−𝑎 𝑡
−(𝑠 − 𝑎)0
∞
=1
𝑠 − 𝑎𝐴𝑠 𝑒−∞ = 0 𝑎𝑛𝑑 𝑒0 = 1
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3. 𝑳 𝐬𝐢𝐧𝒂𝒕 =𝒂
𝒔𝟐+𝒂𝟐
Proof: 𝐿(sin 𝑎𝑡) =𝐿(𝑒𝑖𝑎𝑡 − 𝑒−𝑖𝑎𝑡
2𝑖) =
1
2𝑖𝐿 𝑒𝑖𝑎𝑡 − 𝐿 𝑒−𝑖𝑎𝑡
Since sin 𝜃 =1
2𝑖𝑒𝑖𝜃 − 𝑒−𝑖𝜃
𝑈𝑠𝑖𝑛𝑔 𝐿𝑎𝑝𝑙𝑎𝑐𝑒 𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚 𝐿(𝑒𝑎𝑡) =1
𝑠−𝑎𝑤𝑒 𝑤𝑖𝑙𝑙 𝑔𝑒𝑡
𝐿 sin 𝑎𝑡 =1
2𝑖
1
𝑠 − 𝑖𝑎−
1
𝑠 + 𝑖𝑎=
1
2𝑖
𝑠 + 𝑖𝑎 − (𝑠 − 𝑖𝑎)
𝑠 − 𝑖𝑎 (𝑠 + 𝑖𝑎)
=1
2𝑖
2𝑖𝑎
𝑠2 + 𝑎2=
𝑎
𝑠2 + 𝑎2
4. 𝑳 𝐜𝐨𝐬𝒂𝒕 =𝒔
𝒔𝟐+𝒂𝟐
Proof: 𝐿(cos 𝑎𝑡) =𝐿(𝑒𝑖𝑎𝑡 + 𝑒−𝑖𝑎𝑡
2) =
1
2𝐿 𝑒𝑖𝑎𝑡 + 𝐿 𝑒−𝑖𝑎𝑡
Since cos 𝜃 =1
2𝑒𝑖𝜃 + 𝑒−𝑖𝜃
𝑈𝑠𝑖𝑛𝑔 𝐿𝑎𝑝𝑙𝑎𝑐𝑒 𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚 𝐿(𝑒𝑎𝑡) =1
𝑠−𝑎𝑤𝑒 𝑤𝑖𝑙𝑙 𝑔𝑒𝑡
𝐿 cos 𝑎𝑡 =1
2
1
𝑠 − 𝑖𝑎+
1
𝑠 + 𝑖𝑎=
1
2
𝑠 + 𝑖𝑎 + (𝑠 − 𝑖𝑎)
𝑠 − 𝑖𝑎 (𝑠 + 𝑖𝑎)
=1
2
2𝑠
𝑠2 + 𝑎2=
𝑠
𝑠2 + 𝑎2𝑎𝑠 𝑤𝑒 𝑘𝑛𝑜𝑤 𝑖2 = −1
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5. 𝑳 𝐬𝐢𝐧𝐡𝒂𝒕 =𝒂
𝒔𝟐− 𝒂𝟐
Proof: 𝐿(sinh 𝑎𝑡) = 𝐿(𝑒𝑎𝑡 − 𝑒−𝑎𝑡
2𝑖) =
1
2𝐿 𝑒𝑎𝑡 − 𝐿 𝑒−𝑎𝑡
Since sinh 𝜃 =1
2𝑒𝜃 − 𝑒−𝜃
𝑈𝑠𝑖𝑛𝑔 𝐿𝑎𝑝𝑙𝑎𝑐𝑒 𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚 𝐿(𝑒𝑎𝑡) =1
𝑠−𝑎𝑤𝑒 𝑤𝑖𝑙𝑙 𝑔𝑒𝑡
𝐿 sin 𝑎𝑡 =1
2
1
𝑠 − 𝑎−
1
𝑠 + 𝑎=
1
2
𝑠 + 𝑎 − (𝑠 − 𝑎)
𝑠 − 𝑎 (𝑠 + 𝑎)
=1
2
2𝑎
𝑠2 − 𝑎2=
𝑎
𝑠2 − 𝑎2
6. 𝑳 𝐜𝐨𝐬𝐡𝒂𝒕 =𝒔
𝒔𝟐− 𝒂𝟐
Proof: 𝐿(cosh 𝑎𝑡) =𝐿(𝑒𝑎𝑡 + 𝑒−𝑎𝑡
2) =
1
2𝐿 𝑒𝑎𝑡 + 𝐿 𝑒−𝑎𝑡
Since cos 𝜃 =1
2𝑒𝜃 + 𝑒−𝜃
𝑈𝑠𝑖𝑛𝑔 𝐿𝑎𝑝𝑙𝑎𝑐𝑒 𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚 𝐿(𝑒𝑎𝑡) =1
𝑠−𝑎𝑤𝑒 𝑤𝑖𝑙𝑙 𝑔𝑒𝑡
𝐿 cosh𝑎𝑡 =1
2
1
𝑠 − 𝑎+
1
𝑠 + 𝑎=
1
2
𝑠 + 𝑎 + (𝑠 − 𝑎)
𝑠 − 𝑎 (𝑠 + 𝑎)
=1
2
2𝑠
𝑠2 − 𝑎2=
𝑠
𝑠2 − 𝑎2
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7. 𝑳 𝒕𝒏 =𝒏!
𝒔𝒏+𝟏𝒘𝒉𝒆𝒓𝒆 𝒏 𝒂𝒏𝒅 𝒔 𝒂𝒓𝒆 𝒑𝒐𝒔𝒊𝒕𝒊𝒗𝒆
Proof: 𝑳 𝒕𝒏 = 0∞
𝑡𝑛. 𝑒−𝑠𝑡𝑑𝑡
𝑛𝑜𝑤 𝑢𝑠𝑖𝑛𝑔 𝑠𝑡 = 𝑢 ⇒ 𝑡 =𝑢
𝑠⇒ 𝑑𝑡 =
𝑑𝑢
𝑠We will get
𝐿 𝑡𝑛 = 0
∞ 𝑢𝑛
𝑠𝑛. 𝑒−𝑢
𝑑𝑢
𝑠=
1
𝑠𝑛+1 0
∞
𝑒−𝑢 𝑢𝑛 𝑑𝑢
𝑤𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑎𝑡 0
∞
𝑒−𝑢 𝑢𝑛 𝑑𝑢 = Γ 𝑛 + 1 = 𝑛!
Hence we have 𝐿 𝑡𝑛 =𝑛!
𝑠𝑛+1
-
Example 1: Find the Laplace transform of 𝑠𝑖𝑛3 2𝑡
Solution: we have given 𝑓 𝑡 = 𝑠𝑖𝑛3 2𝑡
And we also know that sin 3𝜃 = 3 sin 𝜃 − 4𝑠𝑖𝑛3 𝜃
From above equation 𝑠𝑖𝑛3 2𝑡 =1
43 sin 2𝑡 − sin 6𝑡
Hence 𝐿 𝑠𝑖𝑛3 2𝑡 =1
43L(sin 2𝑡) − L(sin 6𝑡)
=1
4
6
𝑠2 + 4−
6
𝑠2 + 36𝐴𝑠 𝑤𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑎𝑡 𝐿 sin 𝑎𝑡 =
𝑎
𝑠2 + 𝑎2
=6
4
𝑠2 + 36 − 𝑠2 − 4
𝑠2 + 4 𝑠2 + 36=
48
𝑠2 + 4 𝑠2 + 36
-
Example 2: Find the Laplace transform of 𝑠𝑖𝑛 2𝑡 𝑠𝑖𝑛3𝑡
Solution: we have given 𝑓 𝑡 = 𝑠𝑖𝑛 2𝑡 𝑠𝑖𝑛3𝑡
𝑈𝑠𝑖𝑛𝑔 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 → 𝑠𝑖𝑛 𝐴 𝑠𝑖𝑛𝐵 =1
2cos 𝐴 − 𝐵 − cos(𝐴 + 𝐵)
≫ 𝑠𝑖𝑛 2𝑡 𝑠𝑖𝑛3𝑡 =1
2cos 𝑡 − cos 5𝑡
So 𝐿 𝑠𝑖𝑛 2𝑡 𝑠𝑖𝑛3𝑡 =1
2𝐿(cos 𝑡) − L(cos 5𝑡)
Now using relation 𝐿 cos 𝑎𝑡 =𝑠
𝑠2+𝑎2
We have 𝐿 𝑠𝑖𝑛 2𝑡 𝑠𝑖𝑛3𝑡 =1
2
𝑠
𝑠2+1−
𝑠
𝑠2+25=
12𝑠
𝑠2+1 (𝑠2+25)
-
Example 3: Show that 𝐿1
𝜋𝑡=
1
𝑠𝐺𝑖𝑣𝑒𝑛 𝑡ℎ𝑎𝑡 𝐿 2
𝑡
𝜋=
1
𝑠 ൗ3
2
Solution: Suppose 𝐹 𝑡 = 2𝑡
𝜋𝑡ℎ𝑒𝑛
𝐹′ 𝑡 =1
𝜋𝑡𝑎𝑛𝑑 𝑎𝑙𝑠𝑜 𝑤𝑒 𝑐𝑎𝑛 𝑠𝑒𝑒 𝑡ℎ𝑎𝑡 𝐹 0 = 0
Now we know that 𝐿 𝐹′ 𝑡 = 𝑠𝐿 𝐹 𝑡 − 𝐹(0)
Hence 𝐿1
𝜋𝑡= 𝑠𝐿 2
𝑡
𝜋− 0 = 𝑠.
1
𝑠 ൗ3
2− 0
⇒ 𝐿1
𝜋𝑡=
1
𝑠Example 4: Find the Laplace transform of 𝑡 + 𝑡2 + 𝑡3
Solution: we have given 𝑓 𝑡 = 𝑡 + 𝑡2 + 𝑡3
Now using the relation 𝐿 𝑡𝑛 =𝑛!
𝑠𝑛+1
We have𝐿 𝑓 𝑡 = 𝐿 𝑡 + 𝐿 𝑡2 + 𝐿 𝑡3 =1
𝑠2+
2
𝑠3+
6
𝑠4
-
Example 5: Find the Laplace transform of 𝑡 𝑐𝑜𝑠ℎ 𝑎𝑡
Solution: we have given 𝑓 𝑡 = 𝑡 𝑐𝑜𝑠ℎ 𝑎𝑡
We know that 𝐿 𝑐𝑜𝑠ℎ 𝑎𝑡 =𝑠
𝑠2−𝑎2
Now using the relation 𝐿 𝑡𝑛𝑓 𝑡 = (−1)𝑛𝑑𝑛
𝑑𝑠𝑛 𝐹 𝑠
We will get 𝐿 𝑡 𝑐𝑜𝑠ℎ 𝑎𝑡 = −𝑑
𝑑𝑠
𝑠
𝑠2−𝑎2= −
(𝑠2−𝑎2).1−𝑠.2𝑠
(𝑠2−𝑎2)2
= −(𝑠2−𝑎2 − 2𝑠2
(𝑠2−𝑎2)2
=(𝑠2+𝑎2)
(𝑠2−𝑎2)2
-
12. LAPLACE TRANSFORM OF 𝟏
𝒕𝒇(𝒕)
If 𝑳 𝒇 𝒕 = 𝑭(𝒔) then If 𝑳𝟏
𝒕𝒇(𝒕) = 𝒔
∞𝒇(𝒔) 𝒅𝒔
Proof: As per the Laplace transform
𝐿 𝑓 𝑡 = 𝐹 𝑠 = 0∞
𝑒−𝑠𝑡 𝑓 𝑡 𝑑𝑡
Integrating with respect to s, we get
𝑠
∞
𝐹 𝑠 𝑑𝑠 = 𝑠
∞
0
∞
𝑒−𝑠𝑡𝑓 𝑡 𝑑𝑡 𝑑𝑠
= 0
∞
𝑓(𝑡) 𝑠
∞
𝑒−𝑠𝑡𝑑𝑠 𝑑𝑡 = 0
∞
𝑓(𝑡)𝑒−𝑠𝑡
−𝑡𝑠
∞
𝑑𝑡
= − 0
∞ 𝑓(𝑡)
𝑡𝑒−𝑠𝑡 0
∞𝑑𝑡 = − 0
∞ 𝑓(𝑡)
𝑡𝑒−∞ − 𝑒−𝑠𝑡 𝑑𝑡
= − 0
∞ 𝑓(𝑡)
𝑡0 − 𝑒−𝑠𝑡 𝑑𝑡 =
0
∞
𝑒−𝑠𝑡1
𝑡𝑓(𝑡) 𝑑𝑡 = 𝐿
1
𝑡𝑓(𝑡)
⇒ 𝑳𝟏
𝒕𝒇(𝒕) =
𝒔
∞
𝑭 𝒔 𝒅𝒔 𝑷𝒓𝒐𝒗𝒆𝒅
Similarly: LAPLACE TRANSFORM OF 𝒕. 𝒇(𝒕)
𝑳 𝒕𝒏. 𝒇(𝒕) = (−𝟏)𝒏𝒅𝒏
𝒅𝒔𝒏𝑭(𝒔)
-
13. LAPLACE TRANSFORM OF 𝒕. 𝒇(𝒕)
𝑳 𝒕𝒏. 𝒇(𝒕) = (−𝟏)𝒏𝒅𝒏
𝒅𝒔𝒏𝑭(𝒔)
Example 6: Find the Laplace transform of sin 2𝑡
𝑡
Solution: 𝐿(sin 2𝑡) =2
𝑠2+4
𝐿sin 2𝑡
𝑡= 𝑠
∞ 2
𝑠2+4𝑑𝑠 = 2.
1
2tan−1
𝑠
2 𝑠
∞= tan−1 ∞ − tan−1
𝑠
2=
𝜋
2− tan−1
𝑠
2
=cot−1𝑠
2
Example 7: Find the Laplace transform of the function
𝑓 𝑡 = 𝑡𝑒−𝑡 sin 2𝑡
Solution: 𝐿 sin 2𝑡 =2
𝑠2+4
𝐿 𝑒−𝑡 sin 2𝑡 =2
𝑠+1 2+4= 𝐹 𝑠 𝑠𝑎𝑦
𝐿 𝑒−𝑡 sin 2𝑡 = −𝐹′ 𝑠 =𝑑
𝑑𝑠
2
𝑠+1 2+4=
2.2(𝑠+1)
𝑠+1 2+4 2=
4(𝑠+1)
𝑠+1 2+4 2Ans.
-
14. UNIT STEP FUNCTION
The unit step function is defined as follows:
𝒖 𝒕 − 𝒂 = ቊ𝟎, 𝒘𝒉𝒆𝒏 𝒕 < 𝑎𝟏, 𝒘𝒉𝒆𝒏 𝒕 ≥ 𝒂
𝒘𝒉𝒆𝒓𝒆 𝒂 ≥ 𝟎
-
15. LAPLACE TRANSFORM OF UNIT STEP FUNCTION
𝑳 𝒖 𝒕 − 𝒂 =𝒆−𝒂𝒔
𝒔Proof: Using the definition of Laplace transform, we have
𝑳 𝒖 𝒕 − 𝒂 = 𝟎
∞
𝒆−𝒔𝒕𝒖 𝒕 − 𝒂 𝒅𝒕
Now using the condition of unit step function
𝐿 𝑢 𝑡 − 𝑎 = 0
𝑎
𝑒−𝑠𝑡𝑢 𝑡 − 𝑎 𝑑𝑡 + 𝑎
∞
𝑒−𝑠𝑡𝑢 𝑡 − 𝑎 𝑑𝑡
= 0
𝑎
𝑒−𝑠𝑡0. 𝑑𝑡 + 𝑎
∞
𝑒−𝑠𝑡1. 𝑑𝑡 = 0 +𝑒−𝑠𝑡
−𝑠𝑎
∞
𝐿 𝑢 𝑡 − 𝑎 =𝑒−𝑎𝑠
𝑠
-
Example 8: Convert the following function in terms of unit step
function and then find the Laplace Transform
𝑓 𝑡 = ቊ6, 𝑤ℎ𝑒𝑛 𝑡 < 24, 𝑤ℎ𝑒𝑛 𝑡 ≥ 2
Solution: Given that
𝑓 𝑡 = ቊ6, 𝑤ℎ𝑒𝑛 𝑡 < 24, 𝑤ℎ𝑒𝑛 𝑡 ≥ 2
This further can be written as
𝑓 𝑡 = ቊ6 + 0, 𝑤ℎ𝑒𝑛 𝑡 < 26 − 2, 𝑤ℎ𝑒𝑛 𝑡 ≥ 2
= 6 + ቊ0, 𝑤ℎ𝑒𝑛 𝑡 < 2
−2, 𝑤ℎ𝑒𝑛 𝑡 ≥ 2
= 6 + −2 ቊ0, 𝑤ℎ𝑒𝑛 𝑡 < 21, 𝑤ℎ𝑒𝑛 𝑡 ≥ 2
= 6 − 2𝑢 𝑡 − 2
[Using the condition of unit step function]
𝐿 𝑓 𝑡 = 6𝐿 1 − 2𝐿 𝑢 𝑡 − 2 =6
𝑠− 2
𝑒−2𝑠
𝑠
-
16. PERIODIC FUNCTIONSIf 𝑓 𝑡 be a periodic function with period
𝑇,⇒ 𝑓 𝑡 + 𝑇 = 𝑓(𝑡) then
𝑳 𝒇 𝒕 =𝟎
𝑻𝒆−𝒔𝒕𝒇 𝒕 𝒅𝒕
𝟏 − 𝒆−𝒔𝑻
Proof: As we know
𝐿 𝑓 𝑡 = 0
∞
𝑒−𝑠𝑡 𝑓 𝑡 𝑑𝑡
This can be written as in the following manner
𝐿 𝑓 𝑡 = 0∞
𝑒−𝑠𝑡 𝑓 𝑡 𝑑𝑡 = 0𝑇𝑒−𝑠𝑡 𝑓 𝑡 𝑑𝑡 + 𝑇
2𝑇𝑒−𝑠𝑡 𝑓 𝑡 𝑑𝑡 + 2𝑇
3𝑇𝑒−𝑠𝑡 𝑓 𝑡 𝑑𝑡 + ⋯
Now substituting 𝑡 = 𝑢 + 𝑇, 𝑡 = 𝑢 + 2𝑇, …𝑎𝑛𝑑 𝑑𝑡 = 𝑑𝑢 in second
integral, third integral, and so on respectively, we will get
𝐿 𝑓 𝑡 = 0𝑇𝑒−𝑠𝑡 𝑓 𝑡 𝑑𝑡 + 0
𝑇𝑒−𝑠(𝑢+𝑇) 𝑓 𝑢 + 𝑇 𝑑𝑢 + 0
𝑇𝑒−𝑠(𝑢+2𝑇) 𝑓 𝑢 + 2𝑇 𝑑𝑢 + ⋯
= 0
𝑇
𝑒−𝑠𝑡 𝑓 𝑡 𝑑𝑡 + 𝑒−𝑠𝑇 0
𝑇
𝑒−𝑠𝑢 𝑓 𝑢 + 𝑇 𝑑𝑢 + 𝑒−2𝑠𝑇 0
𝑇
𝑒−𝑠𝑢 𝑓 𝑢 + 2𝑇 𝑑𝑢 + ⋯
= 0
𝑇
𝑒−𝑠𝑡 𝑓 𝑡 𝑑𝑡 + 𝑒−𝑠𝑇 0
𝑇
𝑒−𝑠𝑢 𝑓 𝑢 𝑑𝑢 + 𝑒−2𝑠𝑇 0
𝑇
𝑒−𝑠𝑢 𝑓 𝑢 𝑑𝑢 + ⋯
-
As 𝑓 𝑢 be a periodic function with period 𝑇,⇒ 𝑓 𝑢 + 𝑇 =𝑓 𝑢 + 2𝑇
= ⋯ = 𝑓(𝑢)
Now we can write
= 0
𝑇
𝑒−𝑠𝑡 𝑓 𝑡 𝑑𝑡 + 𝑒−𝑠𝑇 0
𝑇
𝑒−𝑠𝑡 𝑓 𝑡 𝑑𝑢 + 𝑒−2𝑠𝑇 0
𝑇
𝑒−𝑠𝑡 𝑓 𝑢 𝑑𝑡 + ⋯
= 0
𝑇
𝑒−𝑠𝑡 𝑓 𝑡 𝑑𝑡 1 + 𝑒−𝑠𝑇 + 𝑒−2𝑠𝑇 + ⋯
𝑁𝑜𝑤 𝑢𝑠𝑖𝑛𝑔 𝑡ℎ𝑒 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 1 + 𝑥 + 𝑥2 + 𝑥3 + … =1
1−𝑥we have
𝐿 𝑓 𝑡 =0
𝑇𝑒−𝑠𝑡 𝑓 𝑡 𝑑𝑡
1 − 𝑒−𝑠𝑇
-
Example 8: Find the Laplace transform of the waveform
𝑓 𝑡 =2𝑡
3, 0 ≤ 𝑡 ≤ 3.
Solution: 𝐿 𝑓(𝑡) =1
1−𝑒−𝑠𝑇0
𝑇𝑒−𝑠𝑇𝑓 𝑡 𝑑𝑡
𝐿2𝑡
3=
1
1−𝑒−3𝑠0
3𝑒−𝑠𝑡
2
3𝑡 𝑑𝑡 =
1
1−𝑒−3𝑠2
3
𝑡𝑒−𝑠𝑡
−𝑠− 1
𝑒−𝑠𝑡
𝑠2 0
3
=2
3
1
1−𝑒−3𝑠3𝑒−3𝑠
−𝑠−
𝑒−3𝑠
𝑠2+
1
𝑠2=
2
3
1
1−𝑒−3𝑠3𝑒−3𝑠
−𝑠+
1−𝑒−3𝑠
𝑠2
=2𝑒−3𝑠
−𝑠 1−𝑒−3𝑠+
2
3𝑠2
-
17. SOME IMPORTANT FORMULAE OF LAPLACE TRANSFORM
S.No. 𝑓 𝑡 𝐹 𝑠
1 𝑒𝑎𝑡1
𝑠 − 𝑎
2 sin 𝑎𝑡𝑎
𝑠2 + 𝑎2
3 cos 𝑎𝑡𝑠
𝑠2 + 𝑎2
4 sinh 𝑎𝑡𝑎
𝑠2 − 𝑎2
5 cosh 𝑎𝑡𝑠
𝑠2 − 𝑎2
6 𝑡𝑛𝑛!
𝑠𝑛+1
7 ebtsin 𝑎𝑡𝑎
𝑠 − 𝑏 2 + 𝑎2
8 ebtcos𝑎𝑡𝑠 − 𝑏
𝑠 − 𝑏 2 + 𝑎2
9𝑡
2𝑎sin 𝑎𝑡
𝑠
𝑠2 + 𝑎2 2
10 𝑡 cos 𝑎𝑡𝑠2 − 𝑎2
𝑠2 + 𝑎2 2
-
18. INVERSE LAPLACE TRANSFORM
If 𝐹 𝑠 is the Laplace Transform of a function 𝑓 𝑡 , then 𝑓 𝑡 is
known as Inverse Laplace Transform.
𝑓 𝑡 = 𝐿−1 𝐹 𝑠
The Inverse Laplace Transform is very useful to solving the
differential equations without finding the general solution and
arbitrary constants.
-
19.SOME IMPORTANT FORMULAE OF INVERSE LAPLACE TRANSFORM
S.No. 𝐹 𝑠 𝑓 𝑡 = 𝐿−1 𝐹 𝑠
11
𝑠 − 𝑎𝑒𝑎𝑡
2𝑎
𝑠2 + 𝑎2sin 𝑎𝑡
3𝑠
𝑠2 + 𝑎2cos𝑎𝑡
4𝑎
𝑠2 − 𝑎2sinh 𝑎𝑡
5𝑠
𝑠2 − 𝑎2cosh𝑎𝑡
6𝑛!
𝑠𝑛+1𝑡𝑛
7𝑎
𝑠 − 𝑏 2 + 𝑎2ebtsin 𝑎𝑡
8𝑠 − 𝑏
𝑠 − 𝑏 2 + 𝑎2ebtcos 𝑎𝑡
9𝑠
𝑠2 + 𝑎2 2𝑡
2𝑎sin 𝑎𝑡
10𝑠2 − 𝑎2
𝑠2 + 𝑎2 2𝑡 cos𝑎𝑡
111
𝑠
1
-
Example 9: Prove that 1
𝑠1/2= 𝐿
1
𝜋𝑡
Solution: we know that 𝐿−11
𝑠𝑛=
𝑡𝑛−1
(𝑛−1)!=
𝑡𝑛−1
Γ𝑛
Using above relation we can write 𝐿−11
𝑠1/2=
𝑡12−1
Γ1
2
=𝑡−
12
π𝑆𝑖𝑛𝑐𝑒 Γ
1
2= π
⇒ 𝐿−11
𝑠1/2=
1
πt⇒
1
𝑠1/2= 𝐿
1
πt
-
Example 10: Find the inverse Laplace Transform of the
following:
(i) 1
𝑠−3(ii)
1
𝑠2−25(iii)
𝑠
𝑠2+16
(iv) 1
𝑠2+9(v)
1
𝑠−2 2+1(vi)
𝑠−1
𝑠−1 2+4
Solution.
(i) 𝐿−11
𝑠−3= 𝑒3𝑡 𝑠𝑖𝑛𝑐𝑒 𝐿−1
1
𝑠−𝑎= 𝑒𝑎𝑡
(ii) 𝐿−11
𝑠2−25= 𝐿−1
1
5
5
𝑠2− 5 2=
1
5𝑠𝑖𝑛ℎ5𝑡 𝑠𝑖𝑛𝑐𝑒 𝐿−1
𝑎
𝑠2−𝑎2= 𝑠𝑖𝑛ℎ 𝑎𝑡
(iii) 𝐿−1𝑠
𝑠2+16= 𝐿−1
𝑠
𝑠2+ 4 2= cos 4𝑡 𝑠𝑖𝑛𝑐𝑒 𝑠𝑖𝑛𝑐𝑒 𝐿−1
𝑠
𝑠2+𝑎2= 𝑐𝑜𝑠 𝑎𝑡
(iv) 𝐿−11
𝑠2+9= 𝐿−1
1
𝑠2+ 3 2=
1
3sin 3𝑡 𝑠𝑖𝑛𝑐𝑒 𝑠𝑖𝑛𝑐𝑒 𝐿−1
1
𝑠2+𝑎2=
1
𝑎𝑠𝑖𝑛 𝑎𝑡
(v) 𝐿−11
𝑠−2 2+1= 𝑒2𝑡 sin 𝑡 𝑠𝑖𝑛𝑐𝑒 𝑠𝑖𝑛𝑐𝑒 𝐿−1
1
𝑠−𝑏 2+𝑎2= 𝑒𝑏𝑡𝑠𝑖𝑛 𝑎𝑡
(vi)𝐿−1𝑠−1
𝑠−1 2+4= 𝐿−1
𝑠−1
𝑠−1 2+(2)2= 𝑒𝑡 cos 2𝑡 𝑠𝑖𝑛𝑐𝑒 𝑠𝑖𝑛𝑐𝑒 𝐿−1
𝑠−𝑎
𝑠−𝑎 2+𝑏2= 𝑒𝑎𝑡𝑐𝑜𝑠 𝑏𝑡
-
Example 11: Find 𝐿−1𝑠2+3𝑠+8
𝑠3
Solution: Here, we have
𝐿−1𝑠2+3𝑠+8
𝑠3= 𝐿−1
1
𝑠+
3
𝑠2+
8
𝑠3
= 1 +3𝑡
1!+
8
2!𝑡2 𝑠𝑖𝑛𝑐𝑒 𝐿−1
1
𝑠𝑛=
𝑡𝑛−1
𝑛−1 !
= 1 + 3𝑡 + 4 𝑡2
-
20. MULTIPLICATION BY S
𝑳−𝟏 𝒔𝑭(𝒔) =𝒅
𝒅𝒕𝒇 𝒕 + 𝒇 𝟎 𝜹(𝒕)
Example 12: Find the Inverse Laplace Transform of 𝒔
𝒔𝟐+𝟒
Solution: we know that 𝐿−11
𝑠2+𝑎2= sin 𝑎𝑡
Hence 𝐿−11
𝑠2+4= sin 2𝑡
And 𝐿−1𝑠
𝑠2+4=
𝑑
𝑑𝑡(𝑠𝑖𝑛 2𝑡) + sin 0
𝛿 𝑡 = 2 cos 2𝑡 𝐴𝑛𝑠.
-
21. DIVISION BY s (MULTIPLICATION BY 𝟏
𝒔)
𝐿−1𝐹 (𝑠)
𝑠0 =
𝑡𝐿−1 𝐹 (𝑠) 𝑑𝑡 0 =
𝑡𝑓 𝑡 𝑑𝑡
Example 13: Find the Inverse Laplace Transform of
(i) 1𝑠 (𝑠+𝑎)
(ii) 1
𝑠(𝑠2+1)
Solution:
(i) 𝑠𝑖𝑛𝑐𝑒 𝐿−11
𝑠+𝑎= 𝑒−𝑎𝑡
𝐿−11
𝑠 𝑠+𝑎= 0
𝑡𝐿−1
1
𝑠+𝑎𝑑𝑡
= 0𝑡𝑒−𝑎𝑡 𝑑𝑡 =
𝑒−𝑎𝑡
−𝑎 0
𝑡
=𝑒−𝑎𝑡
−𝑎+
1
𝑎=
1
𝑎1 − 𝑒−𝑎𝑡
-
(ii) 𝑤𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑎𝑡 𝐿−11
(𝑠2+1)= sin 𝑡
𝐿−11
𝑠
1
𝑠2+1)= 0
𝑡𝐿−1
1
𝑠2+1= 0
𝑡sin 𝑡 𝑑𝑡 =
= −cos 𝑡 0𝑡 = −cos 𝑡 + 1 = 1 − cos 𝑡
-
22. FIRST SHIFTING PROPERTY
If the inverse Laplace transform of 𝐹 𝑠 𝑖𝑠 𝑓(𝑡) such that
𝐿−1 𝐹 𝑠 = 𝑓(𝑡)
Then 𝐿−1𝐹 𝑠 + 𝑎 = 𝑒−𝑎𝑡 𝐿−1 𝐹 𝑠
Example 14: Find the Inverse Laplace Transform of
(i) 1
(𝑠+4)4(ii)
𝑠
𝑠2+4𝑠 +13(iii)
1
9𝑠2+6𝑠+1
Solution:
(i) 𝑤𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑎𝑡 𝐿−11
𝑠4=
𝑡3
3!
then 𝐿−11
(𝑠+4 )4
= 𝑒−4𝑡𝐿−11
𝑠4𝑢𝑠𝑖𝑛𝑔 𝑓𝑖𝑟𝑠𝑡 𝑠ℎ𝑖𝑓𝑖𝑡𝑖𝑛𝑔 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦
= 𝑒−4𝑡𝑡3
3!=
1
6𝑒−4𝑡𝑡3
-
Solution (ii) 𝐿−1𝑠
𝑠2+4𝑠+13= 𝐿−1
𝑠+2−2
(𝑠+2)2+(3)2
= 𝐿−1𝑠+2
(𝑠+2)2+(3)2− 𝐿−1
2
(𝑠+2)2+(3)2
Using First shifting property ⇒ 𝐿−1𝐹 𝑠 + 𝑎 = 𝑒−𝑎𝑡 𝐿−1 𝐹 𝑠
= 𝑒−2𝑡𝐿−1𝑠
(𝑠)2+(3)2− 𝑒−2𝑡𝐿−1
2
3
3
𝑠2+32
= 𝑒−2𝑡 cos 3𝑡 −2
3𝑒−2𝑡 sin 3𝑡 𝐴𝑛𝑠.
Solution (iii) 𝐿−11
9𝑠2+6𝑠+1= 𝐿−1
1
(3𝑠+1 )2
=1
9𝐿−1
1
(𝑠+1
3)2
=1
9𝑒− Τ𝑡 3𝐿−1
1
𝑠2 𝑈𝑠𝑖𝑛𝑔 𝐹𝑖𝑟𝑠𝑡 𝑠ℎ𝑖𝑓𝑡𝑖𝑛𝑔 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦
=1
9𝑒− Τ𝑡 3𝑡 =
𝑡𝑒− Τ𝑡 3
9
-
23. SECOND SHIFTING PROPERTY
SECOND SHIFTING PROPERTY
𝐿−1 𝑒−𝑎𝑠 𝐹 (𝑠) = 𝑓(𝑡 − 𝑎)𝑢(𝑡 − 𝑎)
Example 15: Obtain Inverse Laplace Transform of
(𝑖) 𝑒−𝜋𝑠
(𝑠+3 ) (𝑖𝑖)
𝑒−𝑠
( 𝑠+1 )3
Solution:
(i) As we know that
𝐿−1 1
𝑠+3= 𝑒−3𝑡
Now using second shifting theorem we can find the inverse
Laplace transform of
𝐿−1𝑒−𝜋𝑠
(𝑠+3)= 𝑒−3( 𝑡−𝜋)𝑢(𝑡 − 𝜋) 𝑠𝑖𝑛𝑐𝑒 𝐿−1 𝑒−𝑎𝑠 𝐹 (𝑠) = 𝑓(𝑡 − 𝑎)𝑢(𝑡 −
𝑎)
-
(ii) 𝐴𝑠 𝑤𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑎𝑡 𝐿−11
𝑠3=
𝑡2
2!
Then
𝐿−11
( 𝑠+1 )3𝑒−𝑡
𝑡2
2! 𝑢𝑠𝑖𝑛𝑔 𝑓𝑖𝑟𝑠𝑡 𝑠ℎ𝑖𝑓𝑖𝑡𝑖𝑛𝑔 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦
Hence 𝐿−1𝑒−𝑠
( 𝑠+1 )3=
𝑒−( 𝑡−1)( 𝑡−1 )2
2!𝑢 𝑡 − 1 𝑢𝑠𝑖𝑛𝑔 𝑠𝑒𝑐𝑜𝑛𝑑 𝑠ℎ𝑖𝑓𝑖𝑡𝑖𝑛𝑔 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦
-
24. INVERSE LAPLACE TRANSFORMS OF DERIVATIVES
𝐿−1𝑑
𝑑𝑠𝐹 (𝑠) = −𝑡𝐿−1 𝐹(𝑠) = −𝑡𝑓 𝑡
⇒ 𝐿−1 𝐹(𝑠) = −1
𝑡𝐿−1
𝑑
𝑑𝑠𝐹 (𝑠)
Example 16: Find 𝐿−1 log𝑠+1
𝑠−1.
Solution: 𝐿−1 log𝑠+1
𝑠−1= −
1
𝑡𝐿−1
𝑑
𝑑𝑠log(
𝑠+1
𝑠−1)
𝑢𝑠𝑖𝑛𝑔 𝑖𝑛𝑣𝑒𝑟𝑠𝑒 𝑙𝑎𝑝𝑙𝑎𝑐𝑒 𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚 𝑜𝑓 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒𝑠
= −1
𝑡𝐿−1
𝑑
𝑑𝑠log 𝑠 + 1 −
𝑑
𝑑𝑠log(𝑠 − 1) = −
1
𝑡𝐿−1
1
𝑠+1−
1
𝑠−1
= −1
𝑡𝑒−𝑡 − 𝑒𝑡 =
1
𝑡𝑒𝑡 − 𝑒−𝑡 𝐴𝑛𝑠.
-
25. INVERSE LAPLACE TRANSFORM OF INTEGRALS
𝐿−1 𝑠
∞𝐹 𝑠 𝑑𝑠 =
𝑓(𝑡)
𝑡=
1
𝑡𝐿−1 𝐹(𝑠)
⇒ 𝐿−1 𝐹(𝑠) = 𝐿−1 𝑠
∞𝐹 𝑠 𝑑𝑠
Example17: Find the Inverse Laplace Transform of 2𝑠
(𝑠2+1)2
Solution: we have to find 𝐿−12𝑠
(𝑠2+1)2
We will solve this using inverse Laplace transform of
integrals
𝐿−12𝑠
(𝑠2+1)2= 𝑡𝐿−1
𝑠
∞ 2𝑠𝑑𝑠
(𝑠2+1)2
𝑡𝐿−1 −1
𝑠2 + 1 𝑠
∞
= 𝑡𝐿−1 −0 +1
𝑠2 + 1= 𝑡𝐿−1
1
𝑠2 + 1= 𝑡 𝑠𝑖𝑛 𝑡
-
26. INVERSE LAPLACE TRANSFORM BY PARTIAL FRACTION METHOD
Example 18: Find the Inverse Laplace Transform of 1
𝑠2−5𝑠+6.
Solution: Let us convert the given function into partial
fractions.
𝐿−11
𝑠2−5𝑠+6= 𝐿−1
1
𝑠−3−
1
𝑠−2
= 𝐿−11
𝑠−3− 𝐿−1
1
𝑠−2= 𝑒3𝑡 − 𝑒2𝑡
Example 19: Find the Inverse Laplace Transform of 𝑠+1
𝑠2−6𝑠+25.
Solution:
𝐿−11
𝑠2−6𝑠+25= 𝐿−1
1
(𝑠−3)2+(4)2= 𝐿−1
𝑠−3+4
(𝑠−3)2+(4)2
= 𝐿−1𝑠−3
(𝑠−3)2+(4)2+ 𝐿−1
4
(𝑠−3)2+(4)2
= 𝑒3𝑡 cos 4𝑡 + 𝑒3𝑡 sin 4𝑡 𝑈𝑠𝑖𝑛𝑔 𝑓𝑖𝑟𝑠𝑡 𝑠ℎ𝑖𝑓𝑖𝑡𝑖𝑛𝑔 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦
-
27. SOLUTION OF DIFFERENTIAL EQUATIONS BY LAPLACE TRANSFORMS
Ordinary linear differential equations with constant
coefficients can be easily solved by the Laplace Transform method,
without finding the general solution and the arbitrary constants.
The method wil be clear from the following examples:
Let us now discuss how the Laplace transform method solves ODEs
and initial value problems. We consider an initial value
problem
𝑦′′ + 𝑎𝑦′ + 𝑏𝑦 = 𝑟 𝑡 , 𝑦 0 = 𝐾0 ,𝑎𝑛𝑑 𝑦′ 0 = 𝐾1
where a and b are constant. Here is the given input (driving
force) applied to the mechanical or electrical system and is the
output (response to the input) to be obtained.
In Laplace’s method we do three steps:
-
Step 1. Setting up the subsidiary equation.
This is an algebraic equation for the transform 𝑌 = 𝐿 𝑦 obtained
by transforming the given differential equation using the Laplace
transform of derivatives,
𝑠2𝑌 − 𝑠𝑦 0 − 𝑦′ 0 + 𝑎 𝑠𝑌 − 𝑦 0 + 𝑏𝑌 = 𝑅(𝑠)
Where 𝑅 𝑠 = 𝐿(𝑟)
Now collecting the Y-terms, we have the subsidiary equation as
follows𝑠2 + 𝑎𝑠 + 𝑏 𝑌 = 𝑠 + 1 𝑦 0 + 𝑦′ 0 + 𝑅 𝑠
Step 2. Solution of the subsidiary equation by algebra.
We divide by and use the so-called transfer function
𝑄(𝑠) =1
𝑠2 + 𝑎𝑠 + 𝑏This gives the solution
𝑌 = 𝑠 + 1 𝑦 0 𝑄(𝑠) + 𝑦′ 0 𝑄(𝑠) + 𝑅 𝑠 𝑄(𝑠)
Note that Q depends neither on r(t) nor on the initial
conditions (but only on a and b).
Step 3. Inversion of Y to obtain 𝒚 = 𝑳−𝟏𝒀
Now take the inverse Laplace transform to get the solution of
differential equations.
-
Example 20: Solve the following equation by Laplace
transform
𝑦′′ − 𝑦 = 𝑡 ; 𝑦 0 = 1 𝑎𝑛𝑑 𝑦′ 0 = 1
Solution: The given equation is𝑦′′ − 𝑦 = 𝑡 ; 𝑦 0 = 1 𝑎𝑛𝑑 𝑦′ 0 =
1
Step 1: Taking the Laplace transform of the given equation, we
get the subsidiary equation
𝐿(𝑦′′) − 𝐿(𝑦) = 𝐿(𝑡)
Now using the condition of Laplace transform of derivatives with
𝐿 𝑦 = 𝑌𝐿 𝑓𝑛(𝑡) = 𝑠𝑛𝐿 𝑓 𝑡 − 𝑠𝑛−1𝑓 0 − 𝑠𝑛−2𝑓′ 0 − 𝑠𝑛−3𝑓′′ 0 − ⋯−
𝑓𝑛−1(0)
We get
𝑠2𝑌 − 𝑠𝑦 0 − 𝑦′ 0 − 𝑌 =1
𝑠2
(𝑠2−1)𝑌 = 𝑠𝑦 0 + 𝑦′ 0 +1
𝑠2= 𝑠 + 1 +
1
𝑠2
𝐴𝑠 𝑔𝑖𝑣𝑒𝑛𝑦 0 = 1 𝑎𝑛𝑑 𝑦′ 0 = 1
-
Step 2: Transfer function is given by
𝑄 𝑠 =1
(𝑠2−1)and hence
𝑌 = 𝑠 + 1 𝑄 𝑠 +𝑄 𝑠
𝑠2=
(𝑠 + 1)
(𝑠2−1)+
1
𝑠2(𝑠2−1)On simplifying
𝑌 =1
(𝑠 − 1)+
1
(𝑠2−1)−
1
𝑠2
Step 3: Now taking the inverse Laplace transform to get the
solution of differential equation
𝑦 𝑡 = 𝐿−1𝑌 = 𝐿−11
(𝑠 − 1)+ 𝐿−1
1
(𝑠2−1)−
1
𝑠2
𝑦 𝑡 = 𝐿−11
(𝑠 − 1)+ 𝐿−1
1
(𝑠2−1)− 𝐿−1
1
𝑠2
𝑦 𝑡 = 𝑒𝑡 + sinh 𝑡 − 𝑡 𝐴𝑛𝑠.
-
28. SELF ASSESSMENT QUESTION
• Self Assessment Question (SAQ) 1: Find the Laplace transform
of 2𝑠𝑖𝑛 2𝑡 cos 4 𝑡
• Self Assessment Question (SAQ) 2: Find the Laplace transform
of 𝑡1
2
• Self Assessment Question (SAQ) 3: Find the Laplace transform
of 𝐹 𝑡 =
ቐ1, 0 ≤ 𝑡 < 1𝑡, 1 ≤ 𝑡 < 2
𝑡2 , 2 ≤ 𝑡 < ∞
• Self Assessment Question (SAQ) 4: Find the Laplace transform
of 1 +𝑠𝑖𝑛 2𝑡
• Self Assessment Question (SAQ) 5: Find the Laplace transform
of 𝑠𝑖𝑛ℎ3 𝑡
-
• Self Assessment Question (SAQ) 6: Find the Laplace transform
of 𝑡 𝑐𝑜𝑠𝑡
• Self Assessment Question (SAQ) 7: Find the Laplace transform
of 1
𝑡𝑠𝑖𝑛2𝑡
• Self Assessment Question (SAQ) 8: Find the Laplace transform
of
𝑓 𝑡 = ቊ𝑡 − 1, 1 < 𝑡 < 20 𝑒𝑙𝑠𝑒𝑤𝑒ℎ𝑟𝑒
• Self Assessment Question (SAQ) 9: Find the Laplace transform
of the periodic function
𝑓 𝑡 = 𝑒𝑡 𝑓𝑜𝑟 0 < 𝑡 < 2𝜋
-
• Self Assessment Question (SAQ) 10: Find the Inverse
Laplace
transform of 1
𝑠−5
• Self Assessment Question (SAQ) 11: Find the Inverse
Laplace
transform of 2𝑠−5
9𝑠2−25
• Self Assessment Question (SAQ) 12: Find the Inverse
Laplace
transform of 𝑠2
𝑠2+𝑎2
• Self Assessment Question (SAQ) 13: Find the Inverse
Laplace
transform of 1
𝑠(𝑠2+𝑎2)
-
• Self Assessment Question (SAQ) 14: Find the Inverse Laplace
transform of 𝑠
(𝑠+7)4
• Self Assessment Question (SAQ) 15: Find the Inverse Laplace
transform of 𝑒−𝑠
(𝑠+2)3
• Self Assessment Question (SAQ) 16: Find the Inverse Laplace
transform by partial fraction method of
1
𝑠2−7𝑠+12
• Self Assessment Question (SAQ) 17: Solve the differential
equation using Laplace transform method
𝑑2𝑦
𝑑𝑥2+ 𝑦 = 0, 𝑤ℎ𝑒𝑟𝑒 𝑦 = 1 𝑎𝑛𝑑
𝑑𝑦
𝑑𝑥= −1 𝑎𝑡 𝑥 = 0
• Self Assessment Question (SAQ) 18: Solve the differential
equation using Laplace transform method
• 𝑦′′ + 4𝑦′ + 4𝑦 = 6𝑒−𝑡 , 𝑤ℎ𝑒𝑟𝑒 𝑦 0 = −2 𝑎𝑛𝑑 𝑦′(0) = 8
-
29. REFERENCES
• George Arfken, H. A. Weber,: Mathematical Methods For
Physicists
• H.K. Dass,: Mathematical Physics, S. Chand Publication
• Erwin Kreyszig,: Advanced Engineering Mathematics, Wiley Plus
Publication
• B.S. Rajput,: Mathematical Physics, Pragati Prakashan
• Satya Prakash,: Mathematical Physics, Pragati Prakashan
-
THANKS
