GOA COLLEGE OF ENGINEERING

“BHAUSAHEB BANDODKAR TECHNICAL EDUCATION COMPLEX"

FARMAGUDI, PONDA- GOA - INDIA

MATHEMATICS-III
UNIT 2: LAPLACE TRANSFORMS
LECTURE 2

DEVELOPED BY MATHEMATICS FACULTY

DEPARTMENT OF SCIENCE & HUMANITIES
GOA COLLEGE OF ENGINEERING
 LAPLACE TRANSFORMS LECTURE 2

TOPICS TO BE COVERED:

➢Definition of Inverse Laplace Transforms.

➢Properties of Inverse Laplace transforms.
➢Computation of inverse Laplace transforms using
properties.
➢Computation of inverse Laplace transforms using
properties and theory of partial fractions.
 Inverse Laplace transform
Definition: If the Laplace transform of f(t) is F(s) then we say that the
inverse Laplace transform of F(s) is f(t) and written as 𝐿−1 𝐹(𝑠) =
𝑓(𝑡).
If the Laplace transform of f(t) and g(t) is F(s) and G(s) and α,β
are constants then
Sr No Property of Laplace transform The equivalent property in terms of inverse
Laplace transform

1 L(αf(t)+βg(t))=αF(s)+βG(s) 𝐿−1 αF(s)+βG(s) = αf(t)+βg(t)

2 L(𝑒 α𝑡 𝑓(𝑡))=F(s-α) 𝐿−1 F(s−α) =𝑒 α𝑡 𝑓(𝑡)= 𝑒 α𝑡 𝐿−1 𝐹(𝑠)

 Inverse Laplace transform
Sr No Property of Laplace transform The equivalent property of inverse Laplace
transform
3 1 𝑠 1 𝑠 𝑠
L(f(at))= F 𝐿−1 F =f(at); 𝐿−1 F = af(at);
𝑎 𝑎 𝑎 𝑎 𝑎
4
L(H(t-a)f(t-a))=𝑒 −𝑎𝑠 𝐹(𝑠) 𝐿−1 𝑒 −𝑎𝑠 𝐹(𝑠) =H(t-a)f(t-a)
5 𝑡 1 −1 1 𝑡
L ‫׬‬0 𝑓 𝑢 𝑑𝑢 = F(s)
𝑠
𝐿
𝑠
F(s) =‫׬‬0 𝑓 𝑢 𝑑𝑢
6 𝑑𝐹 𝑑𝐹 𝑑𝐹
L t f(t) = -
𝑑𝑠
𝐿−1 − = t f(t) ; 𝐿−1 = − t f(t)
𝑑𝑠 𝑑𝑠
𝑛 𝑛
7 𝑑 𝐹 𝑑 𝐹 𝑑𝑛 𝐹
𝐿(𝑡 𝑛 f(t))=(−1)𝑛 𝑛 𝐿−1 𝑛
(−1) 𝑛 = 𝑛
𝑡 f(t); 𝐿−1 =(−1)𝑛 𝑡 𝑛 f(t)
𝑑𝑠 𝑑𝑠 𝑑𝑠 𝑛
 Inverse Laplace transform
Sr No Property of Laplace transform The equivalent property of inverse Laplace
transform
8 𝑡 𝑡
L ‫׬‬0 𝑓 𝑡 − 𝑢 𝑔(𝑢)𝑑𝑢 =F(s)G(s) 𝐿−1 F(s)G(s) =‫׬‬0 𝑓 𝑡 − 𝑢 𝑔(𝑢)𝑑𝑢
9 𝑓(𝑡) ∞ ∞ 𝑓(𝑡)
L =‫𝑢𝑑 𝑢 𝐹 𝑠׬‬ 𝐿−1 ‫= 𝑢𝑑 𝑢 𝐹 𝑠׬‬
𝑡 𝑡
10
 Inverse Laplace transform
1 −1 1
1. L (1)= 𝑠
therefore 𝐿 =1.
𝑠
𝑛! 𝑛! 1 1 𝑛
2. L(𝑡 𝑛 ) = therefore 𝐿−1 = 𝑡𝑛 ; 𝐿−1 = 𝑡
𝑠𝑛+1 𝑠𝑛+1 𝑠 𝑛+1 𝑛!
1 1
3. L(𝑒 𝑎𝑡 )
= therefore 𝐿−1 = 𝑒 𝑎𝑡
𝑠−𝑎 𝑠−𝑎
𝑠 −1 𝑠
4. L(Cosat)= 2 2 therefore 𝐿 2 2 = Cosat
𝑠 +𝑎 𝑠 +𝑎
𝑎 −1 𝑎 −1 1 1
5. L(Sinat)= 2 2 therefore 𝐿 2 2 = Sinat; 𝐿 = Sinat
𝑠 +𝑎 𝑠 +𝑎 𝑠2 +𝑎2 𝑎
𝑠 −1 𝑠
6. L(Coshat)= 2 2 therefore 𝐿 = Coshat
𝑠 −𝑎 𝑠 2 −𝑎2
𝑎 −1 𝑎
7. L(Sinat)= 2 2 therefore 𝐿 = Sinhat;
𝑠 −𝑎 𝑠 2 −𝑎2
 Computation of Inverse Laplace transform
• Find the inverse Laplace transform of the following.
4 3𝑠+8 2𝑠+3 3 2𝑠−5 2 3𝑠 4
a) + b) c) + d)𝑒 −𝑠 e) 𝑒 −𝑠 +𝑒 −2𝑠
𝑠3 𝑠2 +4 𝑠2 +4𝑠+5 (𝑠−2)3 𝑠2 −4𝑠+8 𝑠 2 +1 2
𝑠 +1 𝑠3
Solution:
−1 4 3𝑠+81 −1 2 −1 𝑠 −1 1
a)𝐿 + = 4. 𝐿 + 3𝐿 +8𝐿
𝑠3 𝑠2 +4
2 𝑠3 𝑠2 +4 𝑠2 +4
2 1 By linearity of inverse Laplace transforms
=2𝑡 + 3𝐶𝑜𝑠2t+8. .Sin2t
2
= 2𝑡 2 + 3𝐶𝑜𝑠2t + 4Sin2t
2𝑠+3 2𝑠+3 2𝑠+3 2𝑠+4−1
b) 𝐿−1 = 𝐿−1 = 𝐿−1 = 𝐿−1
𝑠 2 +4𝑠+5 𝑠2 +4𝑠+4+1 (𝑠+2)2 +1 (𝑠+2)2 +1
 Computation of Inverse Laplace transform
2𝑠+4−1 2(𝑠+2)−1 (𝑠+2) 1
= 𝐿−1 =𝐿 −1 =2 𝐿 −1 -𝐿 −1
(𝑠+2)2 +1 (𝑠+2)2 +1 (𝑠+2)2 +1 (𝑠+2)2 +1
−2𝑡 −1 𝑠 −2𝑡 −1 1
=2𝑒 𝐿 2 -𝑒 𝐿
𝑠 +1 𝑠 2 +1 𝐿−1 F(s−α) =𝑒 α𝑡 𝑓(𝑡)= 𝑒 α𝑡 𝐿−1 𝐹(𝑠)
= 2𝑒 −2𝑡 Cost- 𝑒 −2𝑡 𝑆𝑖𝑛𝑡
3 2𝑠−5 1 2𝑠−5
c) 𝐿−1 (𝑠−2)3
+
𝑠 2 −4𝑠+8
=3 𝐿−1
(𝑠−2)3
+ 𝐿−1
𝑠 2 −4𝑠+4+4
1 2𝑠−4−1
=3 𝐿−1 3 +𝐿 −1
(𝑠−2) (𝑠−2)2 +4
−1 1 −1 2(𝑠−2) −1 1
=3𝐿 3 +𝐿 -𝐿
(𝑠−2) (𝑠−2)2 +4 (𝑠−2)2 +4
2𝑡 −1 1 2𝑡 −1 𝑠 2𝑡 −1 1
= 3𝑒 𝐿 + 2𝑒 𝐿 -𝑒 𝐿
𝑠3 𝑠 2 +4 𝑠2 +4
 Computation of Inverse Laplace transform
𝑡 2 1 3 1
= 3𝑒 2𝑡 2 + 2𝑡
2𝑒 Cos2t-
2
𝑒 2𝑡 Sin2t =
2
𝑒 2𝑡 𝑡2 + 2𝑒 2𝑡 Cos2t - 𝑒 2𝑡 Sin2t
2
−1 −𝑠 2
d) 𝐿 𝑒 = H t − 1 f(t − 1) 𝐿−1 𝑒 −𝑎𝑠 𝐹(𝑠) =H(t-a)f(t-a)
𝑠2 +1

−1 2
= 2H t − 1 Sin(t − 1) Since f(t)= 𝐿 =2Sint
𝑠 2 +1
−1 −𝑠 3𝑠 −2𝑠 4
e) 𝐿 𝑒 +𝑒 = 3H(t-1)Cos(t-1)+2H(t-2)(𝑡 − 2)2
𝑠2 +1 𝑠3
−1 3𝑠 −1 4
Since 𝐿 = 3𝐶𝑜𝑠𝑡 𝑎𝑛𝑑 𝐿 = 2𝑡 2
𝑠 2 +1 𝑠3
 Theory of Partial fractions
𝑃1 (𝑥)
• Partial Fraction: A partial fraction is a ratio of two polynomials such that
𝑃2 (𝑥)
deg 𝑃1 (𝑥) < deg 𝑃2 𝑥 .
• A partial fraction can be written as a sum of partial fractions as
follows:
𝑝𝑥+𝑞 𝐴 𝐵
• = +
(𝑎𝑥+𝑏)(𝑐𝑥+𝑑) (𝑎𝑥+𝑏) (𝑐𝑥+𝑑)
𝑝𝑥 2 +𝑞𝑥+𝑟 𝐴 𝐵 𝐶
• 2 = + +
𝑎𝑥+𝑏 (𝑐𝑥+𝑑) (𝑎𝑥+𝑏) (𝑎𝑥+𝑏)2 (𝑐𝑥+𝑑)
𝑝𝑥 2 +𝑞𝑥+𝑟 𝐴𝑥+𝐵 𝐶
• = +
𝑎𝑥 2 +𝑏𝑥+𝑐 (𝑑𝑥+𝑒) 𝑎𝑥 2 +𝑏𝑥+𝑐 (𝑑𝑥+𝑒)
The constants A,B,C are computed so as to maintain equality.
 Computation of Inverse Laplace transforms
1)Find the inverse Laplace transforms of the following.
(𝑠 2 +𝑠+3) (𝑠 2 +3𝑠+3) (2𝑠 2 +𝑠−5)
a) 2 b) c)
(𝑠 +6𝑠+8)(𝑠−2) (𝑠 2 −6𝑠+9)(𝑠+1) (𝑠 2 −4𝑠+13)(𝑠−3)
−1 𝑠 −2𝑠 2𝑠+3
d)Tan 𝑠 e) 𝑙𝑜𝑔𝑒 f)𝑒
𝑠 2 +1 𝑠 2 +4𝑠+3
2)Use Convolution theorem to find the inverse Laplace transform
𝑠
of 2 2
(𝑠 +4)(𝑠 +1)
 Computation of inverse Laplace transform
(𝑠 2 +𝑠+3) (𝑠 2 +𝑠+3) 𝐴 𝐵 𝐶
• 1a) 2 = = + +
(𝑠 +6𝑠+8)(𝑠−2) (𝑠+4)(𝑠+2)(𝑠−2) (𝑠+4) (𝑠+2) (𝑠−2)
(𝑠 2 + 𝑠 + 3) = A(s + 2)(s -2)+ B(s + 4)(s -2)+ C(s + 4)(s +2)
Sub s = -4 we have 15=A(-2)(-6) therefore A=15/12 =5/4.
Sub s=-2 we have 5=B(2)(-4) therefore B= -5/8.
Sub s=2 we have 9=C(6)(4) therefore C=3/8
(𝑠2 +𝑠+3) 5 1 5 1 3 1
= − +
(𝑠2 +6𝑠+8)(𝑠−2) 4 (𝑠+4) 8 𝑠+2 8 (𝑠−2)
(𝑠 2 +𝑠+3) 5 1 5 1 3 1
𝐿−1 2 = 𝐿−1 - 𝐿−1 + 𝐿−1
(𝑠 +6𝑠+8)(𝑠−2) 4 𝑠+4 8 𝑠+2 8 𝑠−2
5 −4𝑡 5 3 2𝑡
= 𝑒 - 𝑒 −2𝑡 + 𝑒
4 8 8
 Computation of inverse Laplace transform
(𝑠 2 +3𝑠+3) (𝑠 2 +3𝑠+3) 𝐴 𝐵 𝐶
• 1b) 2 = = + +
(𝑠 −6𝑠+9)(𝑠+1) (𝑠−3)2 (𝑠+1) (𝑠−3) (𝑠−3) 2 (𝑠+1)
• 𝑠 2 + 3𝑠 + 3=A(𝑠 − 3) (𝑠 + 1)+B(𝑠 + 1)+ C(𝑠 − 3)2
• Sub s=3 ; 21=B (4) B=21/4
• Sub s= -1; 1= C x16 C=1/16
• Sub s=0; 3=-3A+B+9C=-3A+21/4+9x1/16=-3A+(21X4+9)/16=-3A+97/16 ; -3A= 3-97/16 ; A=49/48
(𝑠2 +3𝑠+3) 49 1 21 𝐵 1 𝐶
= + +
(𝑠2 −6𝑠+9)(𝑠+1) 48 (𝑠−3) 4 (𝑠−3)2 16 (𝑠+1)
(𝑠 2 +3𝑠+3) 49 −1 1 21 −1 1 1 −1 1
𝐿−1 = 𝐿 + 𝐿 + 𝐿
2
(𝑠 −6𝑠+9)(𝑠+1) 48 𝑠−3 4 (𝑠−3) 2 16 𝑠+1
49 3𝑡 21 3𝑡 −1 1 1 −𝑡 49 3𝑡 21 3𝑡 1
= 𝑒 + 𝑒 𝐿 2 + 𝑒 = 𝑒 + 𝑒 t+ 𝑒 −𝑡
48 4 𝑠 16 48 4 16
 Computation of inverse Laplace transform
(2𝑠2 +𝑠−5) 𝐴(𝑠−2)+𝐵 𝐶 𝐴(𝑠−2)+𝐵 𝐶 𝐴(𝑠−2)+𝐵 𝐶
• 1c) 2 = + = + = +
(𝑠 −4𝑠+13)(𝑠−3) (𝑠2 −4𝑠+13) (𝑠−3) (𝑠2 −4𝑠+4+9) (𝑠−3) (𝑠−2)2 +9 (𝑠−3)
(2𝑠 2 + 𝑠 − 5)= (𝐴(𝑠 − 2) + 𝐵) (𝑠 − 3)+C ( 𝑠 − 2)2 + 9 ;
Sub s=3; 16= 10C; C=16/10=8/5;
Sub s=2; 5= -B+9C; B=9C-5= 9x8/5-5=(72-25)/5=47/5; B=47/5
Sub s=0; -5 = -3( -2A+B)+13C =6A-3B+13C; A=-5+3B-13C=-5+3X47/5-13X8/5=12/5
2𝑠2 +𝑠−5 12/5(𝑠−2)+47/5 8/5 12 (𝑠−2) 47 1 8 1
= + = + +
(𝑠2 −4𝑠+13)(𝑠−3) (𝑠−2)2 +9 (𝑠−3) 5 (𝑠−2)2 +9 5 (𝑠−2)2 +9 5 (𝑠−3)

2𝑠 2 +𝑠−5 12 𝑠−2 47 −1 1 8 1
𝐿−1 = 𝐿−1 + 𝐿 + 𝐿−1
(𝑠 2 −4𝑠+13)(𝑠−3) 5 (𝑠−2)2 +9 5 (𝑠−2)2 +9 5 𝑠−3
12 2𝑡 𝑠 47 1 8 1 12 2𝑡 47 8
= 𝑒 𝐿−1 + 𝑒 2𝑡 𝐿−1 + 𝐿−1 = 𝑒 Cos3t+ 𝑒 2𝑡 Sin3t+ 𝑒 3𝑡
5 𝑠 2 +9 5 𝑠 2 +9 5 𝑠−3 5 15 5
 Computation of inverse Laplace transform
• 1d) Let 𝐿−1 Tan−1 𝑠 =f(t); then L f(t) = Tan−1 s
𝑑 1
L tf(t) = - Tan−1 s =- = - L 𝑆𝑖𝑛𝑡
𝑑𝑠 𝑠 2 +1
𝑆𝑖𝑛𝑡
Therefore t f(t) = -Sint i:e f(t)= -
𝑡
𝑠 𝑠
1e) Let 𝐿−1 𝑙𝑜𝑔𝑒 =f(t); then L f(t) = 𝑙𝑜𝑔𝑒 = 𝑙𝑜𝑔𝑒 𝑠 - 𝑙𝑜𝑔𝑒 𝑠2 + 1
𝑠 2 +1 𝑠 2 +1
1
= 𝑙𝑜𝑔𝑒 𝑠 - 𝑙𝑜𝑔𝑒 𝑠 2 + 1
2
𝑑 1 1 𝑠
= L tf(t) = - 𝑙𝑜𝑔𝑒 𝑠 − 𝑙𝑜𝑔𝑒 𝑠2 +1 = - = L 1 − L 𝐶𝑜𝑠𝑡
𝑑𝑠 2 𝑠 𝑠 2 +1
1−𝐶𝑜𝑠𝑡
L tf(t) = L(1-Cost) therefore tf(t)= 1-Cost; f(t)=
𝑡
 Computation of inverse Laplace transform
−2𝑠 2𝑠+3
• 1 f) 𝑒
𝑠 2 +4𝑠+3
2𝑠+3 2𝑠+3 𝐴 𝐵 3 1 1 𝐴
2 = = + = +
𝑠 +4𝑠+3 (𝑠+3)(𝑠+1) 𝑠+3 𝑠+1 2 𝑠+3 2 𝑠+1

2𝑠+3 3 −1 1 1 −1 1 3 1
𝐿−1 = 𝐿 + 𝐿 = −3𝑡
𝑒 + 𝑒 −𝑡
𝑠2 +4𝑠+3 2 𝑠+3 2 𝑠+1 2 2
𝐿−1 𝑒 −𝑎𝑠 𝐹(𝑠) =H(t-a)f(t-a)
2𝑠+3 3 1
𝐿−1 𝑒 −2𝑠 =H(t-2) 𝑒 −3(𝑡−2) + 𝑒 −(𝑡−2)
𝑠 2 +4𝑠+3 2 2
 Computation of inverse Laplace transform
𝑠 𝑠 1
• 2) 2 2 = 2
(𝑠 +4)(𝑠 +1) (𝑠 +4) (𝑠 2 +1)
𝑠 1 2SinACosB=Sin(A+B)+Sin(A-B)
−1 −1
𝐿 2 =Cos2t; 𝐿 2 =Sint 2SinASinB=Cos(A-B)-Cos(A+B)
𝑠 +4 𝑠 +1 2CosACosB=Cos(A-B)+Cos(A+B)
−1 𝑠 1 𝑡
𝐿 =‫׬‬0 𝐶𝑜𝑠2 𝑡 − 𝑥 𝑆𝑖𝑛𝑥𝑑𝑥
(𝑠 2 +4) (𝑠2 +1)
1 𝑡
= ‫׬‬0 𝑆𝑖𝑛 𝑥 + 2𝑡 − 2𝑥 + 𝑆𝑖𝑛(𝑥 − 2𝑡 + 2𝑥)𝑑𝑥
2
= ‫׬‬0 𝑆𝑖𝑛 2𝑡 − 𝑥 + 𝑆𝑖𝑛(3𝑥 − 2𝑡)𝑑𝑥 = 2 {𝐶𝑜𝑠 2𝑡 − 𝑥 - 3 Cos (3𝑥 − 2𝑡)}ฬ 𝑥 = 𝑡
1 𝑡 1 1
2 𝑥=0
1 1 1 5 2
= {𝐶𝑜𝑠 t- Cost- Cos2t- Cos2t}= 6 𝐶𝑜𝑠 t- Cos2t
2 3 3 3
 Solving differential equations using
Laplace transforms
𝑑 2 𝑦 𝑑𝑦
• Solve the initial value problem 2 +5 +6y=3𝑒 𝑡 ; y(0)=1,𝑦 / 0 = 2
𝑑𝑡 𝑑𝑡
Solution: Assuming solution exist and has a Laplace transform we
transform the equation to its equivalent Laplace transform equation.
𝑑2 𝑦 𝑑𝑦
L 2 +5 +6y =L(3𝑒 𝑡 )
𝑑𝑡 𝑑𝑡 By Linearity of Laplace transform
𝑑2 𝑦 𝑑𝑦
L + 5L + 6L y =3L(𝑒 𝑡 ) L(𝑓 // ) = 𝑠 2 𝐿 𝑓 − 𝑠𝑓 0 − 𝑓 / (0)
𝑑𝑡 2 𝑑𝑡
3
𝑠2𝐿 𝑦 − 𝑠𝑦 0 − 𝑦 / 0 +5(s𝐿 𝑦 − 𝑦 0 )+ 6L(y)=
𝑠−1
3
L(y)(𝑠 2 +5s+6)- 𝑠𝑦 0 − 𝑦/ 0 -5y(0)=
𝑠−1
 Solving differential equations using
Laplace transforms
2 3
• L(y)(𝑠 +5s+6)- 𝑠 − 2-5= sub y(0)=1,𝑦 / 0 = 2
𝑠−1
3 3+(𝑠−1)(𝑠+7) 𝑠 2 +6𝑠−4
L(y)(𝑠 2 +5s+6)= + s+7= =
𝑠−1 𝑠−1 𝑠−1
𝑠2 +6𝑠−4 𝑠2 +6𝑠−4 𝐴 𝐵 𝐶
L(y)= (𝑠2+5s+6)(𝑠−1)= (s+3)(s+2)(𝑠−1) = 𝑠+3
+
𝑠+2
+
𝑠−1
𝑠 2 + 6𝑠 − 4=A(s+2)(s-1)+B(s+3)(s-1)+C(s+3)(s+2)
Sub s=-3; -13=A(-1)(-4); A=-13/4
Sub s=-2; -12=B(1)(-3); B=12/3=4
Sub s=1; 3=C(4)(3); C=1/4
 Solving differential equations using
Laplace transforms
𝑠 2 −4𝑠+6 13 1 1 1 1
= - + 4 +
(s+3)(s+2)(𝑠−1) 4 𝑠+3 𝑠+2 4 𝑠−1
𝑠 2 −4𝑠+6 13 1 1 1 1
𝐿−1 =𝐿 −1
− + 4 +
(s+3)(s+2)(𝑠−1) 4 𝑠+3 𝑠+2 4 𝑠−1
13 −1 1 −1 1 1 −1 1
y(t)=- 𝐿 +4 𝐿 + 𝐿
4 𝑠+3 𝑠+2 4 𝑠−1
13 −3𝑡 −2𝑡 1 𝑡
y(t) = - 𝑒 +4 𝑒 + 𝑒
4 4
 Solving differential equations using
Laplace transforms
𝑑2 𝑦
• Solve the initial value problem 2 +4y=𝑒 𝑡 ; y(0)=0,y 𝜋
4
=0
𝑑𝑡
Assume solution exists and has a Laplace transform. Let 𝑦 / (0)=a
𝑑2 𝑦 𝑑2 𝑦 1
L +4y =L 𝑒𝑡 ;L + 4L y =
𝑑𝑡 2 𝑑𝑡 2 𝑠−1
2 / 1
𝑠 𝐿 𝑦 − 𝑠𝑦 0 − 𝑦 (0)+4 𝐿 𝑦 = sub y(0)=0, 𝑦 / (0)=a
𝑠−1
2 1 2 1 𝑎𝑠−𝑎+1 𝑎𝑠−𝑎+1
L(y)(𝑠 +4)-0-a= ; L(y)(𝑠 +4)= +a= ; L(y)= 2
𝑠−1 𝑠−1 𝑠−1 (𝑠 +4)(𝑠−1)
𝑎𝑠−𝑎+1 𝐴𝑠+𝐵 𝐶
= 2 +
(𝑠 +4)(𝑠−1) 𝑠 +4
2 𝑠−1
 Solving differential equations using
Laplace transforms
𝑎𝑠 − 𝑎 + 1 = (As + B)(s-1)+C(𝑠 2 +4)
Sub s=1; 1= 5C ; C=1/5
Sub s=0; -a+1= -B+4C=-B+4/5; B= 4/5+a-1
Sub s=2; a+1= 2A+B+8C= 2A+ 4/5+a-1+8/5=2A+7/5+a; A=-1/5
𝑎𝑠−𝑎+1 𝐴𝑠+𝐵 𝐶 1 𝑠 1 1 1
= 2 + = - + (4/5+a-1) +
(𝑠 +4)(𝑠−1) 𝑠 +4
2 𝑠−1 2
5 𝑠 +4 𝑠 2 +4 5 𝑠−1
−1 𝑎𝑠−𝑎+1 1 −1 𝑠 −1 1 1 −1 1
𝐿 =- 𝐿 + (4/5+a-1) 𝐿 + 𝐿
(𝑠 2 +4)(𝑠−1) 5 𝑠 2 +4 𝑠2 +4 5 𝑠−1
1 1 1 𝑡
y(t)= - Cos2t+ (4/5+a-1) 𝑆𝑖𝑛2𝑡+ 𝑒
5 2 5
 Solving differential equations using
Laplace transforms
𝜋
To determine the unknown use the initial condition y 4
=0
𝜋
1 1 1
y(𝜋4)= - Cos2.𝜋4 + (4/5+a-1) 𝑆𝑖𝑛2. 𝜋4+ 𝑒 4 =0
5 2 5
1 1 𝜋 2 𝜋 1
Therefore 0 + (4/5+a-1) + 𝑒 4 =0 ; a= - 𝑒4+
2 5 5 5
Substituting for a in the solution we have
1 1 𝜋 1 𝑡
y(t)= - Cos2t - 𝑒 𝑆𝑖𝑛2𝑡+ 𝑒
4
5 5 5
 Solving integral-differential equations
using Laplace transforms
𝑑𝑦 𝑡 2𝑢
• Solve − 3 ‫׬‬0 𝑒 𝑦 𝑡 − 𝑢 𝑑𝑢 =2𝑒 𝑡 ; y(0)=1.
𝑑𝑡
We assume that the equation has a solution and the solution has a Laplace
transform.
𝑡
𝑑𝑦 𝑡 2𝑢 Note that ‫׬‬0 𝑒 2𝑢 𝑦 𝑡 − 𝑢 𝑑𝑢 is the
L − 3 ‫׬‬0 𝑒 𝑦 𝑡 − 𝑢 𝑑𝑢 = 2𝐿 𝑒 𝑡 convolution of 𝑒 2𝑡 and y(t). Hence its
𝑑𝑡 Laplace transform is L(𝑒 2𝑡 )L(y).
𝑑𝑦 𝑡 2𝑢 2
L − 3𝐿 ‫׬‬0 𝑒 𝑦 𝑡 − 𝑢 𝑑𝑢 =
𝑑𝑡 𝑠−1
2t 2 3 2
sL(y)-y(0)-3L(e )L y = ; sL(y)-1- 𝐿 𝑦 =
𝑠−1 𝑠−2 𝑠−1
3 2 2+𝑠−1 𝑠+1
L(y) 𝑠 − = +1= =
𝑠−2 𝑠−1 𝑠−1 𝑠−1
 Solving integral-differential equations
using Laplace transforms
𝑠 𝑠−2 −3 𝑠+1 𝑠2 −2𝑠−3 𝑠+1
L(y) = ; L(y) =
𝑠−2 𝑠−1 𝑠−2 𝑠−1
(𝑠+1) (𝑠−2) (𝑠+1)(𝑠−2) (𝑠−2)
L(y)= = =
(𝑠−1) (𝑠2 −2𝑠−3) (𝑠−1)(𝑠−3)(𝑠+1) (𝑠−1)(𝑠−3)
To find the inverse Laplace transform let
(𝑠−2) 𝐴 𝐵
= + ; 𝑠 − 2 = A (s-3)+B(s-1)
(𝑠−1)(𝑠−3) (𝑠−1) (𝑠−3)
Sub s=1; -1=-2A; A=1/2; Sub s=3; 1=B2; B=1/2
(𝑠−2) 1 1 1 1
= + ;
(𝑠−1)(𝑠−3) 2 (𝑠−1) 2 (𝑠−3)
−1 1 1 1 1 1
therefore y(t)=𝐿 + = 𝑒 𝑡 + 𝑒 3𝑡
2 (𝑠−1) 2 (𝑠−3) 2
 Solving simultaneous differential
equation using Laplace transforms
𝑑𝑥 2𝑡 𝑑𝑦
• Solve + y=𝑒 ; + x=𝑒 𝑡 x(0)=0, y(0)=1 assume the solution
𝑑𝑡 𝑑𝑡
exists and have Laplace transform.
𝑑𝑥 2𝑡 1 1
L +y =L 𝑒 ; sL(x)-x(0)+L(y)= ; sL(x)+L(y)= -----(1)
𝑑𝑡 𝑠−2 𝑠−2
𝑑𝑦 𝑡 1 𝑠
L +x =L 𝑒 ; sL(y)-y(0)+L(x)= ; sL(y)+L(x)= -----(2)
𝑑𝑡 𝑠−1 𝑠−1
Solving equation (1) and (2) we have
𝑠 3 −2𝑠 2 −𝑠+1 𝑠
L(y) = and L(x) =
(𝑠−1)2 (𝑠+1)(𝑠−2) (𝑠−1)2 (𝑠+1)(𝑠−2)
 Solving simultaneous differential
equation using Laplace transforms
𝑠 3 −2𝑠 2 −𝑠+1 𝐴 𝐵 𝐶 𝐷
• L(y) = = + + +
(𝑠−1)2 (𝑠+1)(𝑠−2) 𝑠−1 2
(𝑠−1) 𝑠+1 𝑠−2
3 1 1 −1
Solving we have A= , B= , C= , D=
4 2 12 3
𝐴 𝐵 𝑐 𝐷 3 1 1 1 1 1 1 1
L(y)= + 2 + + = + 2 - −
𝑠−1 (𝑠−1) 𝑠+1 𝑠−2 4 𝑠−1 2 (𝑠−1) 12 𝑠+1 3 𝑠−2
3 𝑡 1 𝑡 1 −𝑡 1 2𝑡
y(t)= 𝑒 + t𝑒 + 𝑒 - 𝑒
4 2 12 3
𝑠 𝐴 𝐵 𝐶 𝐷
L(x)= 2 = + 2 + +
(𝑠−1) (𝑠+1)(𝑠−2) 𝑠−1 (𝑠−1) 𝑠+1 𝑠−2
 Solving simultaneous differential
equation using Laplace transforms
3 1 1 2
• Solving we have A=- , B=- , C= , D=
4 2 12 3
3 1 1 1 1 1 2 1
• L(x)= - − 2 + +
4 𝑠−1 2 (𝑠−1) 12 𝑠+1 3 𝑠−2
3 𝑡 1 1 −𝑡 2 2𝑡
x(t)= - 𝑒 − t𝑒 𝑡 + 𝑒 + 𝑒 and
4 2 12 3
3 𝑡 1 1 −𝑡 1 2𝑡
y(t)= 𝑒 + t𝑒 𝑡 + 𝑒 - 𝑒 is the solution of the
4 2 12 3
system of equations.
 Special function and their Laplace
transforms
In certain type of physical and engineering problems one has to deal
with an instantaneous force or voltage acting instantaneously at a certain
time or a concentrated load acting at a point. These are modelled by
functions called special/ generalized functions. We define such special
functions.
Consider the Heaviside’s unit step function H(t). The displaced function
H(t-a) represents H(t) displaced to the right by distance a. Then
L(f(t-a)H(t-a))=𝑒 −𝑎𝑠 𝐿 𝑓 𝑡 = 𝑒 −𝑎𝑠 F(s)
L(f(t)H(t-a))=𝑒 −𝑎𝑠 𝐿 𝑓 𝑡 + 𝑎
 Dirac-Delta function
• Dirac-delta Function( Impulse Function):
0 𝑡<𝑎
1
Consider the function ∅(t)=ቐ 𝜖 𝑎 ≤𝑡 ≤𝑎+𝜖
0 𝑡 >𝑎+𝜖
1
1
𝜖 ∅(t)= H(t−a) − H(t−(a+𝜖))
𝜖

a a+𝜖
 Dirac-Delta function and its Laplace
transforms
∞ 𝑎+𝜖 1
‫׬‬0 ∅ 𝑡 𝑑𝑡 = ‫𝑎׬‬ 𝜖
dt = 1 for all values of 𝜖
Taking limit as 𝜖 → 0 the function ∅(t) tends to infinity at t=a and
zero every where, however its integral is unity.
𝛿 𝑡 − 𝑎 = lim ∅(𝑡)
𝜖→0
represents the force acting for a very short duration at time t=a
∞ 𝑎+𝜖
‫׬‬0 𝛿 𝑡 − 𝑎 𝑑𝑡 = lim ‫∅ 𝑎׬‬ 𝑡 𝑑𝑡 = 1
𝜖→0
 Properties of Dirac Delta function
∞
1. ‫׬‬0 𝛿 𝑡 − 𝑎 𝑑𝑡 =1
−𝑠𝑎
2. L(𝛿 𝑡 − 𝑎 ) = 𝑒 for a=0 L(𝛿 𝑡 ) = 1. 𝐿−1 𝑒 −𝑠𝑎 = 𝛿 𝑡 − 𝑎
𝐿−1 1 = 𝛿 𝑡
3. 𝛿 𝑡 − 𝑎 =𝐻/ (t-a)
∞
4. ‫׬‬0 𝑓(𝑡) 𝛿 𝑡 − 𝑎 dt = f(a)
−𝑠𝑎
5. L(f(t) 𝛿 𝑡 − 𝑎 )= 𝑒 f(a) 𝐿−1 𝑒 −𝑠𝑎 𝐹(𝑎) = 𝑓(𝑡)𝛿 𝑡 − 𝑎

6. 𝛿 𝑡 − 𝑎 *𝛿 𝑡 − 𝑏 = 𝛿 𝑡 − (𝑎 + 𝑏)
 Properties of Dirac Delta function
∞ 𝑎+𝜖 𝑎+𝜖 1
• 1 ‫׬‬0 𝛿 𝑡 − 𝑎 𝑑𝑡 = lim ‫∅ 𝑎׬‬ 𝑡 𝑑𝑡 = lim ‫𝑎׬‬ 𝑑𝑡 =1
𝜖→0 𝜖→0 𝜖
∞ −𝑠𝑡 𝑎+𝜖 −𝑠𝑡 1
• 2.L(𝛿 𝑡 − 𝑎 )=‫׬‬0 𝑒 𝛿 𝑡−𝑎 dt= lim ‫𝑒 𝑎׬‬ dt
𝜖→0 𝜖
1 𝑒 −𝑠𝑡 𝑡 =𝑎+𝜖 1
= lim ฬ = lim (𝑒 −𝑠(𝑎+𝜖) − 𝑒 −𝑠𝑎 )
𝜖→0 𝜖 −𝑠 𝑡=𝑎 𝜖→0 −𝑠𝜖
𝑒 −𝑠𝑎 1− 𝑒 −𝑠𝜖 −𝑠𝑎 1−𝑒 −𝑎𝑥
= lim =𝑒 lim
𝑥→0 𝑥
=a
𝑠 𝜖→0 𝜖
For a=0; L(𝛿 𝑡 )= 1
 Properties of Dirac Delta function
H(t−a) − H(t−(a+ε))
• 3 𝛿 𝑡 − 𝑎 = lim ∅(𝑡)= lim = 𝐻/ (t-a)
𝜖→0 𝜖→0 𝜖
∞ 𝑎+𝜖
• 4 ‫׬‬0 𝑓(𝑡)𝛿 𝑡 − 𝑎 𝑑𝑡 = lim ‫∅)𝑡(𝑓 𝑎׬‬ 𝑡 𝑑𝑡
𝜖→0
1 𝑎+𝜖
= lim ‫𝑎׬‬ 𝑓 𝑡 𝑑𝑡 = f(a) by Mean value theorem.
𝜖→0𝜖
∞ −𝑠𝑡 𝑎+𝜖 −𝑠𝑡
• 5 L(f(t) 𝛿 𝑡 − 𝑎 )=‫׬‬0 𝑒 f(t)𝛿 𝑡 − 𝑎 𝑑𝑡 = lim ‫∅)𝑡(𝑓 𝑒 𝑎׬‬ 𝑡 𝑑𝑡
𝜖→0
1 𝑎+𝜖 −𝑠𝑡
= lim ‫׬‬ 𝑒 𝑓(𝑡) 𝑑𝑡 = 𝑒 −𝑠𝑎 𝑓 𝑎 by Mean Value theorem
𝜖→0 𝜖 𝑎
 Properties of Dirac Delta function
• 6 L(𝛿 𝑡 − 𝑎 *𝛿 𝑡 − 𝑏 )=L(𝛿 𝑡 − 𝑎 )L (𝛿 𝑡 − 𝑏 )
= 𝑒 −𝑠𝑎 𝑒 −𝑠𝑎 = 𝑒 −𝑠(𝑎+𝑏) = L(𝛿 𝑡 − (𝑎 + 𝑏 ))
Therefore 𝛿 𝑡 − 𝑎 * 𝛿 𝑡 − 𝑏 = 𝛿 𝑡 − (𝑎 + 𝑏)
 Computation of Laplace transform
Special Functions
• Find the Laplace transforms of 𝑡 2 H(t-1)+Cos2t 𝛿 𝑡 − 𝜋
L(𝑡 2 H(t-1)+Cos2t 𝛿 𝑡 − 𝜋 )=𝑒 −𝑠 𝐿 (𝑡 + 1) 2 +L Cos2t 𝛿 𝑡 − 𝜋
= 𝑒 −𝑠 𝐿 𝑡 2 +2t+1 + 𝑒 −𝜋𝑠 Cos2𝜋
2 1 1
= 𝑒 −𝑠 + 2 2 + + 𝑒 −𝜋𝑠 L(f(t) 𝛿 𝑡 − 𝑎 )= 𝑒 −𝑠𝑎 f(a)
𝑠3 𝑠 𝑠
−2𝑠 𝑠
• Find the inverse Laplace transform of 𝑒 +2 𝑒 −𝑠 + 3𝑒 −3 𝑒 −3𝑠
𝑠2 +4
−1 −2𝑠 𝑠 −𝑠 −3 −3𝑠 𝐿−1 𝑒 −𝑠𝑎 = 𝛿 𝑡 − 𝑎
𝐿 𝑒 +2𝑒 + 3𝑒 𝑒 𝐿−1 1 = 𝛿 𝑡
𝑠2 +4
=H(t-2)Cos2(t-2)+2 𝐿−1 𝑒 −𝑠 +3 𝐿−1 𝑒 −3 𝑒 −3𝑠
𝐿−1 𝑒 −𝑠𝑎 𝑓(𝑎) = 𝑓(𝑡)𝛿 𝑡 − 𝑎
= H(t-2)Cos2(t-2)+2𝛿 𝑡 − 1 +3 𝑒 −𝑡 𝛿 𝑡 − 3
 Solving differential equations involving
Special functions
𝑑2 𝑦 𝑑𝑦
• Solve the equation + 3 - 4y = 2𝑒 𝑡 𝛿 𝑡 − 1 ; y(0)=1,𝑦 / 0 = 1
𝑑𝑡 2 𝑑𝑡
We assume the equation has a solution whose Laplace transform exists.
𝑑2 𝑦 𝑑𝑦
L 2 +3 −4y =L 2𝑒 𝑡 𝛿 𝑡 − 1
𝑑𝑡 𝑑𝑡
𝑠 𝐿 𝑦 − 𝑠𝑦 0 − 𝑦 / (0)+3(s 𝐿 𝑦 − 𝑦 0 )-4L(y)=2 𝑒 1 𝑒 −𝑠
2

• 𝑠 2 𝐿 𝑦 − 𝑠 −1+3(s 𝐿 𝑦 − 1)-4L(y)=2 𝑒 −(𝑠−1)

• L(y)(𝑠 2 +3s-4)-s-1-4= 2 𝑒 −(𝑠−1) ; L(y)(𝑠 2 +3s-4)= 2 𝑒 −(𝑠−1) +s+5
2 𝑒 −(𝑠−1)+s+5 2 𝑒 −(𝑠−1) s+5
• L(y)= 2 = 2 + 2
𝑠 +3s−4 𝑠 +3s−4 𝑠 +3s−4
 Solving differential equations involving
Special functions
𝑒 −(𝑠−1) s+5
y(t)=2𝐿−1 + 𝐿−1
𝑠2 +3s−4 𝑠 2 +3s−4
1 1 𝐴 𝐵 1 1 1 1
= = + =- +
𝑠 +3s−4
2 (s+4) 𝑠−1 s+4 𝑠−1 5 s+4 5 𝑠−1
−1 1 1 −4𝑡 1 𝑡
𝐿 =- 𝑒 + 𝑒
𝑠 2 +3s−4 5 5
s+5 𝐴 𝐵 1 1 6 1
= + =- +
𝑠 +3s−4 s+4 𝑠−1
2 5 s+4 5 𝑠−1
−1 1 1 −4𝑡 6 𝑡
𝐿 =- 𝑒 + 𝑒
𝑠 2 +3s−4 5 5
2 −4(𝑡−1) 2 (𝑡−1) 1 −4𝑡 6 𝑡
y(t)=H(t-1) − 𝑒 + 𝑒 - 𝑒 + 𝑒
5 5 5 5
 Problem on Laplace transforms
1.Find the inverse Laplace transform of the following
2𝑠+3 3𝑠+5 3𝑠 2 +2𝑠+5 𝑠 2 +4𝑠+2
a) 2 𝑏) 2 𝑐) 2 d)
𝑠 +1 𝑠 +6𝑠+10 (𝑠+2)(𝑠 +8𝑠+16) (𝑠−1)(𝑠 2 +6𝑠+13)
2𝑠+1 𝑠−3 −1 𝑠 −2𝑠 1
e) 2 2 f)𝑙𝑜𝑔𝑒 2 g)𝑇𝑎𝑛 h) 𝑒
(𝑠 +1)(𝑠 +4) 𝑠 +4 2 𝑠 2 +4
2) Solve the ordinary differential equation
𝑑2 𝑦 𝑑𝑦 −𝑡 ;
a) − 5 + 6𝑦 = 3𝑒 y(0)=0; 𝑦 / 0 = 1
𝑑𝑡 2 𝑑𝑡
𝑑2 𝑦 𝑑𝑦
b) + 2 − 3𝑦 = 𝑆𝑖𝑛𝑡; y(0)=0; 𝑦 / 0 = 2
𝑑𝑡 2 𝑑𝑡
THANK YOU!
 Problem on Laplace transforms
Solve the integral-differential equations
𝑑𝑦 𝑡
a) + 3𝑦 + 2 ‫=𝑡𝑑𝑦 𝑜׬‬3t ; y(0)=1
𝑑𝑡
𝑑𝑦 𝑡
b) = 1+‫׬‬0 𝑦 𝑡 − 𝑢 𝑆𝑖𝑛𝑢 𝑑𝑢; y(0)=0
𝑑𝑡
Solve the Simultaneous differential equations
𝑑𝑥 𝑡 𝑑𝑦
+ 2𝑦 = 5𝑒 ; + 2𝑥 = 𝑒 2𝑡 ; x(0)=0; y(0)=1
𝑑𝑡 𝑑𝑡
 Gate Questions on Laplace transforms
1
1) The Laplace transform of a f(t) is L(f)= then f(t) is
𝑠 2 +𝜔2
1 1 1 1
A)f(t)= sin(𝜔t) B) f(t) = (1-sin(𝜔t)) C) f(t)= cos(𝜔t) D) f(t) = (1-cos(𝜔t))
𝜔 𝜔2 𝜔 𝜔2
(𝑠+3)
2) If the Laplace transform of the function f(t) is given by then f(0) is
(𝑠+1)(𝑠+2)
A)1 B) 3/2 C) ½ D) 0
3) The Laplace transform of sinh(at) is
𝑎 𝑎 𝑠 𝑠
A) 2 2 𝐵) C) D)
𝑠 −𝑎 𝑠 2 +𝑎2 𝑠 2 −𝑎2 𝑠 2 +𝑎2
4) The Laplace transform of t𝑒 𝑡 𝑖𝑠
𝑠 1 𝑠 𝑠
A) B) C) D)
(𝑠+1)2 (𝑠+1)2 (𝑠−1)2 𝑠−1
 Gate Questions on Laplace transforms
5) If f(t)is a function defined for all t≥0,its Laplace transform F(s) is defined as
∞ ∞ ∞ ∞
(A)‫׬‬0 𝑒 𝑠𝑡 𝑓 𝑡 𝑑𝑡 (B)‫׬‬0 𝑒 −𝑠𝑡 𝑓 𝑡 𝑑𝑡 (C)‫׬‬0 𝑒 𝑖𝑠𝑡 𝑓 𝑡 𝑑𝑡 (D)‫׬‬0 𝑒 −𝑖𝑠𝑡 𝑓 𝑡 𝑑𝑡
6)The Laplace transform of cos(at) is
𝑎 𝑎 𝑠 𝑠
A)𝑠2 −𝑎2 𝐵) C) 𝑠2 −𝑎2 D) 𝑠2 +𝑎2
𝑠 2 +𝑎2
1
7) The Laplace Transform of a function f(t) is then f(t) is
𝑠 2 (𝑠+1)
A)t-1+𝑒 −𝑡 B) t-1+𝑒 −𝑡 C) -1+𝑒 −𝑡 D) 2t+𝑒 −𝑡
1
8) The inverse Laplace transform of is
𝑠 2 +𝑠
A)1+ 𝑒 𝑡 B) 1-𝑒 𝑡 C) 1 + 𝑒 −𝑡 D) 1 − 𝑒 −𝑡
9) F(s ) is the Laplace transform of the function f(t)=2𝑡 2 𝑒 −𝑡 , F(1) is ________ (correct to two decimal
places)
 Gate Questions on Laplace transforms
𝑠+3
10) The inverse Laplace transform of 𝐻(𝑠) = for t>0 is A)3t𝑒 −𝑡 + 𝑒 −𝑡 b)3𝑒 −𝑡 c)2𝑡𝑒 −𝑡 + 𝑒 −𝑡 d) 4t𝑒 −𝑡 + 𝑒 −𝑡
𝑠 2 +2𝑠+1
11) The Dirac-delta function 𝜕 𝑡 − 𝑡0 for t, 𝑡0 ∈ R , has the following property
𝑏 𝑓 𝑡 𝑎 < 𝑡0 < 𝑏
‫ 𝑡 𝜕 𝑡 𝑓 𝑎׬‬− 𝑡0 𝑑𝑡 = ቊ 0
0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
the Laplace transform of the Dirac Delta function L(𝜕 𝑡 − 𝑡0 )=F(s) is (A) 0 (B) ∞ (C) 𝑒 𝑎𝑠 (D) 𝑒 −𝑎𝑠
1−𝑐𝑜𝑠𝑥
12) The value of lim 𝑖𝑠 (A) 1/4 (B) 1/2 (C) 1/3 (D) 1
𝑥→0 𝑥 2