WEBVTT
Kind: captions
Language: en
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welcome to the series of lecture series
on mathematical methods and its applications
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so in the lectures we have seen about laplace
transforms their property inverse laplace
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transform various functions how to find the
laplace inverse and laplace transform etcetera
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so now we will see applications of laplace
transforms what are the various applications
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of laplace transforms so laplace transform
can be used in solving various ordinary
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differential equations which we faced in
various engineering problems like in engineering
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like electrical engineering problems or
in mechanical engineering problems we face
00:00:58.499 --> 00:01:03.789
some ordinary differential equations or partial
differential equations which can be solved
00:01:03.789 --> 00:01:11.990
using laplace transforms some integral equations
some involving discontinuous functions
00:01:11.990 --> 00:01:18.060
which can be solved using laplace transforms
so let us see the application laplace transform
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[pa/part] part one so the first problem is
the differential equation for a current i
00:01:24.960 --> 00:01:33.149
in an electrical circuit containing an inductance
l and a resistance r in series and acted on
00:01:33.149 --> 00:01:40.149
by an electromotive force given by e sin omega
t satisfies the differential equation this
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so what is the differential equations it is
the d l d i by d t plus r i is equal to e
00:01:48.290 --> 00:01:56.790
sin omega t e sin omega t is the e m f electromagnetic
force ok so we have to find out the value
00:01:56.790 --> 00:02:02.670
of the current at any time t if initially
there is no current in the circuit that is
00:02:02.670 --> 00:02:10.290
it is given to us at i at zero is zero
that is i at t equal to zero is zero it is
00:02:10.290 --> 00:02:15.730
given to us so how to solve so we can solve
this problem using our differential equations
00:02:15.730 --> 00:02:22.040
techniques also but we can solve the same
problem using laplace transforms also so how
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can we solve this using laplace transform
let us see
00:02:26.670 --> 00:02:34.510
so first you take laplace both the sides ok
so this l is inductance so lap l into laplace
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transform of d i by d t plus r is constant
laplace transform of i is equals to e is constant
00:02:46.650 --> 00:02:58.689
laplace transform of sin omega t so it is
l ok laplace transform of d i by d t we already
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know that laplace transform of f dash t is
what f dash t is nothing but p f p minus f
00:03:07.439 --> 00:03:19.749
zero where f p is laplace transform of f t
ok now here f t is here it is i dash t
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d i by d t it is i dash derivative of i respect
to t so what what the laplace transform
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of this using that expression is will be it
is p f p minus f zero f zero means i zero
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plus r into suppose laplace transform of i
is f p ok which is equals to e and sin
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omega t is omega upon p square plus omega
square so and assuming laplace transform of
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i s f p you can call it some other function
also f p so it is p into laplace transform
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of i which are call which i am calling as
f p minus i zero or f zero plus r into laplace
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transform of i which are which i am calling
as f p equal to e into laplace transform of
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sin omega t it is sin omega upon p square
plus omega square now i zero is zero is given
00:04:23.889 --> 00:04:34.660
to us ok you can collect f p so f p will be
nothing but l p plus r it is zero and it is
00:04:34.660 --> 00:04:45.699
e omega upon p square plus omega square
so from here f p is nothing but e omega upon
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p square plus omega square into l p plus r
so so what will be this is f p and f p
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is nothing but laplace transform of i so what
will be i basically i is nothing but function
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of t ok i is equal to i of t i is nothing
but function of t here so this implies i t
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will be nothing but laplace inverse of e omega
upon p square omega square into l p plus r
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so now find out the solution of this differential
equation i t we have to find out the laplace
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inverse of this expression this is the only
thing left now ok so how can you find out
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laplace inverse of this so let us focus on
the laplace inverse now laplace inverse of
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this expression so this f p is what is
e omega e omega is constant free from p it
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is p square omega square into l p plus r so
one way is you can make partial fractions
00:06:07.410 --> 00:06:15.930
or you can use theorem so we can make partial
fraction as well so it is a p plus b upon
00:06:15.930 --> 00:06:25.509
p square plus omega square and plus c upon
l p plus r
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so e omega will be nothing but a p plus b
we can find out a b and c now [co/compare]
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compare the coefficients both the sides p
[co/coefficients] coefficients of p square
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here is zero and here it is a l plus c coefficient
of p square now coefficient of p suppose coefficient
00:06:57.500 --> 00:07:09.490
of p will be nothing but it is a r plus b
l and here it is nothing so it is zero and
00:07:09.490 --> 00:07:20.460
constant quantity is b r plus c omega square
which is e omega so using these three expressions
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we can easily find out the value of a b c
sub where you substitute these values over
00:07:26.219 --> 00:07:34.009
here and take the laplace inverse both the
sides so we can find out what will be i t
00:07:34.009 --> 00:07:44.460
i t can we find out ok so what will be i t
i t will be nothing but laplace inverse of
00:07:44.460 --> 00:07:50.100
this expression means this expression so this
laplace inverse this will be nothing but a
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into cos omega t plus b upon omega sin omega
t plus c upon l we can take out common which
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is nothing but e to the power minus r by l
times t ok so these a b c can be find out
00:08:16.700 --> 00:08:21.919
using these three expressions using these
three equations so we can substitute the
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values of a b c from these three equations
over here which will give us the value of
00:08:27.349 --> 00:08:37.099
i t function i t ok
now see the next problem again it is a electrical
00:08:37.099 --> 00:08:43.840
circuit problem in a electrical circuit with
e m f e t resistance r and inductance l the
00:08:43.840 --> 00:08:50.510
current i builds up at the rate given by this
equation ok now here if the if the switches
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is connected at t equal to zero and disconnected
t equal to a find the current i at any time
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t ok so how can we proceed for this problem
so what is a what is the governing equation
00:09:05.520 --> 00:09:17.500
here i d i by d t plus r i equals to e t ok
the same equation it is l c r circuit equation
00:09:17.500 --> 00:09:22.839
i mean e m f resistance and [inductan/inductance
] inductance this is the equation given by
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differential equation given by this expression
now it is given that in a switches is connected
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at equal to zero and disconnected at two a
what was it mean that electromotive force
00:09:32.870 --> 00:09:44.100
which is e t it is some constant quantity
say e when t varying from zero to a and disconnected
00:09:44.100 --> 00:09:52.810
at t equal to a means it is zero when t is
greater than or equal to so this is how electromotive
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force works for this particular problem ok
now to solve to find i at n is ten t we will
00:10:00.680 --> 00:10:08.240
use laplace transforms so it is take laplace
transform both the side it is l it is again
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laplace transform of d i by d t which is given
by p into f p minus i zero plus r into f p
00:10:21.510 --> 00:10:27.880
is equals to is equal to e t e t nothing but
this expression i mean given by this
00:10:27.880 --> 00:10:35.010
hm expression so it is zero to infinity
because you have to find out laplace transform
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of e t so laplace transform of e t is nothing
but zero to infinity e t into e power minus
00:10:41.709 --> 00:10:51.060
p t d t and it is nothing but zero to a e
e to the power minus p t d t otherwise it
00:10:51.060 --> 00:10:59.871
is zero so which is equal to e e to the power
minus p t upon minus p zero to a and which
00:10:59.871 --> 00:11:09.760
is equals to minus e by p e to the power minus
a p minus one so these are this is the right
00:11:09.760 --> 00:11:22.170
hand side and what is f p f p is nothing but
laplace transform of i t this i am assuming
00:11:22.170 --> 00:11:28.230
here ok now you assumed that a t equal to
zero initially there is no current in the
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circuit so i zero is zero ok so i zero is
zero so this implies basically f p p l plus
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r is equals to e upon p one minus e to the
power minus a p because i zero is zero initially
00:11:51.090 --> 00:11:57.120
i am assuming that there is no conductance
circuit it is zero f p is l p plus r and right
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hand side is e by p one minus e to the power
minus a p
00:12:00.829 --> 00:12:09.459
now the only thing is find i t which is
nothing but laplace inverse of f p and f p
00:12:09.459 --> 00:12:20.240
from here is nothing but laplace inverse of
e by a p into p l plus r one minus e to the
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power minus a p so now now we have to find
out the laplace inverse of this expression
00:12:27.110 --> 00:12:34.240
this is the only problem left now so how to
find laplace inverse of this now now again
00:12:34.240 --> 00:12:52.290
what is f p f p is e upon e upon p into
p l plus r minus e into e to the power minus
00:12:52.290 --> 00:13:03.650
a p upon p into p l plus r this is this is
f p and we have to find out laplace inverse
00:13:03.650 --> 00:13:20.040
of this f p ok so first let us find the partial
fraction of this expression so this is
00:13:20.040 --> 00:13:27.610
partial fraction can be find out when you
put is equal to zero so this is r p plus when
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you put equal to this so it is minus l by
r into p l plus r so you can check when you
00:13:40.313 --> 00:13:45.000
take the l c m you can get the right hand
side so this is the partial fraction of this
00:13:45.000 --> 00:13:58.579
expression so what will be f p now it is
e upon r p minus
00:13:58.579 --> 00:14:09.290
minus e into e to the power minus a p upon
r p minus minus plus e into e to the power
00:14:09.290 --> 00:14:21.840
minus a p
now the laplace inverse that is i t will be
00:14:21.840 --> 00:14:30.880
laplace inverse of f p so this will be the
laplace transform of this expression so we
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can further divided into two when t
is more than a so it is one so it is one it
00:14:37.029 --> 00:14:44.100
is one so that will give the i t when t
is greater than a and when t is less than
00:14:44.100 --> 00:14:51.410
a and greater than zero it is zero it is zero
so that we give gives the value of i t
00:14:51.410 --> 00:14:59.490
when t is lying between zero and a zero and
a ok so this will be the i t for this expression
00:14:59.490 --> 00:15:04.259
so this is how we can solve some problems
using laplace transforms
00:15:04.259 --> 00:15:10.440
now we will see some more problems which can
be solved using laplace transforms ok these
00:15:10.440 --> 00:15:16.589
are simple problems just just to illustrate
how can we use laplace transform to solve
00:15:16.589 --> 00:15:22.910
ordinary differential equations so the problem
is simple it is y double dash minus six y
00:15:22.910 --> 00:15:34.050
dash plus five y is equal to e to the power
two t and y zero is one and y dash zero is
00:15:34.050 --> 00:15:45.139
also one it is given ok it is one minus one
so it is one y zero is one and y dash zero
00:15:45.139 --> 00:15:51.870
is minus one ok so now how to solve this
problem is a laplace transforms so first take
00:15:51.870 --> 00:15:57.620
laplace transform both the sides ok so it
will be laplace transform of y double dash
00:15:57.620 --> 00:16:04.180
minus six times laplace transform of y dash
plus five times laplace transform of y is
00:16:04.180 --> 00:16:11.430
equals to laplace transform of e to the power
two t now from the laplace transform derivatives
00:16:11.430 --> 00:16:22.129
we know that the laplace transform of y double
dash is nothing but p square f p minus p f
00:16:22.129 --> 00:16:30.720
zero minus f dash t now you call it f zero
or you can call it y zero there is no problem
00:16:30.720 --> 00:16:43.500
now because here instead of f we are having
y ok so minus six and laplace of y dash will
00:16:43.500 --> 00:16:55.000
be nothing but p f p minus y zero plus five
times laplace of ok laplace of y is f p and
00:16:55.000 --> 00:17:07.760
it is nothing but one upon p minus two
laplace of y i am calling as f p so y zero
00:17:07.760 --> 00:17:15.130
is one and y dash zero is minus one it is
given so f p will be that is p square minus
00:17:15.130 --> 00:17:24.409
six p plus five and y zero is one that
is minus p minus minus plus one and it is
00:17:24.409 --> 00:17:34.380
minus minus plus six into one is six is equals
to one upon p minus two ok it is it is one
00:17:34.380 --> 00:17:40.890
it is minus one so it is p it is plus one
minus minus plus six into one is one so f
00:17:40.890 --> 00:17:52.809
p will be nothing but one upon t minus two
into t square minus six p plus five and it
00:17:52.809 --> 00:18:03.520
is this will go to that side so it is plus
p minus five of seven ok six plus one is seven
00:18:03.520 --> 00:18:14.300
so it is p minus seven upon p square minus
six p plus five so it is nothing but one upon
00:18:14.300 --> 00:18:23.520
p minus two into it is nothing but five and
one so that is p minus one into p minus five
00:18:23.520 --> 00:18:36.010
plus it is p minus seven upon p minus one
into p minus five now what is what will be
00:18:36.010 --> 00:18:48.429
y t y t will be nothing but laplace inverse
of f p so laplace inverse of this can be find
00:18:48.429 --> 00:18:54.409
out using partial fractions ok so you can
use partial fractions to find out the laplace
00:18:54.409 --> 00:19:11.409
of this
00:19:11.409 --> 00:19:19.679
so hm that will give the final answer of
final expression for y t ok
00:19:19.679 --> 00:19:26.890
now the next now we will see the next problem
involving t also in the y y double dash or
00:19:26.890 --> 00:19:35.789
y let us solve this problem so this this problem
can be completed ok you use you take the partial
00:19:35.789 --> 00:19:41.789
fraction make the partial fractions and then
take the laplace inverse of these expressions
00:19:41.789 --> 00:19:53.710
now next problem t y double dash plus two
y dash plus t y equal to zero and it is give
00:19:53.710 --> 00:20:05.450
to you that t y at zero is one and y at pi
is some value zero ok this is given to you
00:20:05.450 --> 00:20:12.330
now how to solve this problem again you will
use laplace transforms you take laplace
00:20:12.330 --> 00:20:20.710
transform both the sides so laplace transform
of t y double dash plus two times laplace
00:20:20.710 --> 00:20:34.179
transform of y dash plus laplace transform
of t y dash t y equal to zero now what is
00:20:34.179 --> 00:20:44.200
laplace transform y double dash i mean f double
dash ok f double dash t is nothing but p square
00:20:44.200 --> 00:20:59.350
f p minus p f zero minus f dash zero and laplace
transform of t f double dash would be what
00:20:59.350 --> 00:21:05.779
will be nothing but minus one to the power
one d by d p of laplace transform f double
00:21:05.779 --> 00:21:15.000
dash t by by the property of multiplication
with t ok so now laplace transform of t into
00:21:15.000 --> 00:21:21.890
y double dash would be what will be minus
one to the power one d by d p of laplace transform
00:21:21.890 --> 00:21:30.230
of y double dash so laplace transform of y
double dash is p square f p minus p into y
00:21:30.230 --> 00:21:41.160
zero minus y dash zero plus two into laplace
transform of y dash laplace transform of y
00:21:41.160 --> 00:21:52.861
dash is nothing but p f p minus y zero plus
laplace transform of t into y dash again is
00:21:52.861 --> 00:21:59.350
nothing but minus one to the power one d by
d p of laplace transform of y which is f p
00:21:59.350 --> 00:22:12.240
it is equal to zero because i am assuming
the laplace transform of y t as f p
00:22:12.240 --> 00:22:18.630
now let us simplify this it is minus derivative
with respect to p so it is p square into f
00:22:18.630 --> 00:22:30.230
dash p plus f p into two p minus derivative
of with respect to p is y zero y zero is
00:22:30.230 --> 00:22:42.100
one and derivative this is zero plus two p
f p and y zero is oneth minus two minus f
00:22:42.100 --> 00:22:56.600
dash p is equal to zero ok it is two f p two
f p y zero is one so two into one is two ok
00:22:56.600 --> 00:23:08.210
now it is nothing but this implies f dash
p into minus p square minus one with f p what
00:23:08.210 --> 00:23:17.049
will be having now it is minus this and plus
this so both will cancel out ok it is minus
00:23:17.049 --> 00:23:27.850
minus plus one minus two so it is minus one
equal to zero so what will be f dash p f dash
00:23:27.850 --> 00:23:41.250
p will be nothing but one upon negative p
square plus one so what will be f p you integrate
00:23:41.250 --> 00:23:52.360
both sides f p will be nothing but minus tan
inverse p plus c now now you find laplace
00:23:52.360 --> 00:24:06.909
inverse both the sides to find out y t
ok so so y t will be nothing but this implies
00:24:06.909 --> 00:24:16.020
so it is f dash p minus p square minus one
so it is ok it is minus minus plus one ok
00:24:16.020 --> 00:24:22.550
it is minus minus plus one ok plus one minus
two is minus one ok and it is one upon p square
00:24:22.550 --> 00:24:28.440
plus one and p square plus one is tan inverse
p plus c ok
00:24:28.440 --> 00:24:36.039
now to calculate this c we will use corollary
of existence theorem for laplace transforms
00:24:36.039 --> 00:24:42.400
we know that as p tend to infinity f p must
tend to zero so take p tend to infinity both
00:24:42.400 --> 00:24:50.179
the sides so p tend to infinity so at p
tend to infinity limit p tend to infinity
00:24:50.179 --> 00:25:00.799
f p must be zero so this is zero equal to
minus tan inverse pi by two tan must infinity
00:25:00.799 --> 00:25:11.690
ok plus c so this implies c so implies c is
pi by two so f p is nothing but minus tan
00:25:11.690 --> 00:25:21.880
inverse p plus pi by two and pi by two minus
tan inverse p is nothing but cot inverse p
00:25:21.880 --> 00:25:28.809
so now we have to find out laplace inverse
of this f p ok so take the derivative both
00:25:28.809 --> 00:25:34.880
the sides or we can take laplace inverse here
itself if we take laplace inverse here so
00:25:34.880 --> 00:25:42.630
it is nothing but minus t f t and it is
nothing but minus of sin t so f t will
00:25:42.630 --> 00:25:52.460
be nothing but sin t by t if we take the
laplace inverse here itself we can we can
00:25:52.460 --> 00:25:57.480
proceed from here also from here to take
the if you want to find out the laplace inverse
00:25:57.480 --> 00:26:05.419
of f p again we have to differentiate it
so we will go back to a same expression ok
00:26:05.419 --> 00:26:11.620
then the laplace inverse of this will be
nothing but sin t upon t so this is nothing
00:26:11.620 --> 00:26:23.200
but y t now y at pi must be zero which is
satisfied y at pi is zero so which is satisfied
00:26:23.200 --> 00:26:28.100
so hence this is the expression their solution
for this problem
00:26:28.100 --> 00:26:34.940
now similarly this problem can be solved
using laplace transforms whenever we involve
00:26:34.940 --> 00:26:41.190
with delta function right hand side so
such problems can also be solved using
00:26:41.190 --> 00:26:47.350
laplace transforms ok you take laplace transform
both the sides and then take laplace inverse
00:26:47.350 --> 00:26:51.590
that will give you the final answer of the
same problems so
00:26:51.590 --> 00:26:51.960
thank you