Production Engineering - Laplace Transformation

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UNIT – IV

Laplace Transformations

Class 1

Section I Introduction The knowledge of Laplace Transformations has in recent years became an essential part of Mathematical background required of engineers and scientists.

This is because the transform methods provide an easy and effective means for the solution of many problems arising in engineering.

This subject originated from the operational methods applied by the English engineer Oliver Heaviside (1850 – 1925) to problems in electrical engineering.

Unfortunately, Heaviside’s treatment was unsystematic and lacked rigour, which was placed on sound mathematical footing by Bromwich and carson during 1916 – 1917.

It was found that Heaviside’s operational calculus is best introduced by means of a particular type of definite integrals called “Laplace Transforms”.

The method of Laplace Transforms has the advantage of directly giving the solution of differential equations with given boundary values without the necessary of first finding the general solution and then evaluating from it the arbitrary constants.

Moreover, the ready tables of Laplace Transforms reduce the problems of solving differential equations to mere algebraic manipulation.

Definition: Integral transform Let K(s, t) be a function of two variables s and t where s is a parameter [s  R or C] independent of t. Then the function f(s) defined by an Integral which 

is convergent. i.e., f ( s ) =

 K (s, t ) F (t ) dt

is called the Integral Transform of

−

the function F(t) and is denoted by [T{F(t)], K(s, t) is kernel of the transformation.

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1  0 ,  t  0 or  0 If kernel K(s, t) is defined as K ( s, t ) =  then t − st e  t  0  

f ( s) =  e − st F (t ) dt

is

called

“Laplace

0

Transform” of the function F(t) and is also denoted by L{ F(t) } or f (s ) . 

 L { F(t) } = f ( s) =  e − st F (t ) dt = f (s ) 0

Definition: Laplace Transformation P

f(P) f is real valued / complex valued function Domain

Range

Let F(t) be a real or complex valued function defined on [0, ). Then the 

function f(s) defined by f ( s) =  e − st F (t ) dt is called Laplace Transformation 0

of F(t) if the integral exists and f(s) = L{ F(t) }.

Note: If L { F(t) } = f (s )  F(t) = L-1 { f (s ) }. Then F(t) is called the inverse Laplace Transform of f (s ) which transforms F(t) into f (s ) is called “The Laplace Transforms Operator”.

Linearity Property of Laplace Transformation: A transformation T is said to be linear if  a1, a2 constants and F1(t), F2(t) be any functions of F. Then L{ a1 F1(t) + a2 F2(t) } = a1 L{ F1(t) + a2 L{ F2(t) } or F1(t), F2(t), F3(t) there exists a1, a2, a3 constants such that L {a1 F1(t) + a2 F2(t) – a3 F3(t)} = a1 L{ F1(t) + a2 L{ F2(t) } – a3 L{ F3(t)}.

Theorem: Laplace transformation is a linear transformation i.e.,  a1, a2 constants L{ a1 F1(t) + a2 F2(t) } = a1 L{ F1(t) + a2 L{ F2(t) }.

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

Proof: By definition, L{ F(t) } = f ( s) =  e − st F (t ) dt 0

 L.H.S.

= L{ a1 F1(t) + a2 F2(t) } 

=

e

− st

[ a 1 F1 (t) + a 2 F2 (t) ] dt

0





0

0

= a 1  e − st . F1 (t) dt + a 2  e − st F2 (t) dt = a1 L{ F1(t) + a2 L{ F2(t) } = R.H.S.  L{ a1 F1(t) + a2 F2(t) } = a1 L{ F1(t) + a2 L{ F2(t) } Hence Laplace Transformation is a linear Transformation.

This completes the proof of the theorem.

Definition: Piece-wise or sectionally continuous A function F(t) is said to be piece-wise or sectionally continuous on a closed interval a  t  b, if it is defined on that interval can be divided into finite number of sub-intervals in each of which F(t) is continuous and has finite left limit and right hand limits. i.e., Lim F(t) = Lim F(t) = Lim F(t) = finite F say  a  t  b. t → -0 t → +0 t→ 0 therefore, F is continuous.

Geometrically: F(t)

0

a

t1 I1

t2 I2

t3 I3

b I4

t

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Figure. piece-wise or sectionally continuous.

Engineering Mathematics - I

UNIT – IV

Semester – 1

Laplace Transformations

By Dr N V Nagendram

Class 2

Definition: Functions of an Exponential order A function F(t) is said to be an exponential order  as t tends to  if there exist a positive real number M a number  and a finite number t0 such that | F(t) | < M e t or | e-t F(t) | < M,  t  t0.

Note: If a function F(t) is of an exponential order . It is also of  such that  > .

Definition: A function of class A A function which is piece-wise continuous or sectionally continuous on every finite interval in the range t  0 and is of an exponential order , as t →  is known as “A function of class A”.

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Theorem: Existence of Laplace Transformation If F(t) is a function which is piecewise or sectionally continuous on every interval (finite) in the range t  0 and satisfies | F(t) |  M. eat,  t  0 where a and M are constants. Then the Laplace Transform exists for every s > a.

Proof: -



0

t

t0

t1



|---------------0-------→|------I1--------→|---------- I2---------------→| 

By definition, L{ F(t) } = f ( s) =  e − st F (t ) dt 0

t0

=

e

− st

F (t ) dt +  e − st F (t ) dt ....................... (1)

0 t0

e

− st



t0

F (t ) dt exists and piece-wise continuous in interval 0  t  t0 so that

0 



− st  e F (t ) dt

e



t0

− st

F (t ) dt

t0



e



− st

.M .e at dt

(since, |F(t)  M eat)

t0 

=

M. e

−( s − a ) t

dt

(since, s > a implies s – a > 0  0)

t0 

=

M

 ( s − a) . e

−( s − a ) t 0

dt

t0 

e

t0

− st

F (t ) dt



M . e −( s − a ) t 0 for any s > a ( s − a)

If t0 is too large then R.H.S. is too small or infinite small therefore L{ F(t) } exists for s > a.

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This completes the proof of the theorem.

Engineering Mathematics - I

UNIT–IV

Semester – 1

By Dr N V Nagendram

Table of General Properties of Laplace Transform 

e

F(s) =

− st

Class 3

f (t ) dt

0

S No. Name

Laplace Transform

Inverse Laplace Transform

01. Definition 02. Linearity

L{f(t)} = f(s) af1(t) +b f2(t)

03. Change of scale

f(at)

04. First shifting Th.

eat f(t)

05. second shifting

 f (t − a) , t  a u(t-a) = {  0 , t  a

e-as f(s)

f (t)

sf(s)-f(0)

f (t)

s2f(s) – s f(0) - f (0)

06. Derivative (multiply by s) 07. Second derivative (multiply by s2) 08. n th derivative (multiply by sn)

f

L-1{f(s)} = f(t) a f1(s)+bf2(s) 1 s f( ) a a f(s-a)

snf(s) – sn-1 f(0) - sn-2 f (0).....– f n-1(0)

(n)(t) t

f (s) s

 f (u) du

09.Integral(division by s)

0

10. Multiple integral (division by 11. Multiply by t 12. Multiply by t2 13. Multiply by tn 14. Division by t

t

t

0

0

n

 ...... f (u ) du



f (s) sn

sn) f (s) f (s) f n(s)

- t f(t) t2 f(t) (-1)n tn f(t)



f (t ) t

 f (u ) du 1

t

15. Convolution

(t − u )n − 1 f (u ) du (n − 1)!

0 t

f(t) * g(t) =

 f (u) g (t − u)du 0

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t

=

 f (t − u) g (u)du

f(s)*g(s)=L(f * g)

0

= L-1(f(s) g(s)} p

16. f-periodic with period p

f(t) = f(t + p)

1 e − su f (u ) du − sp  1− e 0

Table of some Important Laplace Transforms By Dr N V Nagendram ________________________________________________________________ S No. 01.

Laplace Transform L{f(t)} = f(s)

02.

1

03.

t

04.

t2

05.

tn n = 0,1,2,3,...

06.

eat

07.

e-at

08.

t (n -1) e at (n − 1)!

09.

t (k -1) e at ( k )

10.

sin at

11.

cos at

12.

ebt sin at

13.

ebt cos at

14.

sinh at

15.

cosh at

16.

ebt sinh at

Inverse Laplace Transform L-1{f(s)} = f(t) 1 s 1 s2 2! s3 n! s n +1 1 s−a 1 s+a 1 n = 1,2,3,..... ( s − a) n 1 k = 1,2,3,..... ( s − a) k a 2 s + a2 s 2 s + a2 a ( s − b) 2 + a 2 s−b ( s − b) 2 + a 2 a 2 s − a2 s 2 s − a2 a ( s − b) 2 − a 2

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s−b ( s − b) 2 − a 2

17.

ebt cosh at

18.

u(t - a)

19.

f(t – a). u(t – a)

e − as s -as e f(s)

20.

e bt − e at b−a

1 ,a  b ( s − a)( s − b)

Table of some Important Laplace Transforms By Dr N V Nagendram ________________________________________________________________ S No. 21.

22.

23.

Laplace Transform be bt − ae at b−a sin at − at cos at 2a 3 t sin at 2a

Inverse Laplace Transform s , a b ( s − a)( s − ab) 1 2 (s + a 2 ) 2 s 2 (s + a 2 ) 2

24.

sin at + at cos at 2a

s2 (s 2 + a 2 ) 2

25.

1 cos at − at sin at 2

s3 (s 2 + a 2 ) 2

26.

27.

28.

t cos at at cos h at − sinh at 2a 3 t sinh at 2a

s2 − a2 (s 2 + a 2 ) 2 1 2 (s − a 2 ) 2 s 2 (s − a 2 ) 2

(sinh at + at cosh at)/ 2a

s2 (s 2 − a 2 ) 2

(cosh at + 1/2at sinh at)

s3 (s 2 − a 2 ) 2

31.

t cosh at

s2 + a2 (s 2 − a 2 ) 2

32.

t 2 sin at 2a

3s 2 − a 2 (s 2 − a 2 ) 3

33.

t 2 cos at 2

s 3 − 3a 2 s (s 2 + a 2 ) 3

29.

30.

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34.

t 3 cos at 6

s 4 − 6a 2 s 2 + a 4 (s 2 + a 2 ) 4

35.

t 3 sin at 24 a

s3 − a2s (s 2 + a 2 ) 4

e bt − e at t sin t t

36.

37.

Log tan

Log s s

 (1) – log t

38.

Engineering Mathematics - I

UNIT–IV

−1

s −b s −a 1 s

Semester – 1

By Dr N V Nagendram

Class 4

Laplace Transformations and its applications

Definition: Laplace Transform of Periodic function A function f(t) is said to be a periodic function of period T > 0 if F(t) = f(T + t) = f(2T + t) = f(3T + t ) = ..... = f(nT + t).

Sin t , cos t are periodic functions of period 2.

The Laplace transform of a piecewise periodic function f(t) with period p is s

L{ f(t) } =

1 e − st . f (t ) dt ; s > 0 − st  1− e 0 1 1 1 2 2 2 3 1  and (- + 1 ) =(- ) = -2.  2 2

Note: (1) = , ( =

3 3 3 1 ( ) = 2 2 2 2

1 )= 2

3 2

 ,( ) =( +1) = ( ) =

1 2

 ,

5 2

3 2

 ,( ) =( +1)

Ekeeda – Production Engineering

Engineering Mathematics - I

Semester – 1

By Dr N V Nagendram

UNIT–IV Problems Vs Solutions on Laplace transformations Class 5 Problem 01.

02.

03.

04.

05.

06.

07.

08.

09.

10.

11.

12.

Solution 3 f ( s ) = 2 − 5s L{ 3t - 5} s 12 6 8 f ( s) = 4 − 2 + L{ 2t3 – 6t + 8 } s s s 12 5s 12 − 5s f (s) = 2 − 2 = 2 L { 6 sin 2t – 5 cos 2t } s +4 s +4 s +4 3s 20 3s − 20 f ( s) = 2 − 2 = 2 L { 3 cosh 5t – 4 sinh 5t } s − 25 s − 25 s − 25 24 f ( s) = s 4 + 4s 2 + 5 L{ (t2 + 1)2 } s 2 2s f (s) = + 2 L[ cos2 t } s s + 16 72 12 4 10 3s − 2 + 2 L{ 3t4-2t3 + 4 e-3t – 2 sin 5t + 3 cos 2t } f ( s ) = 5 − 4 + s+3 s +5 s +4 s s 2 2a f ( s) = L{ sin2 at } 2 s ( s + 4a 2 ) s+9 f (s) = 2 L{ 2 e3t – e-3t } s −9 5 s f (s) = 2 + 2 L{ sin 5t + cos 3t } s + 25 s + 9 1 1 f ( s) = − L{ e-2t – e-3t } s+2 s+3 t e 0  t  1 1 (1 − e −( s −1) ) f (s) = L{ F(t) } if F(t) =  s − 1 t 1 0

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0 13. L{ F(t) } if F(t) =  4

0t 2 t2

e 0  t  5 14. L{ F(t) } if F(t) =  t 5 3 1 0  t  2 15. L{ F(t) } if F(t) =  t2 t t 0  t  4 16. L{ F(t) } if F(t) =  t4 5 1 17. L{ F(t) } if F(t) = t t

f ( s ) = 4.

20. L{ 1 + 2 t +3

t }

21. L{ cosh at – cos at } 22. L{ cos (at+b) }

)

1 3 1 − e −5( s − 1) + e −5 s s −1 s

f ( s) =

1 e −2 s e −2 s + + 2 s s s

f ( s) = f (s) = f ( s) =

19. L{ e2t + 4 t3 – 2 sin 3t + 3 cos 3t }

(

f ( s) =

3

 1  18. L{  t −  } t 

e −2 s s

1 ( s −1)e −4 s − s2 s2 1 s

  3

−

6

+

+

4  s 5 2 s 3 2 s 12 24 3( s − 2)   1 f ( s) =  + 4 + 2  s +9  s−2 s

8  −1  s 2

1   f ( s ) =  + 3 2 + 3  s s s  2a 2 s   f ( s) =  4 4  s −a   s cos b − a sin b  f (s) =   s2 + a2  

23. L{ (sint – cos t)2 }

f ( s) =

s 2 − 2s + 4 s( s 2 + 4)

24. L{ sin 2t cos 3t }

f ( s) =

2( s 2 − 5) ( s 2 + 1)( s 2 + 25)

25. L{ sin at sin bt }

12

  2abs  f ( s) =  2 2 2 2 2 2   [ s + (a + b) ] [ s + (a − b) ] 

Ekeeda – Production Engineering

Engineering Mathematics - I

Semester – 1

By Dr N V Nagendram

UNIT–IV Problems Vs Solutions on Laplace transformations Class 6 Section II . First shifting / Translation Lemma . Second shifting / Translation Lemma . Change of scale property . Multiplication by tn . Problems Vs Solutions

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Engineering Mathematics - I

UNIT – IV

Semester – 1

By Dr N V Nagendram

Laplace Transformations

Class 6

Section II Lemma: First shifting / Translation Lemma.

If L ( F(t) ) = f (s ) where s >  then L ( eat F(t) ) = f (s )

is a Laplace transformation of F(t) then

transformation of Proof:

f ( s − a) , s >  + a or If f ( s − a)

is a Laplace

eat F(t).

Y

O

f ( s − a)

f (s )

L ( eat F(t) )

L ( F(t) )

X Figure

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

Given f (s ) is a Laplace transformation. Then f (s ) =

e

−s t

F (t ) dt = L { F(t) }

0

So, f ( s − a) =







0

0

0

−( s − a ) t F (t ) dt = e

a t −s t −s t  e e F (t ) dt = eat  e F (t ) dt = eat f (s )



 f ( s − a) =

eat

since, f (s ) =

f (s )

e

−s t

F (t ) dt = L { F(t) }

0

 f ( s − a) is a Laplace transformation of

eat L{ F(t) }.

This completes the proof of the theorem.

Lemma: Second Translation / shifting Lemma. F (t − a) If L { F(t) } = f (s ) and G(t) =  0

ta ta

then, L { G(t) } = e-a s f (s ) .

Proof: −



a

|------------------------ t < a ------------→|0------|--- t > a -----→| F (t − a) −s t = L { F(t) } and given G(t) = e F ( t ) dt  0 0



Given

f (s ) =



So that, f (s ) =

a

−s t  e F (t ) dt =

−s t  e .0 dt +

0

0



e

−s t

ta ta

F (t − a) dt

a

Put t – a = x implies dt = dx , since da/dx = 0 implies a = c. 

=

−s t  e F (t − a) dt = a



−s ( x + a) F ( x) dx = e-a s e a



−s x  e F ( x) dx = e – a s a



e

−s t

F (t ) dt

a

This completes the proof of lemma.

Note: for every x  (a, ) such that x  (0, ) and for every t  (0, ), a = 0

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



e

−as

.e

−s x



F ( x) dx =

a

e

−as

.e − s x F ( x) dx

0

For any t, e-as  L { G(t) } = e-as L { F(t) } 

f (s )

e-as

= L { G(t) }

since,

e

−as

.e

−s x



F ( x) dx =

a

e

−s t

F (t ) dt

a

 L { G(t) } = e-as f (s )  L { G(t) } = e-as L { F(t) } This completes the proof of lemma.

Lemma: Change of scale property 1 a

If L { F(t) } = f (s ) then L { F(at) } = 

Proof: Given

f (s ) =

e

−s t

s f  . a

F (t ) dt = L { F(t) }

0 

Let us consider L { F(at) } =

e

−s t

F (at ) dt

0

Put at = x implies t = x/a so dt = (1/a) dx 

L { F(at) } =

e 0

−s t

1 F (at ) dt = a 1 = a 1 = a



e

x −s   a

F ( x) dx

0 

e

s −x   a

F ( x) dx

0



e 0

s −t   a

b

F (t ) dt

[since,

 a

b

f ( x) dx =

 f (t ) dt a

]

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

s f  a

1 = a

[since, f (s ) =

0

1 s f   or L { F(at) } = a a

1  L { F(at) } = a



e

s −t   a

F (t ) dt

0

This completes the proof of the lemma.

Lemma: Multiplication by tn If L { F(t) } = f (s ) then L {

tn

F(t) } =

(-1)n

dn { f ( s ) } for n = ds n

1,2,3,……………… 

We have f (s ) =

e

−s t

F (t ) dt

0

On differentiation w.r.t. s,

  d d  −s t { f (s ) }  e F (t ) dt  = ds ds  0 

By Leibnitz principle for differentiation under integral sign, 

d d −s t 0 ds (e ) F (t ) dt = ds { f (s ) }  





 −t e −s t

for n = m,

e 0

−s t

[t

m

F (t ) dt =

F (t ) dt = −

0 

−s t

0

te

e

d { f (s ) } ds

d { f (s ) } ds

dm { f (s ) } F (t ) ] dt = ( − 1) ds m m

This completes the proof of lemma on multiplication by tn.

Lemma: Division by t

−s t

F (t ) dt

]

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1 If L { F(t) } = f (s ) then L { F(t) } = t 

Proof: since f (s ) =

e

−s t



 f (s) ds

provided integral exists.

S

F (t ) dt

0

On integrating both sides w.r.t. s from s →  we get, 



 − s t    e F (t ) dt  ds 0 



f ( s) ds

=

 S

S



=

  F (t ) e

−s t

0 S

 − s t  ds dt =  F (t )   e ds  dt here, since t is independent 0 S  

of s 



 e −s t  =  F (t )   dt =  t S 0 Laplace transformation.

1  L { F(t) } = t



e

−s t

0

F (t ) 1 dt = L { F(t) } by definition of t t



 f (s) ds S

This completes the proof of lemma.

Engineering Mathematics - I

UNIT – IV

Semester – 1

Laplace Transformations

Section II Problems Vs Solutions Problem #1: Find L { t sin at }

Problem #2: Find L { t2 sin at }

Problem #3: Find L { t3 e-3t }

Problem #4: Find L { t e-t sin 3t }

By Dr N V Nagendram

Class 7

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 (1 − e t )  Problem #5: Find L  t     (Cos at − Cos bt )   Problem #6: Find L  t  

t e

Problem #7: Find

−2 t

Sin t dt and evaluate at s = 2

0





*Problem #8: Find

0

Sin mt dt t

 e t Sin t  dt  Problem #9: Find L  t 0 

Problem #1: Find L { t sin at } a Solution: Since L { sin at } = s 2 + a 2 2 s.a a d L { t sin at } = − ds { s 2 + a 2 } = − 2 s + a2

(

 L { t sin at } =

− 2 s.a

(s

2

+ a2

)

2

)

2

is required solution.

Problem #2: Find L { t2 sin at } a Solution: Since L { sin at } = s 2 + a 2

Ekeeda – Production Engineering

d2

a

2 L { t2 sin at } = ( − 1) ds 2 { s 2 + a 2 }

d  − 2s.a = ds  2  s + a 2

(

(

) (

(

)

 2.a s 2 + a 2 2 − 2as. 2 s 2 + a 2 . 2s  4 2  = −  s2 + a2  

)

)

  

=

 8.a s 4 + 8a 3 s 2 − 2as 4 − 4s 4 − 4s 2 a 3 − 2a 5  4 s2 + a2 

(

)

  8.a s 4 + 8a 3 s 2 − 2as 4 − 4s 4 − 4s 2 a 3 − 2a 5 = 4 s2 + a2  

 6as 4 + 4s 2 a 3 − 2a 5 4 =  s2 + a2 

  6as 4 + 4s 2 a 3 − 2a 5 = 4 s2 + a2  

 2as 2 + 2a 3 − 8as 2 3 = − s2 + a2 

 = 

(

(

)

)

(

)

 6as 2 − 2a 3  2 3  s + a 2

  

(

)

(

 2as 2 + 2a 3 − 8as 2  = − 3 s2 + a2  

(

 2a(3s 2 − a 2 )   is required solution.

2 L { t2 sin at } = =   s 2 + a 2 

(

)

Problem #3: Find L { t3 e-3t } 1 Solution: since, L { e-3t } = ( s + 3)

L { t3 e-at }

d3  1  = ( − 1) ds 3  s + 3  3

 −1     (s + 3)2    d  − 1. − 2    = - ds  (s + 3)3    d  2    = - ds  (s + 3)3     2. − 3    =  (s + 3)4    =-

d2 ds 2

)

)

  

  

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 6    =  (s + 3)4    L {

t3

e-at

 6    } =  (s + 3)4   

is required solution.

Problem #4: Find L { t e-t Sin 3t } 3 s + 32

Solution: since, L { Sin 3t } =

2

6s d  3  And L { t Sin 3t } = - ds  s 2 + 3 2  = 2 s + 32

(

So, L { t e-t Sin 3t } =

L { t e-t Sin 3t } =

6( s + 1)

(s + 1)

+ 32

2

6( s + 1)

(s + 1)

2

+ 32

)

2



2



2

is required solution.

 (1 − e t )  Problem #5: Find L  t    1 1 Solution: since L { 1- et } = s − s − 1 = f (s )

 (1 − e t )  Now L  t  =  





 S

f ( s) ds =

1

1 

  s − s − 1 ds

= Log s − Log (s − 1)S



S



  s  =  Log  s − 1   S

 1    Log =− 1− 1  s 

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=

 (1 − e t )   s −1 Log     L =  s   t 

 s −1 Log    s 

is required solution.

 (Cos at − Cos bt )   Problem #6: Find L  t  s s − 2 2 Solution : since L { Cos at – Cos bt } = s + a s + b 2 2



 (Cos at − Cos bt )   = L  t 

 S



s   s f ( s) ds =   2 − 2 ds 2 2  s + a s + b   S 

1 1  =  Log ( s 2 + a 2 ) − Log ( s 2 + a 2 )  2 2 S

(s 2 + a 2 ) 1 (s 2 + a 2 ) = Log = 0 − Log 2 (s 2 + b 2 ) (s 2 + b 2 )

(s 2 + a 2 )  (Cos at − Cos bt )  Log =  L  t (s 2 + b 2 )  



Problem #7: Find

t e

−2 t

is required solution.

Sin t dt and evaluate at s = 2

0



Solution : Let

t e

−2 t

Sin t dt

0

d 



1



 2s 

−2 t =  e (t Sin t ) dt = - 1. ds  s 2 + 1  =  s 2 + 1  at s = 2 0





t e 0

−2 t

 2s 

4

 Sin t dt =  2  s + 1  and s = 2 value is 25 

Problem #8: Find

 0

Sin mt dt t

 m  Solution: L { Sin mt } =  s 2 + m 2 



t e 0

−2 t

Sin t dt =

4 25

Is required solution.

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

Now

 0





 0





 0

Sin mt dt = t





  s S

2

m  ds = [ tan −1( s / m)]S 2  +m 

Sin mt s   dt =  − tan −1 ( ) as s → 0 m  t 2

 = 2

if m > 0

 =− 2

if m < 0

Sin mt s   dt =  − tan −1 ( ) m  t 2

is required solution.

 e t Sin t  dt  Problem #9: Find L  t 0  Sin t Solution: since L { t } =

 L { et





  s S

2

ds   = [ tan −1( s)]S = − Tan −1 s = Cot −1 s 2  2 +1 

Sin t t -1 -1 t } =L { e Cot s} = Cot (s – 1)

[ by shifting lemma ]

 e t Sin t  1 dt  = .Cot-1 (s − 1). Hence the solution.  L  t 0  s

Engineering Mathematics - I

UNIT – IV

Semester – 1

By Dr N V Nagendram

Laplace Transformations

Class 8

Section II Problems Vs Solutions S

F(t)

L { F(t) }

Solution

NO 01

Sin t Cos t

L{Sin t Cos t} = ½ L{ 2 sin t cos t} = ½ L { Sin 2t }

1 s + 22 2

, for s>|2|

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02

Cosh2 2t

L{Cosh2 2t }

s2 − 8 s(s 2 − 4 2 )

= ½. L{1+Cosh 4t}

03

04

Sinh at – Sin at

L{Sinh at – Sin at}

Sin (at + b)

2a 3 s4 − a4

L{ Sin (at+b) } =L{Sin at cos b +

Cos at

Sin b} 05

Cos( t +  )

L { Cos( t +  ) }

,s > 0

, for s>|a|

a Cos b + s Sin b , for s2 + a2 s>|a|

s Cos  −  Sin  , s2 + 2

for

s>|| 06

Cos3 3t = 1/4 cos3t + 3/4 cos t

L{ Cos3 3t } = L{ 1/4 cos3t + 3/4 cos t}

07

Cos3 t

L{ Cos3 t }

s( s 3 + 63) , s | 9 | s 2 + 32 s 2 + 9 2

(

)(

( s 3 + 7 s) , s | 3 | s 2 + 12 s 2 + 3 2

( 08

Sin3 2t

L { Sin3 2t }

(Sin 3t = 3 sin t – 4 sin3 t) 09

Sin 2t Sin 3t

L { Sin 2t Sin 3t } =½ L{ Cos t - Cos 5t }

10

Cos 5t Cos 2t

L { Cos 5t Cos 2t } =½ L{ Cos 7t + Cos 3t }

11 F(t) =

2t

0t  1

F(t) = {1

1 t  2 t 2

{2 13 F(t) =

Cos t {0

0  t  2

)

4s , s | 6 | s + 2 s 2 + 62

(

2

2

)(

)

12 s , s | 5 | s + 12 s 2 + 5 2

(

2

)(

)

s( s 3 + 29) , s | 7 | s 2 + 32 s 2 + 7 2

(

)(

)

2 9 (1 − e −5 s ) − e −5 s , s  0 2 5 s

L { F(t) }

1 −s (e + e − 2 s ) s

L { F(t) }

s (e − 2 s − 1) s +1

t  2

e-t[3 cos 5t – 4 sin 5t]

)(

L { F(t) }

t5

{1

0

12

14

0t  5

)

2

L{e-t[3 cos 5t – 4 sin 5t] }

3s − 7

(s + 1)2 + 5 2

,s0

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15

e2t(3 sinh 2t – 5 cosh 2t)

16 − 5s

L{ e2t(3 sinh 2t – 5 cosh 2t)}

(s − 2)2 − 2 2 16

17

e − a t .t n −1 (n − 1) !

e − a t .t n −1 (n − 1) !

L{

e-t cos2 t

L (e-t cos2 t } = L { e-t

,s 0

1 ,s−a ( s + a) n

}

(1 + cos 2t ) 2

}

2s 2 + 3s + 5 ,s 0 2 [(s + 1) + 2 2 ]( s + 1)

18

e-at Sin bt

L { e-at Sin bt }

b ,s  0 ( s + a) 2 − b 2

19

Cosh at – Cos at

L { Cosh at – Cos at }

2s 3 , s 0 [ s 4 + 4a 4 ]

20

( 1 + t e-t )3

L { ( 1 + t e-t )3 }

1 3 6 6 + + + , 2 3 s ( s + 1) ( s + 2) ( s + 3)4 s0

21

t e-4t Sin 3t

L [t e-4t Sin 3t }

3 ,s0 (s + 4) [( s + 4)2 + 32  2

22

(t-1)3[u(t – 1)

L { (t-1)3 [u(t – 1) }

e −1 s 6 4 s

23

eat [ u( t – a ) }

L [eat [ u( t – a ) }

e−a s ( s + 1)

24

e-2t {1- u(t – 1) }

L { e-2t } – L { e-2t (1- u(t – 1) )}

1 − e − ( s + 2) ( s + 2)

25 26

If L { F(t) } =

then

If L [ F(t) } = L[3e2t sin t – 4 e2t cos 4t }=

27 If L { 28

f (s )

1  s

}= Tan-1 

L{

F(t) =

3  Cos(t − 4 )  0 

Sin at t

L { F(t) }=

3 4 3 t 4 t 

Change of scale property.

(

)

2

f ( s − a)

4(15 − s ) ,s0 s − 12 s + 180

L[ F(3t) } =

20 − 4 s ,s0 s − 4 s + 20

(

Sin t t

L { F(t/a) } = a.

}=

)

2

a  s

Tan-1   − s 

 

e 3  ( s 2 + 1)

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29 F(t) =

Cos(t ) 0  t    t  sin t

L { F(t) }

e −S ( S −1) − S ( s 2 + 1)

30

F(t)= t2 + at + b

L { F(t) }

2 a b + + , s3 s 2 s

31

F(t) = t3 + 5 Cos t

L { F(t) }

6 5s + s4 s2 + 1 



32. Evaluate

33. Evaluate

te [Ans. L{ 

− 3t  t e Sin t dt 0

0





− 2t  t e Cos t dt

[Ans. L{ 



−t 4  t e Sin t dt

[Ans. L{

Cos at − cos bt dt 0 t e − t − e −3 t dt 0 t

[Ans. L{

37. Evaluate

e − t Sin 2 t dt 0 t

e

−t

Cost dt

[Ans. L{

Section III





e − t Sin 2 t Log (5) dt }= 0 t 4



[Ans. L{

[Ans. L{

e

−t

Cost dt }=

0

Engineering Mathematics - I

8( s + 1) s ( s + 2s + 17) 2

e − t − e −3 t dt }= Log (3)  0 t

t

0

UNIT – IV

Sin 4 t dt }=





t

−t



 (s 2 + b 2 )  Cos at − cos bt 1  2  Log dt 2  0 }= 2s t  (s + a ) 



36. Evaluate

t e

3 25



t

t

36. Evaluate

5 30

Cos t dt }=

0

0

35. Evaluate

− 2t

Sin t dt }=

0

0

34. Evaluate

t e

− 3t

Semester – 1



1( s + 1) s ( s 2 + 2 s + 2)



By Dr N V Nagendram

Inverse Laplace Transformations

Class 9

Ekeeda – Production Engineering

Having found Laplace Transformation of a new functions let us now determine the inverse Laplace Transformations of given functions of S. We have seen L { F(t) } is an algebraic function which is rational. Hence to find inverse laplace transforms, we have to express the given function of S into partial fractions which will, then to recognize as one of the following standard forms: Sl.

Inverse Laplace Function

Solution

NO 1

2

3

4

5

6

1 L-1 { s }

1

L-1

1 {s−a }

eat

L-1

1 { sn }

t n −1 , (n − 1) ! n=1,2,3..

L-1

1 { ( s − a) n }

t n −1 eat (n − 1) ! , n=1,2,3..

1 L-1 { s 2 + a 2 }

L-1

s { s + a2 }

1 Sin at a Cos at

2

7

1 L-1 { s 2 − a 2 }

1 Sin h at a

8

1 L-1 { ( s − a ) 2 + b 2 }

1 at e Cos bt b

9

s L-1 { s 2 − a 2 }

Cosh at

Ekeeda – Production Engineering

10

11

12

L-1

s−a { ( s − a) 2 + b 2 }

s

L-1 {

(s

2

L-1 {

(s

2

+a 1 +a

eat Cos bt

)

}

1 t Sin at 2a

)

}

1 ( Sin at − at Cos at ) 2a 3

2 2

2 2

Note: Reader is strongly advised to commit these results to memory.

Engineering Mathematics - I

Semester – 1

By Dr N V Nagendram

Ekeeda – Production Engineering

UNIT – IV

Inverse Laplace Transformations

Section III Problems Vs Solutions

 s 2 − 3s + 4   Problem #1: Evaluate L-1  s3    s+2  Problem #02: Evaluate L-1  s 2 − 3s + 4 

Problem #03: Evaluate L-1

 2s 2 − 6s + 5   3  2  s − 6s + 11s − 6 

  4s + 5 Problem #04: Evaluate L-1  ( s − 1) 2 ( s + 2)      5s + 3 Problem #05: Evaluate L-1  ( s − 1) ( s 2 + 2s + 5)   

  s Problem #06: Evaluate L-1  ( s 4 + 4a 4 )   

Problem #07: Evaluate L-1

 3( s 2 − 2) 2    5  2s 

4 s − 18   2s − 5 Problem #08: Evaluate L-1  4 s 2 + 25 + 9 − s 2 

  s Problem #09: Evaluate L-1  ( s + 3) 2 + 4)     3s + 2    s + 2s − 8 

Problem #10: Evaluate

L-1  2

Problem #11: Evaluate

L-1  2

 3s + 7    s − 2s − 3 

Class 10

Ekeeda – Production Engineering

 s 2 − 3s + 4   Problem #1: Evaluate L-1  s3  

 s 2 3s 4   s 2 − 3s + 4  s2   3s  4  = L-1  3 − 3 + 3  = L-1  3  + L-1 − 3  + L-1  3  3 Solution: L-1  s s s   s  s  s  s   =

1  3 4  − L-1  2  + L-1  3  s s  s 

L-1 

t2 = 1 – 3 t + 4 2!



 s 2 − 3s + 4  t2  = 1 – 3 t + 4 2! is required solution. s3  

L-1 

 s+2  Problem #02: Evaluate L-1  s 2 − 3s + 4 

Solution:

 s−2+4   s+2   = L-1  2 2   s − 3s + 4   ( s − 2) + 3 

L-1  2

  s−2 4 = L-1  ( s − 2) 2 + 32 + ( s − 2) 2 + 32        s−2 4 = L-1  ( s − 2) 2 + 3 2  + L-1  ( s − 2) 2 + 3 2      4 2t 2t = 3 e Cos 3t + e Cos 3t

 s + 2  4 2t 2t  L-1  2  = e Cos 3t + e Cos 3t 3 s − 3 s + 4  

is required solution.

Ekeeda – Production Engineering

Problem #03: Evaluate L-1

Solution: L-1

 2s 2 − 6s + 5   3  2  s − 6s + 11s − 6 

 2s 2 − 6s + 5   3  2  s − 6s + 11s − 6 

By using synthetic division method, we can get factors as S3 S= - 1

s

c

1

-6

11

-6

0

1

-5

6

1 S =2

s2

-5

0 1

6

2

0

-6

-3

0

3 2  s − 6 s + 11s − 6 =(S – 1) (S – 2) (S – 3) factor so by partial fractions

 2s 2 − 6s + 5  A B C + +  3 = 2 on solving We get A = ½, B = -1, C = 5/2  s − 6s + 11s − 6  s − 1 s − 2 s − 3

L-1

 2s 2 − 6s + 5   5   1   1   3  + L-1   = L-1   - L-1   2  ( s − 2)   2( s − 3)   2( s −1)   s − 6s + 11s − 6  = ½ et - e-2t + 5/2 e3t

 L-1

 2s 2 − 6s + 5   3 = 2  s − 6s + 11s − 6 

½ et - e-2t + 5/2 e3t is required solution.

Ekeeda – Production Engineering

  4s + 5 Problem #04: Evaluate L-1  ( s − 1) 2 ( s + 2)      4s + 5 Solution: To find L-1  ( s − 1) 2 ( s + 2)  by using partial fractions   4s+5 = A( s – 1 ) ( s + 2 ) + B ( s + 2 ) Put s = 1  9 = 3B  B = 3 and co efficient of s2, 0 = A – 1/3  A = 1/3    1 4s + 5 1 1   L-1  ( s − 1) 2 ( s + 2)  = L-1  3( s − 1) + 3 3( s − 1) 2 − 3( s + 2)     

  1   1  1  -1 = L-1  3( s − 1)  +L-1 3 ( s − 1) 2  − L  3( s + 2)        t t -2t = 1/3 e + 3 t e – 1/3 e   4s + 5  L-1  ( s − 1) 2 ( s + 2)  = 1/3 et + 3 t et – 1/3 e-2t is required solution.  

  5s + 3 Problem #05: Evaluate L-1  ( s − 1) ( s 2 + 2s + 5)      5s + 3 Solution: L-1  ( s − 1) ( s 2 + 2s + 5)    Bs + C 5s + 3 A = + 2 2 ( s − 1) ( s + 2 s + 5) ( s − 1) ( s + 2 s + 5) 2 5s + 3 = A (s + 2s + 5) +(s − 1)( Bs + C ) On solving we get, s = 1  8 = 8A  A = 1

s = 0  3 = 5A − C  C = 2

 1    5s + 3 s 2  L-1  ( s − 1) ( s 2 + 2s + 5)  = L-1  ( s − 1) − ( s 2 + 2s + 5) + ( s 2 + 2s + 5)           1  2 s -1 = L-1  ( s − 1)  − L-1  ( s 2 + 2s + 5)  + L  ( s 2 + 2s + 5)        = et - e- t Cos 2t + 3/2 e-t Sin 2t

  5s + 3  L-1  ( s − 1) ( s 2 + 2s + 5)  = = et - e- t Cos 2t + 3/2 e-t Sin 2t  

Ekeeda – Production Engineering

Is required solution.

  s Problem #06: Evaluate L-1  ( s 4 + 4a 4 )      s Solution: To find L-1  ( s 4 + 4a 4 )    For that by known formula, s4 + 4a4 = (s2+2a2)2-(2as)2=(s2+2a2+2as)(s2+2a22as) As + b s Cs + d + 2 = 2 By partial fractions 2 4 4 ( s + 4a ) ( s + 2a + 2as ) ( s + 2a 2 − 2as )  s = (As + b) (s 2 + 2a 2 − 2as) + (Cs + d) (s 2 + 2a 2 + 2as) Co efficient of s3  A – C = 0 Co efficient of s2  B + D = 0 Co efficient of s  − A + C = 0 and on solving, B =

−1 1 ; D–B= so D = 2a 4a

1 4a

    1   −1 1 1 s  L-1  ( s 4 + 4a 4 )  = 4 a L-1  ( s + a) 2 + a 2 )  + 4 a L-1  ( s − a) 2 + a 2 )       

  − 1 -a t 1 at s  L-1  ( s 4 + 4a 4 )  = 4 a e sin at + 4 a e sin at is required solution.  

Problem #07: Evaluate Solution:

L-1

L-1

 3( s 2 − 2) 2    5  2s 

 3( s 2 − 2) 2  3 1 1   = L-1   −6 L-1  3  + 3 L-1  5  5  2s  s  s   2s  3 t2 t4 = 2 −6 2! + 4!



L-1

 3( s 2 − 2) 2  3 t2 t4   5 = −6 2! + 4!  2s  2

Is required solution.

Ekeeda – Production Engineering

4 s − 18   2s − 5 Problem #08: Evaluate L-1  4 s 2 + 25 + 9 − s 2  4 s − 18   2s − 5 Solution: L-1  4 s 2 + 25 + 9 − s 2 

= L-1

  2   4 

 s 5 1 −  2 s 2 + 5  4  s 2 + 5 2  2  

( )

( )

2

     s 18  1.3   +  2     +L-1  − 4 2 2 3  s − 32   s −3     





5 5 5  1 =  2 Cos 2 t − 4 Sin 2 t  + (− 4 Cosh 3t + 6 Sinh 3t ) 1 5 5 5 = 2 Cos 2 t − 4 Sin 2 t − 4 Cosh 3t + 6 Sinh 3t

5 5 5 4 s − 18  1  2s − 5  L-1  4 s 2 + 25 + 9 − s 2  = 2 Cos 2 t − 4 Sin 2 t − 4 Cosh 3t + 6 Sinh 3t

Is required solution.

  s Problem #09: Evaluate L-1  ( s + 3) 2 + 4)       s +3−3  s Solution: L-1  ( s + 3) 2 + 4)  = L-1  ( s + 3) 2 + 4)      =



 s+3 2 2  − L-1  ( s + 3) + 2 ) 

L-1 

  3  2 2   ( s + 3) + 2 ) 

   s s −1  −3t [Since, L-1  ( s 2 + a 2 )   Cos 2t and L  ( s − 3) 2 + 2 2 )   e Cos 2t      = e-3t Cos 2t - 3/2 e-3t Sin 2t

  s  L-1  ( s + 3) 2 + 4)  = e-3t Cos 2t - 3/2 e-3t Sin 2t is required solution.  

Ekeeda – Production Engineering

3s   Problem #10: Evaluate L-1  s 2 + 2 s − 8 

 3s + 1 − 1  3s   Solution: L-1  s 2 + 2 s − 8  = L-1  ( s + 1) 2 − 32   

 3s + 1    1 = L-1  ( s + 1) 2 − 32  − L-1  ( s + 1) 2 − 32      =3

e-t

1 Cosh 3t − 3. 3 .Sinh 3t.e-t

 e 3t + e −3t = 3 e-t  2 

  − 

 e 3t − e −3t  2 

=

3 e 2t + 3e −4t  e 3t − e −3t −  2 2 

=

3 e 2t − e 2t + 3e −4t + e −4t 2

  e-t 

  e-t 

2t −4t = e +2 e

3s   2t −4t  L-1  s 2 + 2 s − 8  = e + 2 e is required solution.

Ekeeda – Production Engineering

 3s + 7  Problem #11: Evaluate L-1  s 2 − 2 s − 3 

 3( s − 1) + 3 + 7   3s + 7  Solution: L-1  s 2 − 2 s − 3  = L-1  ( s − 1) 2 − 2 2   

 3( s − 1)  10 = L-1  ( s − 1) 2 − 2 2 + ( s − 1) 2 − 2 2     3( s − 1)    1 -1 = L-1  ( s − 1) 2 − 2 2  + 10 L  ( s − 1) 2 − 2 2        2 -1 = 3 et 6 Sinh 2t + 5 L  ( s − 1) 2 − 2 2     e 2t + e −2t = 3 et  2   e 3t + e −t = 3  2   3s + 7   L-1  s 2 − 2 s − 3 

  + 5 Sinh 2t et 

 e 3t − e −t   + 5  2  

  

3t −t = 4 e − e is required solution.

Ekeeda – Production Engineering

Engineering Mathematics - I

UNIT – IV

Semester – 1

By Dr N V Nagendram

Inverse Laplace Transformations

Section III Problems Vs Solutions

 s2 + s − 2  Problem #12: Evaluate L-1  s( s + 3)( s − 2)     s 2 − 10s + 13  Problem #13: Evaluate L-1  ( s − 7)( s 2 − 5s + 6      s Problem #14: Evaluate L-1  (s 2 − 1) 2      2s + 1 Problem #15: Evaluate L-1  ( s − 1) 2 ( s + 2) 2      s Problem #16: Evaluate L-1  ( s − 2)( s 2 + 4)   

  s Problem #17: Evaluate L-1  ( s + 1) 2 ( s 2 + 1)   

Problem #18: Evaluate

L-1

 s3   4 4  s − a 

Problem #19: Evaluate

L-1

 1   3 3 s − a 

Problem #20: Evaluate

L-1 

Problem #21: Evaluate

L-1  2



 s2 + 6 2 2 2   ( s + 1)( s + 3 )   2s − 3    s + 4 s + 13 

  s+2 Problem #22: Evaluate L-1  ( s 2 + 4s + 5) 2   

Class 11

Ekeeda – Production Engineering

  s2 + 5 Problem #23: Evaluate L-1  ( s 2 + 1)( s 2 + 2s + 2    s   Problem #24: Evaluate L-1  s 4 + s 2 + 1

 a( s 2 − 2a 2  Problem #25: Evaluate L-1  s 4 + 4a 4   

Problem #26: Evaluate L-1

Problem #27: Evaluate

L-1

 s2   3  s − 2)   s2   3  (s − 2) 

  s+3 Problem #28: Evaluate L-1  s 2 − 4s + 13)      1 Problem #29: Evaluate L-1  s( s 2 + a 2 )      1 Problem #30: Evaluate L-1  s( s + a) 3   

  s Problem #31: Evaluate L-1  s 2 + a 2 ) 2      1 Problem #32: Evaluate L-1  ( s 2 − a 2 ) 2     s +1  Problem #33: Evaluate L-1 Log ( s − 1)   

Problem #34: Evaluate

L-1

 s2 +1  Log  s( s + 1)  

 −1  2 Problem #35: Evaluate L-1 Tan  s 2 

  

Ekeeda – Production Engineering

Problem #36: Evaluate

L-1

 −1  s   Cot    2  

Engineering Mathematics - I

UNIT – IV

Semester – 1

By Dr N V Nagendram

Inverse Laplace Transformations

Class 11

Section III Problems Vs Solutions

 s2 + s − 2  Problem #12: Evaluate L-1  s( s + 3)( s − 2)     s2 + s − 2  Solution: To find L-1  s( s + 3)( s − 2)     s2 + s − 2   B   C   A By partial fractions, L-1  s( s + 3)( s − 2)  = L-1  s  + L-1  (s + 3)  + L-1  (s − 2)        1 2 4 We get A = 3 ; B = 15 ; C = 5

 s2 + s − 2    2   4 1 L-1  s( s + 3)( s − 2)  = L-1  3s  + L-1 15( s + 3)  + L-1  5( s − 2)         1  2  1  1 1  4 = 3 L-1  s  + 15 L-1  ( s + 3)  + 5 L-1  ( s − 2)      1 2 4 = 3 + 15 e-3t + 5 e2t

 s2 + s − 2  1 4 2  L-1  s( s + 3)( s − 2)  = 3 + 15 e-3t + 5 e2t is required solution.  

Ekeeda – Production Engineering

Problem #13: Evaluate

Solution:



s 2 − 10s + 13   2  ( s − 7)( s − 5s + 6 

L-1 



 ( s − 7)( s − 3) − 8  s 2 − 10s + 13    = L-1  2  ( s − 7)( s − 3)( s − 2)   ( s − 7)( s − 5s + 6 

L-1 

( s − 7)( s − 3) − 8 A B C By partial fractions ( s − 7)( s − 3)( s − 2) = ( s − 7) + ( s − 3) + ( s − 2) 3 2 We get, A = − 5 ; B = 2 ; C = − 5

 ( s − 7)( s − 3) − 8   −2   1  3  1  L-1  ( s − 7)( s − 3)( s − 2)  = L-1  5( s − 7)  + 2 L-1  ( s − 3)  − 5 L-1  ( s − 2)         

 1   1  3  1   ( s − 7)( s − 3) − 8  2  + 2 L-1   − L-1    = − L-1  5  ( s − 7)   ( s − 3)  5  ( s − 2)   ( s − 7)( s − 3)( s − 2) 

L-1 

3 2 = − 5 e7t + 2 e3t − 5 e2t





3 2 s 2 − 10s + 13   == − e7t + 2 e3t − e2t is required solution.

2 5 5  ( s − 7)( s − 5s + 6 

L-1 

Ekeeda – Production Engineering

  s Problem #14: Evaluate L-1  (s 2 − 1) 2   

Solution:

=

L-1

L-1

 ( s + 1) 2 − ( s − 1) 2    s -1   2  2 2 = L 2   (s − 1)   4( s − 1) ( s + 1) 

    ( s + 1) 2 ( s − 1) 2 -1    2 2 − L 2 2  4( s − 1) ( s + 1)   4( s − 1) ( s + 1) 

 1    1 = L-1  4( s − 1) 2  − L-1  4( s + 1) 2       1  1  1  1 -1 -1   2 = 4 L − L  ( s + 1) 2   ( s − 1)  4   1 1 t -t = 4te - 4te

  1 1 s  L-1  2 t et - t e-t is required solution.

2 = 4  (s − 1)  4

Ekeeda – Production Engineering

  2s + 1 Problem #15: Evaluate L-1  ( s − 1) 2 ( s + 2) 2   

  ( s + 2) + ( s − 1)   2s + 1 Solution: L-1  ( s − 1) 2 ( s + 2) 2  = L-1  ( s − 1) 2 ( s + 2) 2     

    ( s + 2) ( s − 1) = L-1  ( s − 1) 2 ( s + 2) 2  + L-1  ( s − 1) 2 ( s + 2) 2           ( s + 2) − ( s − 1)  1 1 = L-1  ( s − 1) 2 ( s + 2)  + L-1  ( s − 1)( s + 2) 2  = L-1  3( s − 1) 2 ( s + 2)  + L-1        ( s + 2) − ( s − 1)   2   3( s − 1)( s + 2)        ( s + 2) ( s + 2) ( s − 1) = L-1  3( s − 1) 2 ( s + 2)  -L-1  3( s − 1) 2 ( s + 2)  + L-1  3( s − 1)( s + 2) 2  -L      

 ( s − 1) 2   3( s − 1)( s + 2) 

1

 1  1  1    1   1 1 1 1 -1 -1 -1 -1       2 = 3L - L ( s − 1)( s + 2) + 3 L - L  ( s + 2) 2     ( s − 1)( s + 2)  3  ( s − 1)  3    1   1  1    1  1 1 1 1 -1 -1 -1 -1       2 = 3L L ( s − 1)( s + 2) + 3 L - L  ( s + 2) 2    ( s − 1)( s + 2)  3     ( s − 1)  3 1 1 t -2t = 3 te - 3 te

  1 1 2s + 1 t -2t is required solution.  L-1  ( s − 1) 2 ( s + 2) 2  = = 3 t e - 3 t e  

Ekeeda – Production Engineering

  s Problem #16: Evaluate L-1  ( s − 2)( s 2 + 4)   

  s Solution: To find L-1  ( s − 3)( s 2 + 4)    A s Bs + C By partial fractions ( s − 3)( s 2 + 4) = ( s − 3) + ( s 2 + 2 2 ) 3 Put s = 3  A = 13 4 Put s = 0  C = 13

3 Put s = 1  B = − 13

 A   s Bs + C  L-1  ( s − 3)( s 2 + 4)  = L-1  ( s − 3) + ( s 2 + 2 2 )  = L-1    

 3 − 3s + 4  +  2 2  13( s − 3) 13( s + 2 ) 

      3 3s 4 = L-1 13( s − 3)  − L-1 13( s 2 + 2 2 )  + L-1 13( s 2 + 2 2 )       

=

L-1

  3   − L-1 13( s − 3) 

  3s  2 2  + L-1 13( s + 2 ) 

  4  2 2  13( s + 2 ) 

  4    1  3 3 s 1 = 13 L-1  ( s − 3)  − 13 L-1  ( s 2 + 2 2 )  + 13 L-1  ( s 2 + 2 2 )        3 3 2 = 13 e3t − 13 Cos 2t + 13 Sin 2t



L-1

  3 3 2 s 3t  = 2 e − 13 Cos 2t + 13 Sin 2t is required solution.  ( s − 2)( s + 4)  13

Ekeeda – Production Engineering

  s Problem #17: Evaluate L-1  ( s + 1) 2 ( s 2 + 1)   

Solution:

L-1

 ( s + 1) 2 − ( s 2 + 1)    s -1   2 2 = L  2( s + 1) 2 ( s 2 + 1)   ( s + 1) ( s + 1)   

=

L-1

  ( s + 1) 2 ( s 2 + 1) −   2 2 2 2  2( s + 1) ( s + 1) 2( s + 1) ( s + 1) 

=

L-1

    ( s + 1) 2 ( s 2 + 1) -1     2 2 2 2 − L  2( s + 1) ( s + 1)   2( s + 1) ( s + 1) 

    1 1 = L-1  2( s 2 + 1)  − L-1  2( s + 1) 2     

    1 1 = L-1  2( s 2 + 1)  − L-1  2( s + 1) 2      1 1 -t = 2 Sin t - 2 t e

  1 1 s -t  L-1  ( s + 1) 2 ( s 2 + 1)  = 2 Sin t - 2 t e is required solution.  

Ekeeda – Production Engineering

Problem #18: Evaluate

Solution:

L-1

L-1

 s3   4 4  s − a 

 s[( s 2 + a 2 ) + ( s 2 − a 2 )   s3  -1  4  4 = L  2( s 2 + a 2 )( s 2 − a 2 )    s − a 

=

L-1

  s[( s 2 + a 2 ) s[( s 2 − a 2 ) +  2 2 2 2 2 2 2 2   2( s + a )( s − a ) 2( s + a )( s − a ) 

=

L-1

    s[( s 2 + a 2 ) s[( s 2 − a 2 ) -1  2   2 2 2 2 2 2 2  + L  2( s + a )( s − a )   2( s + a )( s − a ) 

    s s = L-1  2( s 2 − a 2 )  + L-1  2( s 2 + a 2 )      1 1 = 2 Cosh at + 2 Cos at

 L-1

 s3  1 1  4 4 = Cosh at + 2 Cos at is required solution. s − a  2

Ekeeda – Production Engineering

 1  Problem #19: Evaluate L-1  s 3 − a 3 

    1 1  1  Solution: L-1  s 3 − a 3  = L-1  ( s − a) 3 + 3as( s − a)  = L-1  ( s − a)[( s − a) 2 + 3as]       A    1 Bs + C L-1  ( s − a)[( s − a) 2 + 3as]  = L-1  ( s − a) + [( s − a) 2 + 3as]  by partial fractions     1 Here 1= A[(s-a)2+3as] + (Bs + c) (s-a) ; Put s = a A = 3a 2 ; Put s = 0 C = −2 1 ; Put s = 1 B = 3a 3( a − 1) a 2

 A  Bs + C  1  L-1  s 3 − a 3  = L-1  ( s − a) + [( s − a) 2 + 3as]    =

L-1

  1 s 2 + −   2 [3(a − 1)a 2 ][( s − a) 2 + 3as] (3a)[( s − a) 2 + 3as]   ( s − a)3a

  1 = L-1  ( s − a)3a 2  + L-1  

    2 s    - L-1  2 2 2 [3(a − 1)a ][( s − a) + 3as]   (3a)[( s − a) + 3as] 

 s+a  s+a 2 1 at 1 a  s  = 3a 2 e + 3( a − 1) a 2 L-1  ( s + a) 2 − ( s + a) 2  − 3a L-1  ( s + a) 2 − ( s + a) 2     

 s+a   a  2 1 at 1 1 e -1 -1   2 2 2 2 = 3a + 3( a − 1) a L − L  ( s + a) 2  − 3a L-1  ( s + a)  3( a − 1) a    s+a  2  s  -1   2 + 2 L  ( s + a )  3a  ( s + a)  2 -at 2 1 at 1 1 -at − e + t e-at Cos at = 3a 2 e + 3( a − 1) a 2 e-at − 3(a − 1)a t e 3a 3a



1



1

1

1

2

2

at -at -at -at  L-1  s 3 − a 3  = 3a 2 e + 3( a − 1) a 2 e-at − 3(a − 1)a t e − 3a e + 3a t e Cos at  

Is required solution.

Ekeeda – Production Engineering

Problem #20: Evaluate

Solution:

=



 s2 + 6 2 2 2   ( s + 1)( s + 2 ) 

L-1 

 1 ( s 2 + 1) + ( s 2 + 4) 7   s2 + 6 1 + -1   2 2 2 2 2 2 2 2 2  = L  2 ( s + 1)( s + 2 ) 2 ( s + 1)( s + 2 )   ( s + 1)( s + 2 ) 



L-1 

1 1   1  ( s 2 + 4) ( s 2 + 1) 7 -1   2 2 2 2 2 2 2 2  + L-1  2  + L  2 ( s + 1)( s + 2 )   2 ( s + 1)( s + 2 )   2 ( s + 1)( s + 2 ) 

L-1 

1  1 1 1   7 1 = L-1  2 ( s 2 + 2 2 )  + L-1  2 ( s 2 + 1)  + L-1  2 ( s 2 + 1)( s 2 + 2 2 )         7 [( s 2 + 2 2 ) − ( s 2 + 1)]  1 1 1   1 -1 -1    2 2 + L 2 ( s 2 + 1) + L  2 ( s 2 + 1)( s 2 + 2 2 )   2 (s + 2 )     

=

L-1 

=

L-1 

 7 1 1 1   [( s 2 + 2 2 )]  1 -1 -1     2 2 + L 2 ( s 2 + 1) + L 2.3 ( s 2 + 1)( s 2 + 2 2 )  − L 2 (s + 2 )     

 7 [( s 2 + 12 )]  1 2 2 2   2.3 ( s + 1)( s + 2 )  1 7 1 1  7 1    1 1 = L-1  2 ( s 2 + 2 2 )  + L-1  2 ( s 2 + 1)  + L-1  6 ( s 2 + 1)  − L-1  6 ( s 2 + 2 2 )          1 1 7 7 = 4 Sin 2t + 2 Sin t + 6 Sin t - 12 Sin 2t 1 1 7 7 = 4 Sin 2t + 2 Sin t + 6 Sin t - 6 Sin 2t

1 5 = 3 Sin t - 3 Sin 2t 1 = 3 [5 Sin t - Sin 2t]

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  1 s2 + 6  L-1  2 [5 Sin t - Sin 2t] is required solution.

2 2  = 3  ( s + 1)( s + 2 ) 

 2s − 3  Problem #21: Evaluate L-1  s 2 + 4 s + 13 

Solution:

 2( s + 2) − 7   2s − 3   = L-1  2 2   s + 4 s + 13   ( s + 2) + 3 

L-1  2



 2( s + 2) 7 −  2 2 ( s + 2) 2 + 32   ( s + 2) + 3

=

L-1 

=

L-1 

  2( s + 2)  7 -1   2 2 2 − L 2   ( s + 2) + 3   ( s + 2) + 3 



 2( s + 2)    7 = L-1  ( s + 2) 2 + 3 2  − L-1  ( s + 2) 2 + 3 2      1 = 2 e-2t Cos 3t − 7 e-2t 3 sin 3t

1  2s − 3   L-1  s 2 + 4 s + 13  = 2 e-2t Cos 3t − 7 e-2t 3 sin 3t

is required solution.

  s+2 Problem #22: Evaluate L-1  ( s 2 + 4s + 5) 2        s+2 s+2 Solution: L-1  ( s 2 + 4s + 5) 2  = L-1  ( s + 2) 2 + 12  = e-2t Cos t    

  s+2  L-1  ( s 2 + 4s + 5) 2  = e-2t Cos t is required solution.  

Ekeeda – Production Engineering

Problem #23: Evaluate

Solution:



 s2 + s  2 2  ( s + 1)( s + 2s + 2) 

L-1 

 s 2 + 2s + 2) + ( s 2 + 1) − 3   s2 + s -1  2 2 = L  ( s 2 + 1)( s 2 + 2s + 2)   ( s + 1)( s + 2s + 2)   



L-1 



 ( s 2 + 2s + 2) ( s 2 + 1) 3 + − 2  2 2 2 2 2  ( s + 1)( s + 2s + 2) ( s + 1)( s + 2s + 2) ( s + 1)( s + 2s + 2) 

=

L-1 

=

L-1 

     ( s 2 + 2s + 2) ( s 2 + 1) 3 -1    - L-1  2  2 2 2 2 2 + L  ( s + 1)( s + 2s + 2)   ( s + 1)( s + 2s + 2)   ( s + 1)( s + 2s + 2) 



    1   1 1 = L-1  ( s 2 + 1)  + L-1 [( s + 1) 2 + 1]  - 3 L-1  ( s 2 + 1)( s 2 + 2s + 2)          1 = Sin t + e-t Sin t - 3 L-1  ( s 2 + 1)( s 2 + 2s + 2)   

    s2 + s 1  L-1  ( s 2 + 1)( s 2 + 2s + 2)  = Sin t + e-t Sin t - 3 L-1  ( s 2 + 1)( s 2 + 2s + 2)      Is required solution.

Ekeeda – Production Engineering

Problem #24: Evaluate

s    2  s + s + 1

L-1  4

  1 1   s+ −   s   2 2   -1 -1  2  Solution: L  s 4 + s 2 + 1 = L  2 1 2 3    ( s + 2 )  2         1 1   s+   2 2 − 2 2 = = L-1   2 1 2 3  1  3    ( s 2 + ) 2   ( s + 2 )  2   2 2      

    1 1     s+     2 2 -1 -1    2  2 =L −L  2 1 2 3    2 1 2 3    ( s + 2 )  2    ( s + 2 )  2          

=



e-1/2t

 3  1 2  3    t  − Cos  -1/2t e Sin  2 t  2 2 3 3    

2

 3  1 2  3  s 2       t Cos  -1/2t -1/2t 2 Sin  2 t   2 − 2 3 e 3  s + s + 1 = e    

L-1  4

is required solution.

Ekeeda – Production Engineering

Problem #25: Evaluate

Solution:

 a( s 2 − 2a 2  4 4   s + 4a 

L-1 

 as 2  a( s 2 − 2a 2  2a 2  − -1   4 4 = L s 4 + 4a 4 s 4 + 4a 4    s + 4a  

L-1 

=



as 2  4  −L-1  s + 4a 

L-1  4

 2a 2   4 4   s + 4a 

  s2 a -1 = L  s 4 + (2a 2 ) 2  − a L-1  

  2a  4 2 2  s + (2a ) 

=a Cos 2a2t− a Sin 2a2t

 a( s 2 − 2a 2   L-1  s 4 + 4a 4  =a Cos 2a2t− a Sin 2a2t is required solution.  

Problem #26: Evaluate L-1

Solution: L-1

 s2   3  (s − 2) 

 s2   ( s − 2) 2 + 4( s − 2) + 4    3 = L-1  ( s − 2) 3  (s − 2)   

= L-1

 ( s − 2) 2 4( s − 2) 4  + +   3 3 ( s − 2) ( s − 2) 3   ( s − 2)

= L-1

 ( s − 2) 2   4( s − 2)   4   3 3  + L-1  3  + L-1   ( s − 2)   ( s − 2)   ( s − 2) 

=

L-1

 4   4   1    + L-1  3 2  + L-1   ( s − 2)   ( s − 2)   ( s − 2) 

= e2t + 4 t e2t + 4 t2 e2t

Ekeeda – Production Engineering



L-1

 s2   3  (s − 2) 

= e2t + 4 t e2t + 4 t2 e2t is required solution.

Problem #27: Evaluate

Solution:

=

L-1

L-1

L-1

  s+3  2   s − 4s + 13) 

    s+3 s+3  2  = L-1  2 2   ( s − 2) + 3 )   s − 4s + 13) 

    s−2 5 s−2 + -1    2 2 2 2 2 2  +L-1 = L  ( s − 2) + 3 ) ( s − 2) + 3 )   ( s − 2) + 3 ) 

  5  2 2   3[( s − 2) + 3 )] 

= e2t Cos 3t + 5/3 e2t Sin 3t

  s+3  L-1  s 2 − 4s + 13)  = e2t Cos 3t + 5/3 e2t Sin 3t is required solution.  

  1 Problem #28: Evaluate L-1  s( s 2 + a 2 )      sin at 1 Solution: Since, L-1  ( s 2 + a 2 )  = a  

L-1



  1  2 2 =  s( s + a ) 

L-1

t

sin at 1 1 t dt = 2 (− Cos at )0 = 2 (1 − Cos at ) a a a 0



  1  2 2 =  s( s + a ) 

 1 − Cos at    is required solution. a2  

Ekeeda – Production Engineering

  1 Problem #29: Evaluate L-1  s( s + a) 3      1 Solution: L-1  s( s + a) 3  = L-1    1  Here we have, L-1  s( s − a)  =  

  1 1 L-1  s 2 ( s − a)  = a     1 1 L-1  s 3 ( s − a)  = a 2   

L-1

 L-1

t

 (e

at

0

    1 1  3  = e-at L-1  3 [( s + a) − a]( s + a)   ( s − a) s 

t

e at 1 at e dt 0 = t −a ;

1 − 1) dt = 2 (e at − at − 1) and a

 1  at a 2 t 2 at ( e − at − 1 ) dt 0 = a 3  e − 2 − at − 1   t

  e − at 1  3 = a3  s( s + a)    1 1  3 = 3  s( s + a)  a

 at a 2 t 2   1  a 2t 2  e − − at − 1 = 3 1 − e −at − e −at − at.e −at  2 2    a 

  a 2t 2 1 − e −at − e −at − at.e −at  is required solution.

2  

Problem #30: Evaluate

L-1

  s  2 2 2  s + a ) 

  s Solution: Let F (t) = L-1  s 2 + a 2 ) 2   

Ekeeda – Production Engineering



  1 1  1 1 s  F (t )  Hence, L  t  =  s 2 + a 2 ) 2 ds = − 2  s 2 + a 2 )  = 2  s 2 + a 2  since, after  s 0

1 1 −1  1   F (t )  applying limits. So,  t  = 2 L  s 2 + a 2  = 2a t sin at

  1 s  L-1  s 2 + a 2 ) 2  = 2a t sin at is required solution.  

  1 Problem #31: Evaluate L-1  ( s 2 − a 2 ) 2    Solution:

L-1

=

  1  2 2 2  (s − a )  L-1

  s  2 2 2   s( s − a ) 

t

=  F (t ) dt 0 t

=

t Sin at dt 2a 0



t t 1   − cos at    − cos at   t . −   1. dt    = 2a  a 0 0  a    

1  − cos at Sin at  + 2  = 2a t. a a 

1 = 2a 3 Sin at − Cos at 



L-1

  1 1 Sin at − Cos at  is required solution.  2 3 2 2 = 2 a ( s − a )  

Ekeeda – Production Engineering

 s +1  Problem #32: Evaluate L-1 Log ( s − 1)     s +1  Solution: Let F (t) = L-1 Log ( s − 1)   

So,

d  s + 1  t. F(t) = - L-1  ds  Log ( s − 1)      d  s + 1  = - L-1  ds  Log ( s − 1)      d  1  = - L-1  ds  Log (s + 1) − Log ( s − 1)     

d  1  d  = - L-1  ds Log (s + 1) + L-1  ds  Log ( s − 1)       1   1  = - L-1  s + 1  + L-1  s − 1 

= - e-t + et  t. F(t) = - e-t + et  F(t) =

L-1

 Sinh t s +1  is required solution. Log  =2 t ( s − 1)  

Ekeeda – Production Engineering

 s2 +1  Log  s( s + 1)  

Problem #33: Evaluate

L-1

Solution: t. F(t) = L-1

 d  s 2 + 1  − Log    s ( s + 1)    ds 





d  2 = − L-1  ds Log ( s + 1) − Log s − Log ( s + 1) 

=−

L-1

d d   2  Log ( s + 1) + L-1  Log s  + L-1  ds  ds  

 d   Log (s + 1)  ds 

1   1   2s  = − L-1  s 2 + 1  + L-1  s  + L-1  s + 1

t. F(t) F(t)  F(t) =

(

= − 2 Cos t + 1 + t e-t =

−2 1 Cos t + e-t + t t

1 t e −t + 1− 2 Cos t t

)

is required solution.

Ekeeda – Production Engineering

 −1  2 Problem #34: Evaluate L-1 Tan  s 2 

Solution:

t F(t) =

L-1

= −

=

L-1

  

 d   d   2    2   Tan −1  2   = − L-1  Tan −1  2   −   s    s    ds   ds 

L-1

  1   2  1 + 2   s4 

  − 22  .   s3     

=

L-1

 4s   4 2   s +2 

(

)

= L-1

 ( s 2 + 2 + 2s ) − ( s 2 + 2 − 2s )    4s  2  2 2 = L-1  s 2 + 2) 2 − (2s) 2  s + 2) − (2s)   

= L-1

 ( s 2 + 2 + 2s ) − ( s 2 + 2 − 2s )   2  2  s + 2 + 2s )( s + 2 − 2s) 

= L-1

  ( s 2 + 2 + 2s ) ( s 2 + 2 − 2s) − 2  2  2 2 s + 2 + 2s )( s + 2 − 2s)   s + 2 + 2s )( s + 2 − 2s)

(

)

(

(

)

)

(

) (

)

    ( s 2 + 2 − 2s) ( s 2 + 2 + 2s) -1    2  2 2 2 − L  s + 2 + 2s )( s + 2 − 2s)   s + 2 + 2s )( s + 2 − 2s) 

(

)

(

)

    1 1 1 1     = L-1  s 2 + 2 − 2 s  − L-1  s 2 + 2 + 2 s  = L-1  ( s − 1) 2 + 1 − L-1  ( s + 1) 2 + 1    

= et Sin t − e- t t F(t) = 2 Sinh t sin t

 e t − e −t Sin t = 2  2 

  Sin t = 2 sinh t sin t 

Ekeeda – Production Engineering

 2 Sinht Sin t   is required solution.  F(t) =  t 

 −1  s   Problem #35: Evaluate L-1 Cot  2     −1  s   Solution: Let t F(t) = L-1 Cot  2   

=

= t F(t)  F(t) =

L-1

−L-1

 d  −1  s   − 1  Cot     2    ds   2   2 2  s + 2 

= −Sin 2t − Sin 2t is required solution. t

Ekeeda – Production Engineering

Engineering Mathematics - I

UNIT – IV

Semester – 1

By Dr N V Nagendram

Inverse Laplace Transformations

Class 12

Section III Convolution theorem: t

If L-1{ f (s ) } = F(t) and L-1{ g (s ) } = G(t) then L-1{ f ( s ) g ( s ) } =

 f (u). g (t − u). du = F * G is 0

called the convolution or falting of F and G.

u u= t t→

t=u

O

u=0

t

Let, (t) =

 f (u). g (t − u). du 0



L{ (t) }

=

e

− st

0



=

e 0 u

t   f (u ). g (t − u ). du dt = 0 

− st

f (u ). g (t − u ). dt du

 t

e 0 0

− st

f (u ). g (t − u ). du dt

t

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

=

e

− st

0



=

e

− su

0

 − s ( t − u )  f (u ) e . g (t − u ). dt  du here put t-u =v then dv = dt 0 

 − sv  f (u ) e . gv. dv  du = 0 



e

− su

f (u ). g ( s) du = f ( s ). g ( s ) = F * G.

0

t

 L-1{ f ( s ) g ( s ) } =

 f (u). g (t − u). du = F * G 0

This completes the proof of the theorem.

Exercises: Try yourself…..

By Dr N V Nagendram

  s Problem #01 Evaluate L-1  ( s 2 + a 2 ) 2    2   s Problem #02 Evaluate L-1  ( s 2 + a 2 )( s 2 + b 2 )      1  2 -1 Problem #03 Evaluate L  s ( S + 2)  

1 [Ans. 2 a t Sin at

]

1 [Ans. 2 a t Sin at

]

1 − 2t [Ans. − 4 (e + 2t − 1)

]

  1  3 Problem #04 Evaluate L-1  s( S + 2)  

  s Problem #05 Evaluate L-1  (a 2 s 2 + b 2 )      1 Problem #06 Evaluate L-1  s 2 ( s 2 + a 2 )    Problem #07 Evaluate L-1 Problem #08 Evaluate L-1

  1  3 2   s ( s + 1)   s   2  ( s + a) 

 2as  Problem #09 Evaluate L-1  ( s 2 + a 2 ) 2    2  s  Problem #10 Evaluate L-1  ( s + a) 3    Problem #11 Evaluate

L-1

  1 + s   Log   s  

1 1 − 2t 2 1 [Ans. 8 − 4 e (t + t + 2 ) ]

1 b [Ans. a 2 Cos  a  t

]

1 1 [Ans. a 2 ( t- a Sin at)

]

t2  [Ans.  2 + Cos t − t   

]

-at [Ans. e (1 – at)

] ]

[Ans.

(

1 -at 2 2 [Ans. 2 e a t − 4at + 2

)

]

1 -at [Ans. t (1 - e )

]   s + a  Problem #12 Evaluate L-1  Log  s + b   

1 -b t - e-at) [Ans. t (e

]

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Problem #13 Evaluate

L-1

Problem #14 Evaluate L-1 Problem #15 Evaluate

L-1

Problem #16 Evaluate

L-1

Problem #17 Evaluate L-1 Problem #18 Evaluate

L-1

   s +1  Log   ( s + 2)( s + 3)   1  s 2 + b 2   2  Log  2  2 ( s + a )      a 2  1 − 2  Log  s      s 2 + 1    Log  2  ( s − 1 )     −1  2   Tan    s  

Cot (s + 1) −1

  s − 1  Problem #19 Evaluate L-1 s. Log  s + 1   

Problem #1 Evaluate

Solution:

[Since,

L-1

L-1

L-1

-t -2t -3t [Ans. e − e − e

]

1 [Ans. t (Cos at – Cos bt) ] 2 [Ans. t (1 – Cosh at)

]

2 t [Ans. t (e – Cos t)

]

Sin 2t t

]

[Ans.

e − t Sin t [Ans. t 2 [Ans. t 2 ( Sin t − t Coh t )

]

  s  2 2 2  (s + a ) 

    s s 1 .

-1   2  2 2 2 2 2 = L 2   (s + a ) (s + a )   (s + a ) 

    1 1 s . . = Sin at ] = Cos at -1  2   2 2 2  and L  (s + a )   (s + a )  a

By convolution theorem, L-1

  s 1 . 2  2 2 2 =  (s + a ) (s + a ) 

t

 Cos a(t − u) 0

t

1 1 [ Sin at − Sin(2au − at ) du .Sin au. du = 2 a 0 a

1 1 t = 2 a [[u Sin at − 2a Cos(2au − at ) ]0

L-1

  1 1 1 s [[u Sin at − Cos(2au − at ) ]t0 = t Sin at  2 2 2 = 2a 2a  (s + a )  2a

  1 s  L-1  ( s 2 + a 2 ) 2  = 2 a t Sin at is required solution.  

]

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Problem #2 Evaluate L-1

Solution: To find

Since,

L-1

L-1

  s2  2 2 2 2   ( s + a )( s + b ) 

  s2  2 2 2 2   ( s + a )( s + b ) 

    1 s . . = Cos bt = Cos at -1   2  2 2  2 and L ( ( s s + + b a ) )    

by convolution theorem, L-1

    s s2 s . 2  2 2 2 2 2 2  = L-1  2   (s + a ) (s + b )   ( s + a )( s + b ) 

t

=  Cos a(t − u ) Cos bu. du = 0

t

1 [Cos[a(t − u ) + bu ] + Cos [a(t − u ) − bu ]]du 2 0

1 [ Sin at + (b − a)t ] − Sin at   Sin [at − (b + a)t ] − Sin at  −   = 2  b−a b+a   

1 = 2(b − a 2 ) (2 b Sin bt − 2a Sin at ) 2

1 = (b 2 − a 2 ) ( b Sin bt − a Sin at )

  1 s2 ( b Sin bt − a Sin at ) is required solution.  L-1  2 2 2 2 2 2 =  ( s + a )( s + b )  (b − a )

Ekeeda – Production Engineering

Engineering Mathematics - I

UNIT – IV

Semester – 1

By Dr N V Nagendram

Application of Laplace Transform

Class 13

Section III Application of Laplace Transform to differential equations with constant coefficients: Laplace transform is especially suitable to obtain the solution of linear nonhomogeneous ordinary differential equations with constant coefficients, when all the boundary conditions are specified for the unknown function and its derivatives at a single point.

Consider the initial value problem

d2y dy + a + by = r (t ) …………………………..

2 dt dt

(1) y(t = 0) = k0, y1(t=0) = k1 …………………………………………………………………

(2) Where a,b,k0,k1 are all constants and r(t) is a function of t.

Method of Solution to differential equation (D.E) by Laplace Transform (L.T.):

1. Apply Laplace transform on both sides of the given differential equation (1), resulting in a subsidiary equation as

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[s2Y – s y(0) - y(0)] +a[sY-y(0)]+bY = R(s) …………………………………………. (3) where Y = L{y(t)} and R(s) = L{r(t)}.

Replace y(0), y(0) using given initial conditions (2).

2. Solve (3) algebraically for Y(s), usually to a sum of partial fractions.

3. Apply inverse Laplace transform to Y(s) obtained in 2. This yields the solution of O.D.E. (1) satisfying the initial conditions (2) as y(t) = L-1[Y(s)].

Problem #1 Solve the differential equation by using Laplace transformation y-2y-8y = 0, y(0) = 3, y(0) = 6.

Solution: On applying Laplace transform (s2Y - 3s - 6) – 2( sY – 3 ) - 8Y = 0 On solving, Y(s) =

3s 3s = s − 2 s − 8 ( s − 4)( s + 2) 2

By using partial fractions, Y(s) =

2 1 + ( s − 4) ( s + 2)

 1  -1  1  Applying 1 L.T. y(t) = L-1(Y(s)) = 2 L-1   = 2 e4t + e-2t.  +L  ( ( s s + − 2 4 ) )    

y(t) = 2 e4t + e-2t is required solution.

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Problem #2 Solve the differential equation by using Laplace transformation y+2y+5y = e-t Sin t, y(0) = 0, y(0) = 1.

Solution: Using L.T. On applying Laplace transform we get the given 1 differential equation as, (s2Y - 0 - 1) + 2( sY – 0 ) = L{ e-t Sin t } = ( s + 1) 2 + 12 s 2 + 2s + 3 On solving Y = 2 ( s + 2s + 5)( s 2 + 2s + 2) Cs + D As + B s 2 + 2s + 3 By partial fractions, 2 = 2 + 2 2 ( s + 2s + 5)( s + 2s + 2) ( s + 2 s + 5) ( s + 2 s + 2) s 2 + 2s + 3 = (As + B) ( s 2 + 2s + 2 )+ (Cs + D ) ( s 2 + 2s + 5 ) s 2 + 2s + 3 = s3(A + C) + s2 (2A + 2C + B + D ) + s (2A + 5C + 2B + 2D )+ 2B + 5D Equating coefficients of s on either side, A + C = 0 ; 2A + 2C + B + D = 1; 2A + 5C + 2B + 2D = 2; 2B + 5D = 3 1 2  A = 0, B = , C = 0, D = 3 3 1 2 As + B Cs + D 1 1 Y(s)= 2 + 2 = + 2 2 ( s + 2 s + 5) ( s + 2 s + 2) 3 ( s + 1) + 2 ) 3 ( s + 1) 2 + 12 )

Applying I.T. y(t) = L-1{Y}=

 2 -1   1 -1  1 1 L  + L  2 2  2 2  3  ( s + 1) + 2 )  3  ( s + 1) + 1 ) 

using first shifting theorem, y(t)=

1 -t -1  1  2 -t  1  e L  2  + e L-1  2 2  3 s + 2   s + 1)  3

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y(t)=

e −t [ Sin t + Sin 2t ] is required solution.

3

Problem #3 Solve the differential equation by using Laplace transformation y+n2y = a Sin (nt+2), y(0) = 0, y(0) = 0.

Solution: y+n2y = a Sin (nt+2) = a [Sin nt Cos 2 + Cos nt Sin 2] Applying L.T. L{ y} + n2 L { y } = a cos 2. L{Sin nt} + a Sin 2 L{Cos nt} Using L.T. On applying Laplace transform we get the given differential equation as, n n [s2Y – sy(0) - y(0)] + n2Y = 2 .a.Cos 2 + 2 .a. Sin 2 2 s +n s + n2 Solving Y, Y(s) =

n n .a. Cos 2 + 2 .a. Sin 2 2 2 (s + n ) (s + n 2 ) 2 2

Applying I.T. we get

    1 1 y(t) = n. a. Cos 2. L-1  2 + a. Sin 2. L-1  2 …… (1) From 2 2 2 2  (s + n )   (s + n )  I.L.T. tables, we know that 2nd term in R.H.S.

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  t. Sin nt s L-1  2 = ……………………………………………………………… (2) 2 2 2n  (s + n )  to find first term in R.H.S.

  -1 1  1 t 1 1 s L-1  2 =L = . t. Sin t . nt dt = 3 .[−nt Cos nt + Sin nt ] ….(3) .  2 2 2 2 2 2n  s ( s + n )  2n 0  (s + n )  Thus substituting (2) and (3) in (1), we get y(t)

= a.n. Cos 2.

a 2n 2 a = 2n 2 a = 2n 2 a  y(t) = 2n 2

=

Sin nt 1 .[−nt Cos nt + Sin nt ] + a Sin 2 3 2n 2n

[ − nt .Cos 2 . Cos nt + Cos 2. Sin nt + nt. Sin 2 . Sin nt] [ Sin nt .Cos 2 . − nt ( Cos nt.Cos 2 − Sin nt. Sin 2) ] [ Sin nt .Cos 2 . − nt Cos (nt + 2) ] [ Sin nt .Cos 2 . − nt Cos (nt + 2) ] is required solution.

Problem #4 Find the general solution of the differential equation by using Laplace transformation y - 3y +3y - y =t2 et, y(0) = 1, y(0) = 0,y(0) = 2.

Solution: Since the initial conditions are arbitrary assume y(0) = a, y(0) =b, y(0) =c.

Then Using L.T. On applying Laplace transform we get the given differential 2 equation as, (s2Y – as2 – bs - c) - 3( s2Y – as -b ) + 3(sY – a) - Y = ( s − 1) 3 as 2 + (b − 3a) s + (3a − 3b + c) 2 + 3 ( s − 1) ( s − 1) 3 c3 c1 c2 2 By partial fractions Y = where c1, c2, c3 are + + + 3 2 1 ( s − 1) ( s − 1) ( s − 1) ( s − 1) 6 constants depending on a, b, c.

Y=

Applying I.T. and using first shifting theorem y(t) = c1

t2 t t5 t e e + c2 t et + c3 et + 2 60

is required solution.

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Problem #5 Solve the differential equation by using Laplace transformation y - 3y +3y - y =t2 et, y(0) = 1, y(0) = 0,y(0) = - 2.

Solution: Applying L.T. to D.E. L{ y - 3y +3y - y} =L{ t2 et } 

L{ y} - 3L{ y} +3 L { y} - L{ y} =L{ t2 et }

[s3Y – s2y(0) - sy(0) - y(0) ] – 3 [s2Y – sy(0) - y(0)] + 3[sY – y(0)] – Y =

2 ( s − 1) 3

Using the initial condition y(0) = 1, y(0) = 0,y(0) = - 2, and solving for Y (s3 – 3a2 + 3s – 1)Y – s2 + 3s – 1=

2 ( s − 1) 3

( s − 1) 2 − ( s − 1) − 1 2 s 2 − 2s + 1 − s s 2 − 3s + 1 2 2 + + + = = 3 3 3 6 6 ( s − 1) ( s − 1) 6 ( s − 1) ( s − 1) ( s − 1) ( s − 1) 1 1 1 2 − − + Y= 2 3 ( s − 1) ( s − 1) ( s − 1) ( s − 1) 6

Y=

On applying I L.T. we get

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 1   1   1   1  y(t) = L-1 { Y } = L-1  − L-1  +2 L-1   − L-1  6 2 3  ( s − 1)   ( s − 1)   ( s − 1)   ( s − 1)  y(t) =

L-1

{Y}=

et

−t

et

t 2et t 5et − + is required solution.

2 60

Engineering Mathematics - I

UNIT – IV Section III

Semester – 1

By Dr N V Nagendram

Application of Laplace Transform

Class 14

Exercise Try yourself……..

Use L.T. to solve each of the following I.V.P. consisting of a D.E. with I.C. 01. y  − y = 0 , general solution 02. y  − y = e 3t , y(0) = 2 03. y  + y  = 0, y (0) = A, y (0) = B 04. y  + y  = 2e t , y(0) = 0, y (0) = 2 05. y  − 6 y  + 9 y = 0, y (0) = 2, y (0) = 9 06. y  + 4 y = 9t , y (0) = 0, y (0) = 7

[Ans. Assume y(0)=0=A=constant;y=Aet] [Ans. y=(3et +e3t)/2 ] -t [Ans.G.S. y=C + De , C=A + B,D= − B ] [Ans.G.S. y=et+ Cos t + Sin t [Ans.G.S. y=(3t + 2)e3t 9 19 [Ans.G.S. y= t + Sin 2t 4 4

] ] ]

07.

y  + 7 y  + 10 y = 4e −3t , y (0) = 0, y (0) = −1

[Ans.G.S. y=e-2t-2e-3t+e-5t

]

08.

y  − 8 y  + 15 y = 9te 2t , y (0) = 5, y (0) = 10

[Ans.G.S. y=4e2t+3te2t+3e3t-2e5t

]

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09. y  + y  = t Cos 2t , y (0) = 0, y (0) = 0 y  + n 2 y = a Sin (nt +  ), y (0) = 0, y (0) = 0 y  + y  = Sin t Sin 2t , y (0) = 1, 11. y (0) = 0

10.

[Ans.G.S. y= [Ans.

y=

9 5 1 Sin 2t − Sin t − t Cos 2t 4 9 3

]

a [ Sin nt Cos  − nt Cos(nt +  ) ] 2n 2

[Ans.G.S. y=

15 t 1 Cos t + Sin t + Cos 3t ] 16 4 16

1 e −t −2t     12. y + y = e Sin t , y(0) = 0, y (0) = 0 [Ans.y= ( Sin t − Cos t ) + ( Sin t + Cos t ) ] 8 8 y  + 4 y  + 5 y  + 2 y = 10 Cos t , 13.

[Ans.y=−e-2t + 2e-t -2te-t – Cos t +2 Sin t] y (0) = 0, y (0) = 0 , y (0) = 3 −1

1 e 2t 3 3 3 et (9 Cos t+5 Sin t) − 14. y  − y = e , y(0) = 0, y (0) = 0, y (0) = 0 [ e t + 3 18 2 2 2 2 ( iv)  y − 16 y = 30 Sin t , y (0) = 0, y (0) = 2, 15.

[Ans.G.S. y= 2(Sin 2t – Sin t y ( ) = 0, y ( ) = −18 t

Engineering Mathematics - I

UNIT – IV

Semester – 1

] ]

By Dr N V Nagendram

Application of Laplace Transform

Class 14

Section IV Application of Laplace Transform to system of simultaneous differential equations: Laplace transform can also be used to solve a system or family of m simultaneous equations in m dependent variables which are functions of the independent variable t. consider a family of two simultaneous differential equations in the two dependent variables x, y which are functions of t.

a1

(1)

d 2x d2y dx dy + a + a3 + a4 + a5 x + a6 y = R1 (t ) 2 2 2 dt dt dt dt

………………………………………..

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d 2x d2y dx dy b1 2 + b2 2 + b3 + b4 + b5 x + b6 y = R2 (t ) dt dt dt dt (2)

………………………………………..

Initial conditions x(0) = c1; y(0) = c2; x(0) = c3; y(0) = c4 …………………………

(3) Here a1, a2, a3, a4, a5, a6,b1, b2, b3, b4, b5, b6 and c1, c2, c3, c4 are all constants and R1(t), R2(t) are functions of t.

Method of solution to system of differential equation (D.E.): I. Apply Laplace transform on both sides of each of the two differential equation (1) and (2) above. This reduces (1) and (2) to two algebraic equations in X(s) and Y(s) where X(s) = L{ x(t) } and Y(s) = L{ y(t) }. a1[s2X-sx(0) - x(0)] + a2[ s2Y – sy(0) - y(0)] + a3[sX – x(0) ] + a4[sY- y(0) ]+ a5 X + a6 Y = Q1(s)…..(4) b1[s2X-sx(0) - x(0)] + b2[ s2Y – sy(0) - y(0)] + b3[sX – x(0) ] + b4[sY- y(0) ]+ b5 X + b6 Y = Q2(s)…

(5)

use the initial conditions (3) and substitute for x(0) , y(0) , x(0) , y(0).

II. Solve (4) and (5) for X(s) and Y(s).

III. The required solution is obtained by taking the inverse Laplace Transform of X(s) and Y(s) as x(t) = L-1{ X(s) } and y(t) = L-1 { Y(s) }. dx dy d 2x dx dy − 3x + = 1; conditions x(0) = −3 − + 2 y = 14t + 3 ; 2 dt dt dt dt dt 0; y(0) = 6.5; x(0) = 0;

Problem #1 Solve

Solution: Taking Laplace Transformation of the given differential equation we have [s2X-sx(0) - x(0)] -3 [sX – x(0) ] - [sY- y(0) ]+ 2Y = 14

1 1 +3 2 s s

1 s Use given initial conditions (I.C.) x(0) = 0; y(0) = 6.5=13/2; x(0) = 0; 2 + 13s 28 + 6s − 13s 2 S(s - 3)X +(2 – s)Y = and (s – 3 )X + sY = 2 2s 2s 3 2 13s + 15s − 6s − 28 4(7 − 6s) On solving we get Y(s) = and X(s) = 2 2 s( s − 3)( s + 2)( s − 1) 2s ( s + s − 2)

[sX – x(0) ] – 3X + [sY- y(0) ] =

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Taking inverse Laplace Transform y(t) =

L-1{Y(s)}

=

L-1

13s 3 + 15s 2 − 6s − 28    2 2  2s ( s + s − 2) 

C D  A B + = L-1  2 + +  s s −1 s + 2 s 7 5 1 5  = L-1  2 + − +  s s − 1 2( s + 2)  s

 y(t) = 7t + 5 – et +

5 -2t e 2

  4(7 − 6s) Similarly, x(t) = L-1{X(s)} = L-1   by partial fractions  s( s − 3)( s + 2)( s − 1)  2 1 1 1  = L-1  − − −   s 2( s − 3) 2( s − 1) ( s + 2)  1 1 = 2− et − e3t− e-2t 2 2

 x(t) =2−

1 t 1 3t -2t e − e −e 2 2

Problem #2 Solve 2

Is required solution.

dx dy dx dy + + 2 x + y = e t ; conditions x(0) = 2; + − x − y = e −t ; dt dt dt dt

y(0) = 1; Solution: Taking Laplace Transformation of the given differential equation we have 1 s +1 1 sX(s) – x(0) + sY(s) - y(0)+ 2 X(s) + Y(s) = s −1

2[sX(s) - x(0)] + sY(s)- y(0) –X(s)- Y(s) =

Use given initial conditions (I.C.) x(0) = 2; y(0) = 1.

5s + 6 3s − 2 (2s - 1)X(s) +(s – 1)Y(s) = and (s + 2 )X(s) + (s+1)Y(s) = s +1 s −1 3 2 s − 13 1 s − 12 s − s + 14 + 2 On solving we get Y(s) = = 2 2 2 2 ( s + 1) ( s − 1) ( s + 1)( s − 1 )

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and X(s) =

2s + 8 ( s 2 + 1)

 s − 13 1  Taking inverse Laplace Transform y(t) = L-1{Y(s)} = L-1  2 + 2   ( s + 1) ( s − 1)  = Cos t – 13 Sin t + Sinh t  y(t) = Cos t – 13 Sin t + Sinh t

 2s + 8  Similarly, x(t) = L-1{X(s)} = L-1  2  by partial fractions  ( s + 1)  = 2 Cos t + 8 Sin t  x(t) =2 Cos t + 8 Sin t Is required solution.

Engineering Mathematics - I

UNIT – IV

Semester – 1

By Dr N V Nagendram

Application of Laplace Transform

Exercise Try yourself ……..

Class 15

Solve the following system of equations

dx dy = 2x − 3y ; = y − 2 x , x(0) = 8, y(0) = 3 dt dt [Ans. x(t) = 5 e-t + 3 e4t; y(t) = 5 e-t – 2 e4t]

01.

d2y dx d 2 x dy −t + + 3x = 15e ; 2 − 4 + 3 y = 15 Sin 2t , 02.

2 dt dt dt dt x(0) = 35,y(0) = 27; x(0) = -48,y(0) = -55.

[Ans. x(t) = 30 Cos t – 15 Sin 3t + 3 e-t + 2 Cos 2t;

Ekeeda – Production Engineering

y(t) = 30 Cos t – 60 Sin t - 3 e-t + Sin 2t] d 2x d2y 2t 03. ; = = − x − 2 y , x(0) = y(0) = 1; x(0) = y(0) = 0 2 x + 3 y + e dt 2 dt 2 1 1 2 [Ans. x(t) = (3e t + 7e −t ) − (19 Cos t − 2 Sin t ) + e 2t ; 4 10 5 −1 t 1 1 (3e + 7e −t ) + (19 Cos t − 2 Sin t ) − e 2t ] y(t) = 12 10 5 dy dx = Sin t − x , x(0) = 1, y(0) = 0 = y + et ; 04. dt dt (e t + Cos t + 2 Sin t − t Cos t ) ( t Sin t − e t + Cos t − Sin t ) [Ans. x(t) = ; y(t) = ] 2 2 dx dx dy dy + 4 + 3 y = 0 , x(0) = 3, y(0) = 0 + + 2x = 1 ; dt dt dt dt −t −6t / 11 (5 − 2e − 3e (e −t − 3e −6t / 11 ) ) [Ans.x(t)= ; y(t) = ] 10 5

05. 3

dx dx d2y + 2 x − y = 1 , x(0) = y(0)= y(0) = 0 06. + 2 2 = e −t ; dt dt dt [Ans. x(t)=1+ e-t – e-at-e-bt; y(t)=1+e-t– be-at− a e-bt ] 1 1 Note: a = (2 − 2 ) ; b = (2 + 2 ) 2 2

dx dy d 2x dx − − 2 x + 2 y =1 − 2t ; + 2 + x = 0 , x(0) = 0, x(0) = 0 2 dt dt dt dt -t [Ans. x(t) = 2−2e (1+t); y(t) = 2−t-2e-t (1+t)]

07.

dx dy dx dy + 4 + x − y = 3e t ; + + 2 x + 2 y = e t , x(0) =1, y(0) =0 dt dt dt dt 1 [Ans. x(t) = e-2t - t et; y(t) = et +tet ] 3

08. 2

dx dy − 6 x + 3 y = 8e t ; − y − 2 x = 4e t ,x(0) =-1, y(0) = 0 dt dt − 2 t 2 4t [Ans. x(t) = -2 et + e4t; y(t)= e + e ] 3 3

09.

dy dx + x = Cos t , x(0) = 2, y(0) = 0 = y Sin t ; dt dt [Ans. x(t)=e-t + et = 2Cosh t; y(t)=Sin t – 2 Sinh t]

10.