WEBVTT
Kind: captions
Language: en
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welcome to the lecture series on mathematical
methods and its applications so let us discuss
00:00:25.280 --> 00:00:30.840
some more application of fourier transforms
so we have already discussed some application
00:00:30.840 --> 00:00:38.730
in last lectures fourier series fourier
integral fourier transforms now now this
00:00:38.730 --> 00:00:45.590
this we have already discussed that how
we can find out fourier derivative of
00:00:45.590 --> 00:00:51.140
sin or cosine transform of derivatives this
we have already discussed that if we want
00:00:51.140 --> 00:00:57.500
to find out fourier cosine of f dash x
so that will be nothing but omega fourier
00:00:57.500 --> 00:01:04.010
sin of f x minus f zero and similarly if you
want to find out fourier sin of f dash
00:01:04.010 --> 00:01:08.690
x it will nothing but minus omega fourier
cosine of f dash f x
00:01:08.690 --> 00:01:14.760
so when you replace f by f f dash in the
first expression and f by f dash in the second
00:01:14.760 --> 00:01:18.700
expression you will get back to these two
expression so what are what are these two
00:01:18.700 --> 00:01:23.670
expression so let us this we have already
derived in the last lectures so what is fourier
00:01:23.670 --> 00:01:30.070
cosine of f double dash x fourier cosine of
f double dash will be nothing but x dash we
00:01:30.070 --> 00:01:43.670
have already derived it is minus omega square
fourier cosine of f x minus f dash zero and
00:01:43.670 --> 00:01:53.320
fourier sin of f double f double dash x is
nothing but omega square fourier fourier
00:01:53.320 --> 00:02:07.310
fourier sin of f x ok it is minus omega
square fourier sin of this plus omega f zero
00:02:07.310 --> 00:02:11.670
so these two results these we have already
discussed in the last lectures now what how
00:02:11.670 --> 00:02:18.500
they are important and why i am discussing
this these things over here let us see now
00:02:18.500 --> 00:02:29.380
in this problem for the equation u t
equal to del u by del t is equal to two del
00:02:29.380 --> 00:02:35.540
square u by del x square this equation we
have to solve now x is greater than zero and
00:02:35.540 --> 00:02:44.980
t is greater than zero now x is varying
from zero to infinity so that means we have
00:02:44.980 --> 00:02:52.170
either to apply a fourier cosine or fourier
sin transform because if x is this x is not
00:02:52.170 --> 00:02:56.490
varying from minus infinity to plus infinity
if x is varying if any one of variable is
00:02:56.490 --> 00:03:00.230
varying from minus infinity to plus infinity
then we have to apply fourier transform
00:03:00.230 --> 00:03:07.080
b ut this x is varying only in half range
from zero to infinity so that means we have
00:03:07.080 --> 00:03:17.670
to apply either fourier cosine or or fourier
sin ok now now which we have to apply
00:03:17.670 --> 00:03:23.569
fourier sin or cosine how we will decide we
will decide by seeing the conditions if we
00:03:23.569 --> 00:03:33.180
apply a fourier cosine in f double dash here
f double dash if this this term we are
00:03:33.180 --> 00:03:38.599
we are applying fourier transform fourier
sin or cosine transform respect to x ok so
00:03:38.599 --> 00:03:43.460
we have a double derivative so when we have
a double derivative we apply suppose fourier
00:03:43.460 --> 00:03:53.290
cosine so in this we need f dash zero
so here here we need here we need del u
00:03:53.290 --> 00:04:04.050
by del x del u by del x at t equal to zero
at x equal to zero sorry ok we need the derivative
00:04:04.050 --> 00:04:09.190
condition so which is not here it is only
u equal to zero when x equal to zero the first
00:04:09.190 --> 00:04:17.570
condition so if we now observe fourier sin
transform they are we need only f zero ok
00:04:17.570 --> 00:04:26.630
when when x is zero ok so so in this
way we conclude that we have to apply fourier
00:04:26.630 --> 00:04:34.560
sin transform in this expression ok so basically
whenever we have a half range we have to
00:04:34.560 --> 00:04:39.930
apply either fourier sin or fourier cosine
when the initial conditions when in the
00:04:39.930 --> 00:04:46.710
initial condition we have the partial derivative
at x equal to zero so then we have to apply
00:04:46.710 --> 00:04:51.380
fourier cosine otherwise we have to apply
fourier sin transform
00:04:51.380 --> 00:04:56.510
so in this particular problem we have to apply
fourier sin transform so take fourier sin
00:04:56.510 --> 00:05:02.500
transform both the sides because we have the
condition when x equal to zero u equal
00:05:02.500 --> 00:05:08.390
to zero if we have derivative condition at
x equal to zero then we apply fourier sin
00:05:08.390 --> 00:05:14.460
fourier cosine transform now take fourier
sin transform both the side what will be the
00:05:14.460 --> 00:05:21.810
fourier sin transform of del u by del t
it will be nothing but zero to infinity del
00:05:21.810 --> 00:05:32.330
u by del t of sin omega x into d omega ok
so that will be nothing but d by d t because
00:05:32.330 --> 00:05:40.820
this term is free from t so zero to infinity
u into sin omega x into d omega which is nothing
00:05:40.820 --> 00:05:51.550
but d by d t of fourier sin transform of u
x t ok so now when you take fourier sin transform
00:05:51.550 --> 00:06:00.659
both the side it will be nothing but d by
d t of fourier sin transform of u x t which
00:06:00.659 --> 00:06:09.240
is equal to two times minus omega square apply
this result so fourier transform of sin transform
00:06:09.240 --> 00:06:18.460
of u x t plus omega into zero because u is
zero when x is zero
00:06:18.460 --> 00:06:28.779
now let fourier sin transform of u x t as
suppose you take it for f of s it is omega
00:06:28.779 --> 00:06:36.860
t ok because we are applying fourier transform
with respect to x so that is it is d upon
00:06:36.860 --> 00:06:45.610
f x upon d t which is minus minus plus two
omega square f s which is equals to zero so
00:06:45.610 --> 00:06:54.340
that implies d f s upon f s is equals to minus
two omega square d t when w e integrate both
00:06:54.340 --> 00:07:05.310
sides f x is nothing but you obtain as k times
e k power minus two omega square d
00:07:05.310 --> 00:07:17.130
so again this k will be nothing but function
of omega so now we have and the next condition
00:07:17.130 --> 00:07:24.210
where when t equal to zero u equal to e k
power minus x where this condition so now
00:07:24.210 --> 00:07:36.870
apply this condition over here so what we
obtain so u u x zero is e k power minus
00:07:36.870 --> 00:07:44.630
x so apply fourier sin transform both the
sides so fourier sin transform of this expression
00:07:44.630 --> 00:07:51.080
will be nothing but fourier sin transform
of e k power minus x which is nothing but
00:07:51.080 --> 00:08:01.610
zero to infinity e k power minus x sin
omega x into d omega into d x sorry it will
00:08:01.610 --> 00:08:08.560
be d x so it will be equal to one upon one
plus omega square and it is minus e k power
00:08:08.560 --> 00:08:18.500
minus x to integrate by parts and minus e
k power minus x omega cos omega x from zero
00:08:18.500 --> 00:08:22.930
to infinity
so this is nothing but one upon one plus omega
00:08:22.930 --> 00:08:30.500
square so when x is tend to infinity both
will ten to zero when x is tends to zero it
00:08:30.500 --> 00:08:35.630
will zero it will be nothing but minus minus
plus so it is omega so that will be nothing
00:08:35.630 --> 00:08:44.360
but omega upon one plus omega square ok now
when you take when you take t equal to zero
00:08:44.360 --> 00:08:51.720
here so fourier transform fourier sin transform
of u x zero will be nothing but f s omega
00:08:51.720 --> 00:08:56.540
zero so this is nothing but basically f x
omega zero
00:08:56.540 --> 00:09:04.910
so when you take f x omega zero here f x when
you substitute t equal to zero here basically
00:09:04.910 --> 00:09:11.240
so it will be k omega and k omega is nothing
but from here will be nothing but omega upon
00:09:11.240 --> 00:09:22.350
omega upon one plus omega square ok so therefore
therefore what will be f s f s omega t it
00:09:22.350 --> 00:09:30.430
will be omega upon one plus omega square e
k power minus two omega square t now take
00:09:30.430 --> 00:09:36.690
inverse fourier sin transform both the
side to find out the solution so this implies
00:09:36.690 --> 00:09:46.940
u x t will be nothing but two upon pi integral
zero to infinity omega upon one plus omega
00:09:46.940 --> 00:09:55.500
square e k power minus two omega t and it
is sin omega x d omega
00:09:55.500 --> 00:10:05.750
so this will be the final solution of this
problem ok so here is a here is a solution
00:10:05.750 --> 00:10:11.740
the same solution i have discussed here so
this is a solution two upon pi f zero to infinity
00:10:11.740 --> 00:10:16.360
omega upon one plus omega square e k power
minus two omega square d sin omega x into
00:10:16.360 --> 00:10:26.779
d omega ok now in the next problem it is
again in half range now here the half range
00:10:26.779 --> 00:10:33.839
is in y y is varying from zero to infinity
so that means we have to apply fourier sin
00:10:33.839 --> 00:10:40.600
or fourier cosine transform with respect to
y first we have to decide that for which
00:10:40.600 --> 00:10:46.189
variable we have to apply fourier sin or cosine
transform and that depends on the range here
00:10:46.189 --> 00:10:51.220
x is in zero to l and y is varying from zero
to infinity so we have to apply fourier sin
00:10:51.220 --> 00:10:57.899
or cosine transform with respect to y number
one number two we have we have to apply
00:10:57.899 --> 00:11:02.110
fourier sin or fourier cosine that will decide
by the conditions
00:11:02.110 --> 00:11:08.709
now here we will apply fourier sin or
cosine transform with respect to y and the
00:11:08.709 --> 00:11:13.769
partial derivative with respect to y when
y is zero is zero it is given to us that means
00:11:13.769 --> 00:11:20.949
we will apply a fourier cosine transform here
because when you see because when you see
00:11:20.949 --> 00:11:27.360
this expression in fourier cosine transform
the second derivative here we have the
00:11:27.360 --> 00:11:33.899
derivative at t equal to at x equal to
zero ok so that means we have to apply fourier
00:11:33.899 --> 00:11:40.430
cosine transform when when we have the
derivative term in the initial conditions
00:11:40.430 --> 00:11:47.449
in the conditions ok
so here del u by del y at y equal to
00:11:47.449 --> 00:11:53.339
zero is given that means we have to apply
fourier cosine transform so now in this problem
00:11:53.339 --> 00:11:59.750
apply fourier cosine transform both the sides
so what is the problems basically now the
00:11:59.750 --> 00:12:07.670
problems here is del square u upon del x square
plus del square u upon del y square equal
00:12:07.670 --> 00:12:19.060
to zero so you will apply fourier cosine transform
that is clear and this respect to variable
00:12:19.060 --> 00:12:32.930
y that you have to apply apply fourier cosine
transform
00:12:32.930 --> 00:12:39.740
with respect to y
so what we will obtain so x will be three
00:12:39.740 --> 00:12:47.389
so it is d square f upon d x square now let
us suppose fourier cosine transform of u x
00:12:47.389 --> 00:12:57.720
t u x y will be nothing but suppose it is
f c x omega because we are apply with respect
00:12:57.720 --> 00:13:09.930
to y ok so x we will treat as it is so it
will be f c x omega plus now to find out
00:13:09.930 --> 00:13:17.279
now when we apply fourier cosine transform
here so we will apply that result that
00:13:17.279 --> 00:13:23.810
result that will be nothing that is nothing
but minus omega square fourier sin fourier
00:13:23.810 --> 00:13:34.490
sin cosine transform of x omega plus and minus
it is del u by del y when y equal to at x
00:13:34.490 --> 00:13:44.480
comma zero equal to zero ok this condition
we have applied for fourier cosine transform
00:13:44.480 --> 00:13:57.160
so what we obtain finally oh sorry so basically
now this this term is zero so what we obtained
00:13:57.160 --> 00:14:06.519
d square f c upon d x square minus omega
square f c will be zero because this is zero
00:14:06.519 --> 00:14:14.540
ok so this gives d square minus omega square
f c is equal to zero where d is nothing but
00:14:14.540 --> 00:14:22.569
d by d x so f c will be nothing but c one
e k power or you can break in terms of hyperbolic
00:14:22.569 --> 00:14:36.360
functions so it is sin hyperbolic omega
x ok plus c two sin hyperbolic omega y and
00:14:36.360 --> 00:14:42.620
again c one and two are the arbitrary functions
and f c is a functions of x and omega so c
00:14:42.620 --> 00:14:49.920
one and c two are the nothing but functions
of omega
00:14:49.920 --> 00:14:55.160
ok
so it is c one sin hyperbolic omega ok
00:14:55.160 --> 00:15:02.829
it is sorry sorry it is cos ok so it is cos
hyperbolic omega y now now use a initial
00:15:02.829 --> 00:15:09.230
use the conditions conditions are u zero
y equal to e k power minus two y is given
00:15:09.230 --> 00:15:17.380
to you ok you apply a fourier cosine transform
both the sides so it is fourier cosine transform
00:15:17.380 --> 00:15:23.809
of e k power minus two y so which is given
as zero to infinity e k power minus two y
00:15:23.809 --> 00:15:33.439
and it is cos omega y into d y so when
you integrate it what we will obtain it is
00:15:33.439 --> 00:15:39.750
one upon four plus omega square and again
you integrate so it is minus two minus two
00:15:39.750 --> 00:15:50.220
y it is cos omega y minus minus plus e k power
minus two y omega sin omega y from zero to
00:15:50.220 --> 00:15:54.920
infinity
at infinity both will tend to zero and at
00:15:54.920 --> 00:16:00.839
zero this is zero and this will be one and
minus minus plus so it is two upon four plus
00:16:00.839 --> 00:16:06.350
omega square so this we will obtain here as
a fourier cosine transform of this term and
00:16:06.350 --> 00:16:11.069
again when you substitute x equal to zero
both the sides this will be nothing but is
00:16:11.069 --> 00:16:23.940
equal to f c of zero omega so when you take
f c of zero omega here f c of zero omega
00:16:23.940 --> 00:16:35.080
here so sin hyperbolic zero is zero and
cos hyperbolic zero is one that is nothing
00:16:35.080 --> 00:16:46.249
but c two ok it is nothing but it is it
is also actually x it is also x ok we are
00:16:46.249 --> 00:16:50.379
putting x equal to zero
so when you put x equal to zero it is zero
00:16:50.379 --> 00:16:57.779
and it is one total c two and c two will be
nothing but two upon four plus omega square
00:16:57.779 --> 00:17:04.930
ok so in this way c two can be obtained now
for c one we apply another condition another
00:17:04.930 --> 00:17:14.480
condition is u at l y is zero now again
take fourier cosine transform both the sides
00:17:14.480 --> 00:17:20.991
so fourier cosine transform of u l y will
be nothing but fourier cosine transform
00:17:20.991 --> 00:17:28.240
of zero which is nothing but zero so and this
is also equal to when you substitute x equal
00:17:28.240 --> 00:17:35.559
to l both the sides this is nothing but f
c l omega so when you find f c l omega from
00:17:35.559 --> 00:17:45.670
here f c l omega so that will be nothing but
c one sin hyperbolic omega l plus c two sin
00:17:45.670 --> 00:17:56.549
c two cos hyperbolic omega l which is equal
to zero from here
00:17:56.549 --> 00:18:04.010
now now we know we know c two c one
can be find out so what will be c one c one
00:18:04.010 --> 00:18:16.690
will be nothing but minus c two cos hyperbolic
w l upon sin hyperbolic w l and c two is nothing
00:18:16.690 --> 00:18:24.159
but this term so it is minus two upon four
plus omega square into cos hyperbolic omega
00:18:24.159 --> 00:18:33.169
l upon sin hyperbolic omega l so in this way
once we obtain c one and c two which is here
00:18:33.169 --> 00:18:37.840
then the only thing is you take inverse fourier
cosine transform both the sides to find out
00:18:37.840 --> 00:18:41.559
the final solution ok that is the only thing
remain now
00:18:41.559 --> 00:18:50.950
so what is f c omega t f c omega t here
is nothing but ok f c x omega sorry because
00:18:50.950 --> 00:18:57.950
f c is a functions of x in omega so x and
omega is nothing but c one sin hyperbolic
00:18:57.950 --> 00:19:05.450
omega x plus c two cos hyperbolic omega x
here c one and c two are given by these two
00:19:05.450 --> 00:19:11.000
expressions so the case are inverse cosine
transform both the sides so u x y will be
00:19:11.000 --> 00:19:20.470
nothing but two upon pi integral zero to infinity
integral zero to infinity and that will be
00:19:20.470 --> 00:19:33.680
f c x omega into e k power eta eta omega
y into d omega so that will be two upon pi
00:19:33.680 --> 00:19:39.530
integral zero to infinity and f c is nothing
but c one times c one is this term so this
00:19:39.530 --> 00:19:47.120
is minus two upon four plus omega square cos
hyperbolic omega l upon sin hyperbolic omega
00:19:47.120 --> 00:19:59.720
l c one at this term into sin hyperbolic omega
x hyperbolic omega x ok and plus c two c two
00:19:59.720 --> 00:20:11.690
is two upon two upon four plus omega square
into cos cos hyperbolic omega x and whole
00:20:11.690 --> 00:20:18.470
multiplied by e k power eta omega y in o d
y d omega
00:20:18.470 --> 00:20:26.380
so we can simplify this expression and that
will give the final solution of final solution
00:20:26.380 --> 00:20:38.100
u x you which is a solution of this partial
differential equation ok which is a solution
00:20:38.100 --> 00:20:47.360
of this partial differential equation ok so
hence we can solve such type of problems ok
00:20:47.360 --> 00:20:54.260
so whenever the half range is given to us
you first you first make sure that respect
00:20:54.260 --> 00:20:59.309
to which variable you will have to apply fourier
transform x or y and that will be decided
00:20:59.309 --> 00:21:05.260
by the conditions given to us ok now you apply
either fourier sin or fourier cosine transforms
00:21:05.260 --> 00:21:09.950
and for for finding the arbitrary constants
c one and c two you have the conditions given
00:21:09.950 --> 00:21:15.460
the problem and final take initial sin or
cosine transform to find out the final solution
00:21:15.460 --> 00:21:23.419
of the given partial differential equation
now so this is the this is what i have done
00:21:23.419 --> 00:21:29.330
here also this is when you simplify this when
you simplify this expression so we will get
00:21:29.330 --> 00:21:37.159
this form of answer this one can easily
solve ok so this is the final answer which
00:21:37.159 --> 00:21:48.060
we can obtained after solving this ok ok
it is ok ok sorry it is a one one thing
00:21:48.060 --> 00:21:54.580
i have missed here it is not e k power
eta y it is basically because we are
00:21:54.580 --> 00:22:00.620
taking inverse co inverse cosine transform
it will be nothing but cos omega y and
00:22:00.620 --> 00:22:11.270
it is d y ok d omega so similarly it it
is not it will be cos omega y into d y
00:22:11.270 --> 00:22:16.760
into d omega because we are taking inverse
fourier cosine transform it will be cosine
00:22:16.760 --> 00:22:24.700
term will be here ok so that we have missed
so in this way we will get the final answer
00:22:24.700 --> 00:22:30.990
ok
now the last problem the equation for the
00:22:30.990 --> 00:22:36.289
vibration of the string is given as this and
initially the string is at rest and the initial
00:22:36.289 --> 00:22:41.040
displacement is f x now here x is ranging
from minus infinity to plus infinity so that
00:22:41.040 --> 00:22:48.919
means we have to apply fourier transform ok
so initially string is at rest these are
00:22:48.919 --> 00:22:52.789
the conditions given to us so simply apply
fourier transform both the sides to solve
00:22:52.789 --> 00:22:59.940
this problem so what what is a condition what
is the problem del square u upon del t square
00:22:59.940 --> 00:23:08.370
is equals to c square del square u upon del
x square ok and initially the finite infinity
00:23:08.370 --> 00:23:20.899
string is at rest so that means del by
del t of u at x comma zero is zero and
00:23:20.899 --> 00:23:27.080
initial displacement is f x that means
x comma zero is fx
00:23:27.080 --> 00:23:33.010
so these are the conditions given to you and
x is varying from minus infinity to plus
00:23:33.010 --> 00:23:38.279
infinity you have to solve this equation so
again we will apply fourier transform technique
00:23:38.279 --> 00:23:45.010
so you will take fourier transform both the
sides respect to ok because x is varying from
00:23:45.010 --> 00:23:51.519
minus infinity to plus infinity so that means
it is d square upon d t square of f which
00:23:51.519 --> 00:24:00.519
is equals to c square eta omega square
f here f is nothing but fourier transform
00:24:00.519 --> 00:24:12.820
of u x t which is a function of basically
omega and t ok
00:24:12.820 --> 00:24:20.160
so when you simply d square f upon d t square
and it is minus one so it is plus c square
00:24:20.160 --> 00:24:27.559
omega square f equal to zero so it is d square
plus c square omega square into f equal to
00:24:27.559 --> 00:24:35.779
zero so where d is nothing but d by d t
so when you simply this so f we can obtain
00:24:35.779 --> 00:24:48.690
as c one e k power oh it is cos because
so it is cos c omega t plus c two sin c omega
00:24:48.690 --> 00:24:58.980
t again c one and c two are the functions
of omega ok
00:24:58.980 --> 00:25:11.039
now we will apply this conditions so we will
take the first condition is del u upon
00:25:11.039 --> 00:25:19.620
d t at x comma zero is zero when you take
fourier transform both the sides of this expression
00:25:19.620 --> 00:25:32.049
we will obtain fourier transform of del
u by del t at x comma zero is zero so that
00:25:32.049 --> 00:25:40.590
means d by d t of capital f at x comma
zero is zero a capital if it is f is nothing
00:25:40.590 --> 00:25:48.690
but fourier transform of u x t ok its a
it is fourier transform of u x t so that means
00:25:48.690 --> 00:25:55.029
when you take d by x by d t that is zero at
t equal to zero
00:25:55.029 --> 00:26:00.880
so when you apply this condition over here
so the derivative of this will be sin minus
00:26:00.880 --> 00:26:06.779
sin ok when t equal to zero it will be zero
and it will be cos when t equal to zero that
00:26:06.779 --> 00:26:15.990
will be one so we will left with c two
c omega which is c two c omega which is
00:26:15.990 --> 00:26:26.769
equal to zero ok c two c omega which is equal
to zero so this implies of course this implies
00:26:26.769 --> 00:26:34.279
c two equal to zero because c one c and omega
are constant so c cant be zero and omega cant
00:26:34.279 --> 00:26:37.320
be zero c two is the only thing which can
be zero
00:26:37.320 --> 00:26:47.029
so f is nothing but c one cos c omega t ok
now we have one more condition to find out
00:26:47.029 --> 00:26:58.330
the value of c one that is u x zero equal
to f x ok now now u x zero is f x so take
00:26:58.330 --> 00:27:04.500
fourier transform both the sides so fourier
transform of u x zero will be nothing but
00:27:04.500 --> 00:27:11.419
fourier transform of f x and that will
be nothing but minus infinity to plus infinity
00:27:11.419 --> 00:27:20.309
f x into e k power minus eta omega x into
d omega into d x sorry so that will be nothing
00:27:20.309 --> 00:27:27.789
but f of because you are putting t equal to
zero so that will be o f omega zero
00:27:27.789 --> 00:27:35.429
so when you take f omega zero so that will
be c one and that c one will be nothing but
00:27:35.429 --> 00:27:42.779
this expression so that will be nothing but
minus [infinity] plus infinity f x e k power
00:27:42.779 --> 00:27:55.139
minus eta omega x into d x ok so so now
now we have the value of c one and c two in
00:27:55.139 --> 00:28:00.500
our hand so we will simply take inverse fourier
transform to find out the find out the final
00:28:00.500 --> 00:28:10.090
solution so what is f f is nothing but c one
cos c omega t where c is given by this expression
00:28:10.090 --> 00:28:15.639
take inverse fourier transform both the sides
so it is u x t will be nothing but one upon
00:28:15.639 --> 00:28:28.880
two pi integral minus [infinity] to plus infinity
f omega t e k power eta omega x into d omega
00:28:28.880 --> 00:28:34.220
that will give that will give the final answer
so it is nothing but one upon two pi integral
00:28:34.220 --> 00:28:41.269
minus infinity to plus infinity and f omega
f omega t is c one into cos omega cos c
00:28:41.269 --> 00:28:47.840
omega t and c one is given by this expression
so we can substitute c one also it is f psi
00:28:47.840 --> 00:29:04.929
e k power minus eta omega psi and cos c
omega t e k power eta omega x d psi d omega
00:29:04.929 --> 00:29:10.860
ok so that will be here now when you further
simplify it so it is one upon two pi integral
00:29:10.860 --> 00:29:17.350
minus infinity to plus infinity integral minus
infinity to plus infinity f psi e k power
00:29:17.350 --> 00:29:33.210
minus eta psi minus x into omega cos c omega
t into d psi into d omega
00:29:33.210 --> 00:29:41.690
so again when you write e k power minus
eta k eta as cos theta minus eta sin theta
00:29:41.690 --> 00:29:48.770
then cos theta cos theta will be will contain
omega and which is even in omega it is also
00:29:48.770 --> 00:29:56.940
even in omega even into even remain even however
when you have a sin term ok now this can be
00:29:56.940 --> 00:30:06.480
written as cos omega psi minus x minus eta
sin omega psi minus x and this into this is
00:30:06.480 --> 00:30:12.649
a even function in omega and this into this
is an odd function in omega so this will be
00:30:12.649 --> 00:30:17.809
zero in terms of omega so we will left with
only one term that is one upon pi integral
00:30:17.809 --> 00:30:26.640
minus infinity to plus infinity minus infinity
to plus infinity as psi cos omega psi minus
00:30:26.640 --> 00:30:37.720
x cos c cos c omega t d psi into d omega
so that will be the final solution of the
00:30:37.720 --> 00:30:44.200
given differential equation so in this
way we can probe we can solve several problems
00:30:44.200 --> 00:30:48.980
several several partial differential equation
using fourier series fourier integrals or
00:30:48.980 --> 00:30:55.539
fourier transforms so i hope in this course
you have learnt lot of things about mathematical
00:30:55.539 --> 00:30:59.929
matters lot of techniques to solve ordinary
differential equation partial differential
00:30:59.929 --> 00:31:06.159
equation laplace transforms its applications
fourier series fourier transform and fourier
00:31:06.159 --> 00:31:12.120
integrals their applications that all we
have seen so i hope you might have enjoyed
00:31:12.120 --> 00:31:17.210
the course so i wish you all the best for
your bright career ahead
00:31:17.210 --> 00:31:19.049
thank you very much