WEBVTT
Kind: captions
Language: en
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so welcome to the series of lecture series
on mathematical methods and its applications
00:00:25.029 --> 00:00:30.600
so on the last lecture we have done some
applications of laplace transforms we have
00:00:30.600 --> 00:00:37.320
seen that how we can solve an ordinary equation
ordinary differential equation using laplace
00:00:37.320 --> 00:00:44.780
transforms we have also solved some circuits
problem and solved them using laplace
00:00:44.780 --> 00:00:50.510
transforms ok now again we will see some more
application laplace transform in this lecture
00:00:50.510 --> 00:00:57.650
now suppose we have the first problem
such problems are basically called integral
00:00:57.650 --> 00:01:06.480
equations where where y is unknown function
involved under sign of integral so in in these
00:01:06.480 --> 00:01:11.780
problems y is unknown function and it involve
under sign of integral so such equations are
00:01:11.780 --> 00:01:17.460
called integral equations so laplace transform
is also useful to solve integral equations
00:01:17.460 --> 00:01:21.040
so to illustrate this let us discuss this
examples
00:01:21.040 --> 00:01:30.590
so first problem first problem is y dash
minus four y plus three times integral zero
00:01:30.590 --> 00:01:43.530
to t y tau d tau and equal to one equal to
t and it is given to you that y zero is so
00:01:43.530 --> 00:01:48.289
how can we solve this equation using laplace
transform so this is an integral equation
00:01:48.289 --> 00:01:53.689
because y is an unknown function involved
in under sign of integral also ok so how can
00:01:53.689 --> 00:01:58.829
we solve this take laplace transform both
the sides so laplace transform of y dash is
00:01:58.829 --> 00:02:16.730
p f p p f p minus f zero or y zero minus four
times f p plus three times now integral
00:02:16.730 --> 00:02:23.490
zero to two y t y tau d tau the laplace
of this is nothing but f p by p we already
00:02:23.490 --> 00:02:31.030
know this so this is f p y p which is equal
to the laplace of p is nothing but one by
00:02:31.030 --> 00:02:47.090
p square so here i am calling laplace of this
y t as f p ok now it is f p you can take
00:02:47.090 --> 00:02:54.930
common f p coefficient f p will be p from
here minus four from here and plus three upon
00:02:54.930 --> 00:03:02.519
p from here y zero is one given to us so you
can substitute y zero as one here so it will
00:03:02.519 --> 00:03:14.379
be equal one by p square plus one so this
implies f p into p square minus four p plus
00:03:14.379 --> 00:03:26.060
three upon p is equals to one plus p square
upon p square so p p cancel out this implies
00:03:26.060 --> 00:03:39.220
this implies f p is nothing but one plus p
square upon upon p into now this is p minus
00:03:39.220 --> 00:03:50.480
one p minus three the factor of this so this
will be the f p so to find out y t we will
00:03:50.480 --> 00:03:53.219
make we will take laplace inverse both the
sides
00:03:53.219 --> 00:03:58.909
so to find out the laplace inverse of this
f p we will make use of partial fractions
00:03:58.909 --> 00:04:08.980
so this is nothing but upon p a upon p
plus b upon p minus one plus c upon p minus
00:04:08.980 --> 00:04:18.870
three so to calculate to calculate a we
will substitute p equal to zero in this
00:04:18.870 --> 00:04:26.750
expression so when when we take p equal to
zero it is nothing but one upon one upon three
00:04:26.750 --> 00:04:35.450
so one upon three will be the value of a
for b you can substitute p equal to one p
00:04:35.450 --> 00:04:46.550
is one it is two upon it is two upon one into
minus two that is minus one and when p is
00:04:46.550 --> 00:04:53.860
three when p is three it is nine plus one
ten upon three into two two fives are ten
00:04:53.860 --> 00:05:06.410
it is five by three so these are the values
of a b and c ok so what will be y t y t will
00:05:06.410 --> 00:05:12.630
be nothing but laplace inverse of f p and
which is nothing but laplace inverse of this
00:05:12.630 --> 00:05:17.990
expression so laplace inverse of this expression
[is] nothing but one by three minus e to the
00:05:17.990 --> 00:05:25.810
power t plus five by three e to the power
three t so that will be the laplace inverse
00:05:25.810 --> 00:05:30.610
of this expression ok
so you can check when y zero equal to one
00:05:30.610 --> 00:05:36.680
when t equal to zero when t equal to zero
this value is one by three minus one that
00:05:36.680 --> 00:05:44.080
is minus two by three minus two by three
plus five by three is one so this condition
00:05:44.080 --> 00:05:49.880
is also satisfied so this is the solution
of this integral equation now suppose you
00:05:49.880 --> 00:05:54.940
want to solve the second problem is it
is again a integral equation so how can
00:05:54.940 --> 00:06:02.710
we solve this problem let us see again using
laplace transforms this problem is y x is
00:06:02.710 --> 00:06:22.080
equal to three x square plus integral zero
to x y t sin s minus t d t so this is the
00:06:22.080 --> 00:06:26.930
problem ok now how can we solve this integral
equation using laplace so again take laplace
00:06:26.930 --> 00:06:36.930
both the sides so laplace of y x suppose it
is y p suppose laplace y is y p three x square
00:06:36.930 --> 00:06:45.940
so laplace x square or t square is nothing
but two upon p q plus
00:06:45.940 --> 00:06:51.590
now to find out laplace of this expression
it is nothing but the the convolution of
00:06:51.590 --> 00:07:00.270
y with sin the convolution of y with sin function
is this expression so by the convolution theorem
00:07:00.270 --> 00:07:06.460
we already know that convolution theorem
of laplace transform it is zero to t
00:07:06.460 --> 00:07:18.970
f u g t minus u d u is nothing but laplace
of this is nothing but f p into g p where
00:07:18.970 --> 00:07:25.380
f p is the laplace of function f and g p is
a laplace of function g small g so this is
00:07:25.380 --> 00:07:29.460
by the convolution theorem of laplace transforms
so we again apply a convolution theorem over
00:07:29.460 --> 00:07:35.340
here in this expression so we will get laplace
transform of y t we have already assumed that
00:07:35.340 --> 00:07:43.640
it is y p into laplace transform of sin function
is nothing but one upon p square plus one
00:07:43.640 --> 00:07:52.340
this is by the convolution theorem of laplace
transforms now y p you can collect the
00:07:52.340 --> 00:08:06.530
terms containing y p
so it is y p into into it is p square upon
00:08:06.530 --> 00:08:19.180
p square plus one which is equals to six upon
p q now this implies y p is equal to six p
00:08:19.180 --> 00:08:27.550
square plus one upon p to the power one so
this would be equal to six one by p cube plus
00:08:27.550 --> 00:08:36.640
one by p to the power five so to calculate
to calculate y t is nothing but laplace inverse
00:08:36.640 --> 00:08:44.070
of y p and it is nothing but laplace inverse
of this expression so laplace inverse of this
00:08:44.070 --> 00:08:50.649
expression is nothing but six into laplace
inverse of one by p cube is t square upon
00:08:50.649 --> 00:08:55.660
two and laplace inverse of one by t to the
power five is nothing but t to the power four
00:08:55.660 --> 00:09:03.360
upon factorial four so this is nothing but
three t square plus t to the power four upon
00:09:03.360 --> 00:09:13.660
four so this is the this is the solution of
this integral equation ok
00:09:13.660 --> 00:09:19.339
now let us solve next problem again
it is a integral equation based on the same
00:09:19.339 --> 00:09:29.689
concept ok so in the next problem is that
is zero to x y t upon under root x minus
00:09:29.689 --> 00:09:41.740
t d t and it is equal to one plus two x minus
x square so we have to find that y which satisfy
00:09:41.740 --> 00:09:47.310
this expression basically we have to find
that y which satisfy this expression so this
00:09:47.310 --> 00:09:53.150
is an integral equation basically so how
can we solve this again take laplace transforms
00:09:53.150 --> 00:09:57.550
both the sides now to find out the laplace
transforms of this expression we will again
00:09:57.550 --> 00:10:03.420
use convolution theorem for laplace transforms
because it is a convolution of under root
00:10:03.420 --> 00:10:11.000
x with under root t with y t ok because a
because it involve two functions basically
00:10:11.000 --> 00:10:19.870
zero to x y t into x minus t to the power
minus half so it a con it is a convolution
00:10:19.870 --> 00:10:27.390
of this function with function ok convolution
of y t and t so we will take the laplace transform
00:10:27.390 --> 00:10:33.079
both the sides so laplace transforms of
this by the convolution theorem is laplace
00:10:33.079 --> 00:10:38.829
transform y t which are which suppose is y
p and laplace transform of t to the power
00:10:38.829 --> 00:10:47.050
minus half is equal to or f to the power minus
half both are same you can take t k power
00:10:47.050 --> 00:10:53.790
minus half or s k power minus half
now laplace transform of one is one by p two
00:10:53.790 --> 00:11:01.740
by p square minus it is laplace transform
of t square t laplace transform of t square
00:11:01.740 --> 00:11:13.779
is two by p cube so it is y p laplace transform
of t to the power minus half is gamma half
00:11:13.779 --> 00:11:21.850
upon p to the power half which is one by p
plus two by p square minus minus two by p
00:11:21.850 --> 00:11:31.210
cube gamma half is under root pi so y p is
nothing but one by under root p time one by
00:11:31.210 --> 00:11:39.439
under root pi times one by under root pi plus
two upon p to the power three by two minus
00:11:39.439 --> 00:11:48.760
two upon p to the power five by two ok
now y t is nothing but laplace inverse of
00:11:48.760 --> 00:11:53.730
this expression so take laplace inverse both
the sides so laplace inverse of this will
00:11:53.730 --> 00:11:58.999
be nothing but it is one by under root pi
now laplace inverse of one by under root p
00:11:58.999 --> 00:12:03.709
again recall the formula of t to the power
laplace of t to the power n is nothing but
00:12:03.709 --> 00:12:11.410
gamma n plus one upon p to the power n plus
one so from here laplace inverse of one upon
00:12:11.410 --> 00:12:24.130
p to the power n plus one is nothing but gamma
is nothing but gamma n plus one upon oh
00:12:24.130 --> 00:12:31.089
it is nothing but t to the power n upon gamma
n plus one ok
00:12:31.089 --> 00:12:37.970
so laplace inverse of this so you substitute
n s minus half to get under root p so it will
00:12:37.970 --> 00:12:46.670
be t to the power minus half upon gamma half
plus two two into one upon p to the power
00:12:46.670 --> 00:12:52.610
three by two so you will substitute n s half
it is nothing but t to the power half upon
00:12:52.610 --> 00:13:02.660
gamma three by two minus again two into it
will be t to the power three by two upon
00:13:02.660 --> 00:13:10.540
gamma five by two so you can we can simplify
this very easily so this will be nothing but
00:13:10.540 --> 00:13:18.529
this will be nothing but one by under root
pi the first expression is one upon under
00:13:18.529 --> 00:13:28.490
root pi into under root t plus two under root
t and it is one by two under root pi minus
00:13:28.490 --> 00:13:36.050
two into three t to the power three by two
upon three by two one by two under root pi
00:13:36.050 --> 00:13:42.649
so this we can easily simplify so this
after simplification we get one upon pi under
00:13:42.649 --> 00:13:48.939
root pi can be taken out common so it is one
under root t minus plus four times under
00:13:48.939 --> 00:13:56.959
root t minus it is four eight upon three
times eight upon three times at t to the power
00:13:56.959 --> 00:14:04.149
t to the power three by two so this will be
the this will be the final answer ok for this
00:14:04.149 --> 00:14:07.810
problem
now similarly we can solve the last problem
00:14:07.810 --> 00:14:13.860
of this slide again we will take laplace
transform both the sides in the left hand
00:14:13.860 --> 00:14:21.879
side we have d y by d t so the laplace transform
of d y by d t will be p f p minus y zero y
00:14:21.879 --> 00:14:27.930
zero given as one and the when the right hand
side we will apply convolution theorem for
00:14:27.930 --> 00:14:34.480
laplace transforms so this is that is how
we can solve integration equations using
00:14:34.480 --> 00:14:42.490
laplace transforms ok now next is we can also
solve simultaneous ordinary equations using
00:14:42.490 --> 00:14:49.180
laplace transforms so let us see some problems
based on it so the first problem is
00:14:49.180 --> 00:14:56.129
simultaneous differential equation is d x
upon d t is equals to two x minus three y
00:14:56.129 --> 00:15:11.100
and d y by d t as given as d y d t is y minus
two x and given as x x at zero is eight and
00:15:11.100 --> 00:15:19.759
y at zero is three y at zero is three
so let us solve this problem again again we
00:15:19.759 --> 00:15:25.249
will take laplace transform of both the expressions
so laplace when we take laplace transform
00:15:25.249 --> 00:15:32.540
of first expression so it will be nothing
but it is d by d s by d t so it will
00:15:32.540 --> 00:15:47.120
be nothing but p into x p minus x zero
ok is equals to two x p minus three y p so
00:15:47.120 --> 00:15:54.290
that will be laplace of first expression what
is x zero x zero is eight so what this expression
00:15:54.290 --> 00:16:08.929
implies this implies p minus two times x p
ok plus three y p is equal to eight so that
00:16:08.929 --> 00:16:12.250
will be the first equation which is obtained
here
00:16:12.250 --> 00:16:17.350
now apply laplace transform in the second
equation when you apply laplace transform
00:16:17.350 --> 00:16:24.819
both the sides here we will be having p into
y p so this y p is nothing but laplace transform
00:16:24.819 --> 00:16:38.879
of y t ok now minus y zero which is equals
to y p minus two x p so again this implies
00:16:38.879 --> 00:16:45.610
you compare take the coefficients so it is
two into x p two times x p and the coefficient
00:16:45.610 --> 00:16:58.129
of y p is coefficient of y p is p minus one
which is equals to y zero y zero is three
00:16:58.129 --> 00:17:02.449
so these are the two equations which we obtain
p minus one times y p
00:17:02.449 --> 00:17:07.589
so these are two equations which we obtain
over here now we will find the these are the
00:17:07.589 --> 00:17:15.829
two linear equations in x p and y p ok two
linear equations x p and y p so we will solve
00:17:15.829 --> 00:17:23.150
them for x p and y p and then take the laplace
inverse to find out x t and y t that is a
00:17:23.150 --> 00:17:30.840
solution of this simultaneous differential
equation so how can we solve this so to find
00:17:30.840 --> 00:17:37.250
out the solution it is nothing but hm i
mean simultaneous linear equations to into
00:17:37.250 --> 00:17:47.190
unknowns so we can solve it like this it is
x p upon minus y p upon and it is one upon
00:17:47.190 --> 00:17:55.960
so it is the determinant of eight three and
three p minus one this is by the cramer rule
00:17:55.960 --> 00:18:03.190
we are solving it then minus y p y p coefficient
will be eight three c one c two and it is
00:18:03.190 --> 00:18:11.410
p minus two and two and here the right hand
side will be nothing but p minus one ok p
00:18:11.410 --> 00:18:24.950
minus two it is p minus two and three two
and p minus one
00:18:24.950 --> 00:18:35.330
ok
so this implies x p upon it is eight p
00:18:35.330 --> 00:18:45.990
minus eight minus nine is minus eight minus
nine is minus seventeen minus y p upon it
00:18:45.990 --> 00:18:56.700
is sixteen sixteen plus six sixteen plus six
is twenty two minus three p it is one upon
00:18:56.700 --> 00:19:09.570
it it is nothing but p square p square
minus three p plus two minus six ok so this
00:19:09.570 --> 00:19:19.360
is x p and y p so what are x p and y p
now so x p is nothing but eight p upon [sen/minus]
00:19:19.360 --> 00:19:25.740
minus minus seventeen upon it is nothing but
p square minus three p minus four which can
00:19:25.740 --> 00:19:34.180
be is the factorize it is nothing but p minus
four into p plus one which is p square minus
00:19:34.180 --> 00:19:44.090
three p minus four of course and y p is nothing
but y p is nothing but three p minus twenty
00:19:44.090 --> 00:19:54.020
two upon again the same quantity in the denominator
now this is nothing but we can make use of
00:19:54.020 --> 00:20:04.930
partial fractions again so it is p minus four
it is upon minus four it is plus p plus one
00:20:04.930 --> 00:20:13.360
so we will substitute p as four first so it
is thirty two it is thirty two minus seventeen
00:20:13.360 --> 00:20:21.410
thirty two minus seventeen is five fifteen
so fifteen upon when you substitute four five
00:20:21.410 --> 00:20:28.170
three so it is three this partial fraction
you can also compute compare the coefficients
00:20:28.170 --> 00:20:33.490
both the sides as you already know now you
substitute p as minus one so it is minus eight
00:20:33.490 --> 00:20:40.380
minus seventeen upon minus five so it is nothing
but minus twenty five upon minus five which
00:20:40.380 --> 00:20:49.030
is five so it is five ok
now we will use a partial fraction again here
00:20:49.030 --> 00:20:57.070
it is minus p minus four plus p plus one so
when you substitute p as four it is twelve
00:20:57.070 --> 00:21:04.290
minus twenty two upon five so it is minus
ten upon five it is minus two and when you
00:21:04.290 --> 00:21:09.620
substitute p as minus one it is minus three
minus twenty two upon minus five that is minus
00:21:09.620 --> 00:21:21.010
twenty five upon minus five it is five ok
now to find x t x t which is nothing but laplace
00:21:21.010 --> 00:21:31.650
inverse of x p so we will make laplace inverse
of this expression three upon p minus four
00:21:31.650 --> 00:21:39.360
plus five upon p plus one so it is nothing
three e to the power four t plus five e to
00:21:39.360 --> 00:21:49.530
the power minus t again to calculate y t we
will take laplace inverse of y p which is
00:21:49.530 --> 00:21:54.390
nothing but laplace inverse of this expression
and it is nothing but minus two e to the power
00:21:54.390 --> 00:22:01.830
four t plus five e to the power minus t so
these are solutions of this differential equation
00:22:01.830 --> 00:22:07.430
you can check also f zero is eight yeah f
zero is eight and y zero is three yeah y zero
00:22:07.430 --> 00:22:13.690
is three and the it will also satisfy the
these equations you can easily verify ok
00:22:13.690 --> 00:22:19.160
now let us discuss one more example based
on this which is a second order simultaneous
00:22:19.160 --> 00:22:28.110
differential equation so this also we can
solve using using laplace transforms so
00:22:28.110 --> 00:22:35.050
how can we solve let us see so it is nothing
but second derivative of x respect to t
00:22:35.050 --> 00:22:48.640
which is equals to given as x plus y and
d two y by d x two is nothing but it is minus
00:22:48.640 --> 00:22:56.610
x minus y ok and the initial conditions are
given to you ok
00:22:56.610 --> 00:23:02.890
now again take laplace transform both the
sides first for the first equation when you
00:23:02.890 --> 00:23:12.640
take laplace both the sides so it will be
nothing but p square x p minus p into x zero
00:23:12.640 --> 00:23:24.241
minus x dash zero ok and it is equal to x
p plus y p course x p here is laplace transform
00:23:24.241 --> 00:23:32.440
x t and y p is the laplace transform of y
t ok now x zero is given to you as two so
00:23:32.440 --> 00:23:41.540
this value is two and x dash zero is zero
so this implies p square plus minus one times
00:23:41.540 --> 00:23:54.750
x p minus y p will be equal to x zero is two
so it is two and x dash zero is zero so this
00:23:54.750 --> 00:24:02.700
is the first equation which we obtained
now apply laplace transform in this equation
00:24:02.700 --> 00:24:07.460
both the sides when you take laplace transform
both the sides so what we will obtain from
00:24:07.460 --> 00:24:15.650
here so when you take laplace transform both
the sides here so it is p into y p minus p
00:24:15.650 --> 00:24:28.100
into y zero minus y dash zero is equals to
minus x p minus y p so this implies now y
00:24:28.100 --> 00:24:38.410
zero is minus one given to us and y dash zero
is zero so it is x p plus p plus one times
00:24:38.410 --> 00:24:52.060
y p y zero is minus one so it is minus it
is minus one it is like p so it is minus p
00:24:52.060 --> 00:25:03.470
so this is the second equation which we obtained
ok so it is it is x p x p will come here
00:25:03.470 --> 00:25:11.850
it is plus one times y p and this y zero is
y zero is minus one minus minus plus p we
00:25:11.850 --> 00:25:16.980
may come here it is minus p
now in this again verify this equation this
00:25:16.980 --> 00:25:26.260
is f p square minus one time x p and y
p x zero is two it is two p ok and minus
00:25:26.260 --> 00:25:33.460
y p so this equation is obtained now here
it is p square y p p square y p minus p
00:25:33.460 --> 00:25:39.340
y zero minus y dash zero equal to this so
it is nothing but p square y so what are two
00:25:39.340 --> 00:25:47.570
equations which which we obtained it is
p square minus one times x p minus y p
00:25:47.570 --> 00:25:59.650
is equals to two p and here it is x p plus
p square plus one times y p is equals to minus
00:25:59.650 --> 00:26:05.360
p so again we will these are simultaneous
linear equations so we solve for x p and y
00:26:05.360 --> 00:26:16.790
p so it is x p upon a determinant of two p
minus p minus one p square plus one is equals
00:26:16.790 --> 00:26:27.660
to minus y p upon p determinant of p square
minus determinant of two p minus p and
00:26:27.660 --> 00:26:36.310
p square minus one upon one and one upon it
is determinant of p square minus one minus
00:26:36.310 --> 00:26:47.910
one one p square plus one ok
so this implies x p upon this expression is
00:26:47.910 --> 00:27:02.650
two p q plus two p minus p ok this is minus
y p upon this expression is two p minus minus
00:27:02.650 --> 00:27:15.410
plus p cube minus p ok and this is one upon
p to the power four upon one plus one so clearly
00:27:15.410 --> 00:27:27.520
x p will be nothing but it is p cancels
it is nothing but two p square plus one
00:27:27.520 --> 00:27:41.070
ok upon p cube so it is two upon plus one
by p cube and x p y p is nothing but it is
00:27:41.070 --> 00:27:54.110
negative of it is negative of one plus
one plus p square it is two p minus p is
00:27:54.110 --> 00:27:59.890
p p p cancels and it is one plus p square
upon p cube so it is nothing but minus one
00:27:59.890 --> 00:28:13.200
by p cube minus one by p ok
so now x t now x t is nothing but x t we will
00:28:13.200 --> 00:28:21.970
laplace inverse of x p and y t is laplace
inverse of y t so what will be x t x t
00:28:21.970 --> 00:28:30.770
will be nothing but laplace inverse of x p
which is equals to two plus now one by p cube
00:28:30.770 --> 00:28:39.710
one by p cube is t square by two and y t is
will be nothing but minus of two by minus
00:28:39.710 --> 00:28:49.630
of t square by two minus one so these are
the solutions which we can obtain from here
00:28:49.630 --> 00:28:57.750
ok so hence using laplace transforms we
ca also solve hm integral equations and
00:28:57.750 --> 00:29:02.620
simultaneous ordinary differential equation
so in the next lecture we will also see that
00:29:02.620 --> 00:29:07.650
how can we use laplace transform to solve
partial differential equations so
00:29:07.650 --> 00:29:08.090
thank you