WEBVTT
Kind: captions
Language: en
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so we were studying some of the properties
of laplace transforms we have already seen
00:00:26.720 --> 00:00:32.680
that what is shifting property and how
can we solve the questions related to shifting
00:00:32.680 --> 00:00:39.390
property and change of scale property also
we have seen in the last lecture and some
00:00:39.390 --> 00:00:46.640
problems based on change of scale
property also we have seen
00:00:46.640 --> 00:00:54.672
now let us see some more properties of laplace
transforms so this is laplace transform of
00:00:54.672 --> 00:01:02.540
derivatives so if function f t is continuous
for all t greater than and equal to zero and
00:01:02.540 --> 00:01:12.470
is of exponential order say alpha and f dash
t is piecewise continuous on zero to infinity
00:01:12.470 --> 00:01:20.070
then laplace of f dash t exist for p greater
than alpha and laplace of f dash t which is
00:01:20.070 --> 00:01:29.020
a derivative of f respect to t is given by
p f p minus f zero we are f p is nothing but
00:01:29.020 --> 00:01:38.979
laplace of f t so how can we proof this so
laplace of f dash t so what is laplace of
00:01:38.979 --> 00:01:49.950
f dash t it is zero to infinity e k power
minus p t f dash t d t
00:01:49.950 --> 00:01:56.569
now now you can apply integration by
a parts you can use integration by a parts
00:01:56.569 --> 00:02:03.369
to find out the integral of this so first
this second this let us suppose so first
00:02:03.369 --> 00:02:12.560
as it is integral of second which is f t from
zero to infinity minus integral derivative
00:02:12.560 --> 00:02:26.849
of this integral of second d t and it is from
zero to infinity now when t is tending to
00:02:26.849 --> 00:02:37.159
infinity this value tends to zero why because
f t is of exponential order ok since f t is
00:02:37.159 --> 00:02:44.900
an exponential order so there will exists
some k and alpha such that mot of f
00:02:44.900 --> 00:02:54.489
t will be less than or equals to k e k
power n alpha t so mode of e k power minus
00:02:54.489 --> 00:03:01.799
p t into f t will be less than equals to k
into e k power alpha t into e k power minus
00:03:01.799 --> 00:03:06.460
p t
because it is a non negative quantity we can
00:03:06.460 --> 00:03:12.620
multiply both the sides if will not change
the inequality so this is equal to k into
00:03:12.620 --> 00:03:22.879
e k power minus p minus alpha times t and
as as t tending to infinity so since p is
00:03:22.879 --> 00:03:29.189
greater than alpha we are assuming so this
value will tend to zero so hence the limit
00:03:29.189 --> 00:03:40.049
hence this quantity will tends to zero
as t tends to infinity and this is because
00:03:40.049 --> 00:03:47.219
of the property of exponential order so
this will tends to zero so as t tend to infinity
00:03:47.219 --> 00:03:55.209
this will tends to zero and as t tends to
zero this is nothing but minus f zero because
00:03:55.209 --> 00:04:03.450
e k power zero is one so this is minus minus
plus p times zero to infinity e k power minus
00:04:03.450 --> 00:04:16.400
p t f t d t and what is what is the integral
the integral is nothing but laplace of f t
00:04:16.400 --> 00:04:24.790
so it is minus f zero plus p into laplace
of f t and laplace of f t we are calling as
00:04:24.790 --> 00:04:32.750
f p though it is minus f zero plus p into
f p
00:04:32.750 --> 00:04:39.640
so this is how we can obtain this result that
laplace of f dash d t will be nothing but
00:04:39.640 --> 00:04:48.760
p f p minus f zero now on the same lines if
f t and f dash t continuous and are of exponential
00:04:48.760 --> 00:04:55.540
order and f double dash t is piecewise continuous
then laplace of f double dash t is given by
00:04:55.540 --> 00:05:02.790
this expression so this also we can prove
it is very easy to prove this once we obtain
00:05:02.790 --> 00:05:16.821
for f dash t now what is laplace of f double
dash t so the first expression which we have
00:05:16.821 --> 00:05:24.240
just obtained you simply replace f by f dash
in that expression you simply replace f by
00:05:24.240 --> 00:05:37.970
f dash so what we will obtained this is
p into laplace of f dash t minus f dash zero
00:05:37.970 --> 00:05:44.730
because you see that this result the first
result holds for every p i mean every f
00:05:44.730 --> 00:05:52.020
so this will hold for f dash also so we replace
f by f dash in the first expression so we
00:05:52.020 --> 00:05:58.020
will we get this result this expression now
this is equals to p into what is laplace of
00:05:58.020 --> 00:06:07.580
f dash t now you again apply the result
it is this minus f zero minus f dash zero
00:06:07.580 --> 00:06:18.030
so this is nothing but p square f p minus
p f zero minus f dash zero so hence we obtain
00:06:18.030 --> 00:06:23.220
the next result of laplace of f double
dash t
00:06:23.220 --> 00:06:34.680
now we can generalize this result for n th
order derivative for laplace of f of i mean
00:06:34.680 --> 00:06:41.760
n th or derivative of f respect to t will
be given by this expression this can be
00:06:41.760 --> 00:06:52.380
proved very easily because in order to prove
this we can use mathematical induction
00:06:52.380 --> 00:06:59.900
when n equal to one that is laplace of f dash
t we have just proved that laplace of f dash
00:06:59.900 --> 00:07:05.270
t is p f p minus f zero when you substitute
n equal to one in this expression so we will
00:07:05.270 --> 00:07:13.220
get by the same expression that is p f p minus
f dash zero p f p minus f zero sorry ok
00:07:13.220 --> 00:07:29.740
now let us assume it to be true for n equal
to k so assume it to be true for n equal to
00:07:29.740 --> 00:07:38.650
k so what we have assumed we have assumed
that laplace of f k th derivative is k th
00:07:38.650 --> 00:07:43.870
derivative with respect to t is nothing but
you can simply replace n by k
00:07:43.870 --> 00:08:04.780
so it is p k power k f p minus p k minus one
f f zero minus p k minus two f dash zero and
00:08:04.780 --> 00:08:16.020
so on minus k minus one derivative of at t
equal to zero now we have to show we have
00:08:16.020 --> 00:08:23.000
to show the same result for n equal to k plus
one this is mathematical induction we first
00:08:23.000 --> 00:08:30.400
prove it for n equal to one assume it to be
true for n equal to k and try to prove the
00:08:30.400 --> 00:08:36.190
result for n equal to k plus one if you obtain
the result for k equal to n equal to k
00:08:36.190 --> 00:08:43.010
plus one this means the result hold for
all n this is the concept of mathematical
00:08:43.010 --> 00:08:58.020
induction now for n equal to k plus one
laplace of f k plus one t is nothing but
00:08:58.020 --> 00:09:09.320
it is zero to infinity it is zero to infinity
e k power minus p t k plus one th derivative
00:09:09.320 --> 00:09:15.070
respect to t into d t
again by part first second so it is nothing
00:09:15.070 --> 00:09:25.240
but first as it is integral of second zero
to infinity minus integral derivative of first
00:09:25.240 --> 00:09:34.840
it is minus p e k power minus p t integral
of second it is k th derivative t d t so again
00:09:34.840 --> 00:09:45.080
since all the derivatives up to n minus
one that is up to k here so are of exponential
00:09:45.080 --> 00:09:52.120
order some exponential order so this will
tends to zero as t tends to infinity this
00:09:52.120 --> 00:09:59.950
can be proved in the same lines and when t
equal to zero it is nothing but minus k th
00:09:59.950 --> 00:10:06.740
derivative at t equal to zero minus minus
plus p and it is nothing but laplace of k
00:10:06.740 --> 00:10:17.480
th derivative of t this so this will be the
expression now the laplace of this we have
00:10:17.480 --> 00:10:24.601
already assumed this equal to this expression
now this is minus f k th derivative at t equal
00:10:24.601 --> 00:10:31.860
to zero plus p into laplace of k th derivative
of t we have assumed equal to this so it is
00:10:31.860 --> 00:10:46.090
p f p k f p minus p k minus one f zero minus
p k minus two f dash zero and so on minus
00:10:46.090 --> 00:10:54.130
k minus one derivative at t equal to zero
so this is nothing but p k plus one from this
00:10:54.130 --> 00:11:15.570
expression f p minus p k f zero and so on
minus f k th derivative at zero and this is
00:11:15.570 --> 00:11:23.720
at infinity it is zero and at zero it is
minus k and when you multiply by p it is
00:11:23.720 --> 00:11:34.810
p k plus one into f k it is k so p k ok p
times yeah it is p times sorry so p times
00:11:34.810 --> 00:11:44.490
f k minus one and minus f k derivative at
zero so what we have obtained that if we
00:11:44.490 --> 00:11:52.270
replace in this expression and by k
plus one so the result holds this we have
00:11:52.270 --> 00:12:00.260
shown that if we replace in this expression
and by k plus one so we hold this the result
00:12:00.260 --> 00:12:07.280
hold hence the results holds for all n so
hence the hence proved this result
00:12:07.280 --> 00:12:18.320
now the next is how can we solve some problems
using derivatives using this concept so
00:12:18.320 --> 00:12:24.550
just for illustration let us try to find out
laplace of this simple problems using derivatives
00:12:24.550 --> 00:12:33.130
laplace of derivatives so we already know
that laplace of f dash t is nothing but p
00:12:33.130 --> 00:12:43.880
f p minus f zero so what is the f p f p here
f p is nothing but laplace of f t so we
00:12:43.880 --> 00:13:00.070
have to find laplace of laplace of t e
k power two t so so here f t is t e k power
00:13:00.070 --> 00:13:10.930
two t so we if we apply this result though
laplace of f dash t so first we find f dash
00:13:10.930 --> 00:13:20.410
t what is derivative of f respect to t it
is first as it is derivatives of second plus
00:13:20.410 --> 00:13:28.589
second as it is derivative of first
so laplace of use this result laplace of f
00:13:28.589 --> 00:13:37.190
dash t that is two t e k power two t plus
e k power two t is nothing but p f p p f p
00:13:37.190 --> 00:13:46.000
is laplace of f t e t e k power two t f p
is nothing but laplace of f t minus f zero
00:13:46.000 --> 00:13:53.870
so what is f zero it is zero so it is nothing
but two times laplace of t e k power two t
00:13:53.870 --> 00:14:06.210
plus laplace of e k power two t equal to p
laplace of p e k power two t so it is nothing
00:14:06.210 --> 00:14:18.690
but this implies two minus p laplace of t
e k power two t is equals to minus of laplace
00:14:18.690 --> 00:14:29.899
of e k power two t which is nothing but minus
of one upon p minus two so laplace of p e
00:14:29.899 --> 00:14:37.399
k power two t will be nothing but this will
go in the denominator and negative if we adjust
00:14:37.399 --> 00:14:43.589
though this is nothing but one upon p minus
two square ok so this will be the laplace
00:14:43.589 --> 00:14:49.970
of this expression
so this we can directly find of course because
00:14:49.970 --> 00:14:55.960
we know the laplace of t that is one upon
p square and using shifting property if we
00:14:55.960 --> 00:15:01.750
replace p by p minus two though we we get
back of this expression but it is just
00:15:01.750 --> 00:15:07.440
for illustration that how can we solve some
problems using laplace of derivatives also
00:15:07.440 --> 00:15:12.880
so similarly we can solved the second problem
also it is also based on laplace transform
00:15:12.880 --> 00:15:18.940
of derivatives that also we can find out otherwise
also we can find out the laplace on sin square
00:15:18.940 --> 00:15:25.690
by writing sin square as one minus cos two
t whole divided by two and then use linearity
00:15:25.690 --> 00:15:30.760
property of laplace transform we can find
out laplace transform of sin square t
00:15:30.760 --> 00:15:37.620
now let us try to solve few problems again
of laplace inverse based on laplace
00:15:37.620 --> 00:15:43.830
transform derivatives so we get back the
same result again now this is laplace transform
00:15:43.830 --> 00:15:52.440
of derivatives so what is what does it imply
this implies f dash t is nothing but laplace
00:15:52.440 --> 00:16:04.920
inverse of p f p minus f zero and here f p
is nothing but f p is nothing but laplace
00:16:04.920 --> 00:16:16.240
of f t so this expression we can derive from
laplace transform of derivatives now it
00:16:16.240 --> 00:16:25.730
is given towards at laplace inverse of laplace
inverse of p upon p square plus one whole
00:16:25.730 --> 00:16:33.520
square is equal to one by two t sin t it is
given to us in the problem
00:16:33.520 --> 00:16:50.220
so suppose it is f t say so if it is f t so
what is f dash t one by two t cos t plus
00:16:50.220 --> 00:17:07.130
sin t so if i use this expression so f dash
t is one by two t cos t plus sin t will be
00:17:07.130 --> 00:17:16.370
equal to laplace inverse of p into f p f p
is laplace transform of f t and that is nothing
00:17:16.370 --> 00:17:27.199
but that is nothing but p into p upon p square
plus on whole square may if laplace inverse
00:17:27.199 --> 00:17:33.649
of this expression is f t so laplace of f
t will be p upon p square plus on whole square
00:17:33.649 --> 00:17:43.730
and minus f zero f zero from here is nothing
but zero so so hence we get back to laplace
00:17:43.730 --> 00:17:48.940
inverse of p square upon p square plus
one whole square which is nothing but one
00:17:48.940 --> 00:17:55.929
by two t cos t plus sin t so in this way we
have solved first part of this problem
00:17:55.929 --> 00:18:03.649
now the second part can be solved now what
is laplace inverse of p square upon p square
00:18:03.649 --> 00:18:11.779
plus one whole square we have just obtained
that it is equals to one by two t cos t plus
00:18:11.779 --> 00:18:22.559
sin t it is it is this expression we have
just obtained half whole expression now
00:18:22.559 --> 00:18:28.480
this we have obtained in the first part in
order to obtain the second part laplace
00:18:28.480 --> 00:18:38.769
inverse of p square upon p square plus one
whole square can be rewritten as laplace inverse
00:18:38.769 --> 00:18:45.500
of p square plus one minus one upon p square
plus on whole square where does subtract to
00:18:45.500 --> 00:18:53.759
one in the numerator this is nothing but laplace
inverse of one upon p square plus one minus
00:18:53.759 --> 00:19:03.299
one upon p square plus one whole square now
this is nothing but laplace inverse of one
00:19:03.299 --> 00:19:12.429
upon p square plus one minus laplace inverse
of one upon p square plus one whole square
00:19:12.429 --> 00:19:19.139
and this is nothing but sin t minus laplace
inverse of one upon p square plus one on whole
00:19:19.139 --> 00:19:24.830
square but laplace inverse of this is equal
to this
00:19:24.830 --> 00:19:30.940
this we have obtained in the first part of
this problem so this is equal to one by two
00:19:30.940 --> 00:19:40.919
t cos t plus sin t so what should be the laplace
inverse of one upon p square plus one whole
00:19:40.919 --> 00:19:47.710
square so this implies laplace inverse of
one upon p square plus one whole square is
00:19:47.710 --> 00:19:52.289
nothing but you can put this in the left and
right hand side and this here this will be
00:19:52.289 --> 00:20:01.830
nothing but half of sin t minus one by
t two cos t so that is nothing but one by
00:20:01.830 --> 00:20:11.799
two sin t minus t cos t so this will be the
laplace inverse of one upon p square plus
00:20:11.799 --> 00:20:18.640
one whole square
so so hence we can say that the laplace
00:20:18.640 --> 00:20:25.999
transform derivatives can also be used to
find out the laplace inverse of some problems
00:20:25.999 --> 00:20:32.230
now the next property if laplace transform
f t is f p then laplace transform of f t by
00:20:32.230 --> 00:20:40.539
t is nothing but integral p to infinity f
u d u where this is f u is nothing but again
00:20:40.539 --> 00:20:47.429
laplace transform of f t so provided this
limit exist because function must be piecewise
00:20:47.429 --> 00:20:58.470
continuous for existence of laplace transform
so f t by t will be will be piecewise continuous
00:20:58.470 --> 00:21:05.139
if laplace if limit t tend in to zero plus
f t by t exist so that must that is required
00:21:05.139 --> 00:21:13.529
in this problem so what is how to find this
so take this side integral p to infinity f
00:21:13.529 --> 00:21:19.940
u d u the take the right hand side and we
will obtain the left hand side so what is
00:21:19.940 --> 00:21:26.580
the what is f u what is capital f u it is
nothing but laplace transform of f t so it
00:21:26.580 --> 00:21:40.220
is p to infinity e k power minus u t instead
of p we have u f t d t of course integral
00:21:40.220 --> 00:21:48.570
from zero to infinity it is d t and whole
multiplied by d u so this we are having in
00:21:48.570 --> 00:21:53.799
this problem
now we will this is the double integral so
00:21:53.799 --> 00:22:00.370
now we will make use of change of order of
integration how can we proceed for changing
00:22:00.370 --> 00:22:08.869
th order of integration suppose this is line
of t this is u now t is varying from zero
00:22:08.869 --> 00:22:14.570
to infinity this is the limit for t t is varying
from zero to infinity so this is first and
00:22:14.570 --> 00:22:23.700
forth (( )) ok t is varying from zero to
infinity ok and u is varying from p to infinity
00:22:23.700 --> 00:22:32.489
so suppose this is u equal to p line and
u is varying from p to infinity that is from
00:22:32.489 --> 00:22:39.700
this to this side so this is the region this
is the required region this is the required
00:22:39.700 --> 00:22:50.250
region of integration ok now we have to change
the order of integration that means it is
00:22:50.250 --> 00:22:58.710
e k power minus u t f t and d u and then d
t so first you want the limits for u and then
00:22:58.710 --> 00:23:04.529
you want the limit for u t so we have to make
a strip parallel to u axis because you want
00:23:04.529 --> 00:23:07.899
d u first so we make a strip parallel to u
axis
00:23:07.899 --> 00:23:16.080
so u axis this way the strip here now here
u is p and goes up to infinity so u is varying
00:23:16.080 --> 00:23:22.139
from p to infinity and t is varying from zero
to infinity so e is varying from zero to infinity
00:23:22.139 --> 00:23:30.769
so that is how we can change this order of
integration ok so now this is equals to t
00:23:30.769 --> 00:23:38.830
equal to zero to infinity now we have the
u term only in this expression so we can integrate
00:23:38.830 --> 00:23:45.020
this respect to u so it is nothing but e k
power minus u t upon minus t because their
00:23:45.020 --> 00:23:58.389
integration respect to u ok so this is this
p to infinity f t d t and when u tending
00:23:58.389 --> 00:24:06.360
to infinity it is tending to zero and when
u tending to p it is nothing but zero to infinity
00:24:06.360 --> 00:24:23.019
e k power minus p t upon t f t d t because
when t tend to infinity tend into zero i i
00:24:23.019 --> 00:24:30.679
mean sorry it is limit for u so u tend to
infinity tend to zero but when u is p so it
00:24:30.679 --> 00:24:38.059
is e k power minus p t upon t so it is nothing
but laplace transform of f t by t so it is
00:24:38.059 --> 00:24:46.690
nothing but laplace transform of f t by t
so hence we obtain this result that laplace
00:24:46.690 --> 00:24:56.129
transform f t by t is nothing but p to
infinity f u d u now how this result can
00:24:56.129 --> 00:25:03.480
be use to find out laplace transform of some
function of laplace inverse let us see by
00:25:03.480 --> 00:25:09.529
solving few problems based on this so let
us find out laplace transform of sin hyperbolic
00:25:09.529 --> 00:25:16.100
t upon t very simple problem so we know this
is result ok laplace transform of f t by t
00:25:16.100 --> 00:25:32.799
is something p to infinity f u d u we are
we are laplace transform of f t is f p so
00:25:32.799 --> 00:25:39.279
we know this result so using this result we
will try to solve this problems so laplace
00:25:39.279 --> 00:25:49.390
transform of sin hyperbolic upon t so first
first we have to find out laplace transform
00:25:49.390 --> 00:25:56.059
of numerator quantity here numerator is here
f t is sin hyperbolic t so what a laplace
00:25:56.059 --> 00:26:04.950
transform of sin hyperbolic t this you already
know this is nothing but one upon t square
00:26:04.950 --> 00:26:15.539
minus one sin hyperbolic t a upon a square
a upon p square minus a square and division
00:26:15.539 --> 00:26:21.440
by t is here for division by t we have integral
of p to infinity f u d u it is integral p
00:26:21.440 --> 00:26:29.091
to infinity
so here this is f p so what will be f u one
00:26:29.091 --> 00:26:41.820
by u square minus one into d u so now we can
integrate this using partial fraction
00:26:41.820 --> 00:26:50.669
or so it is one one u minus one into u plus
one d u the difference of these two difference
00:26:50.669 --> 00:26:57.330
of this and this and this two which is a quotient
quantity though you we can multiply a two
00:26:57.330 --> 00:27:04.289
and divide by two in the numerator so one
by two p to infinity and two can written as
00:27:04.289 --> 00:27:16.460
u plus one minus u minus one
so this is nothing but one by two integral
00:27:16.460 --> 00:27:30.369
p to infinity one upon u minus one minus one
upon u plus one d u
00:27:30.369 --> 00:27:37.359
this is the partial fraction of this so this
is equals to one by two now integral of this
00:27:37.359 --> 00:27:47.789
is l n u minus one and integral of this is
l n u plus one from p to infinity so this
00:27:47.789 --> 00:27:58.769
is one by two l n u minus one upon u plus
one from p to infinity at infinity it is tending
00:27:58.769 --> 00:28:09.139
to one log one is zero and at p it is tending
to l n p minus one upon p plus one so because
00:28:09.139 --> 00:28:16.739
of negative sin integral will change
this log u plus one upon u minus p plus one
00:28:16.739 --> 00:28:26.480
upon p minus one will come because we have
upper limit minus lower limit so because of
00:28:26.480 --> 00:28:32.669
that negative these two will interchange
so it is p plus one upon p minus one so this
00:28:32.669 --> 00:28:40.659
will be the laplace transform of the first
problem the second problem can be solved
00:28:40.659 --> 00:28:47.179
on the same lines first you find laplace transform
of the numerator you see numerator what
00:28:47.179 --> 00:28:53.379
is f t here in the numerator it is e k power
minus two t sin three t what a laplace transform
00:28:53.379 --> 00:29:01.860
of sin t sin three t it is three upon p square
plus nine and for e k power minus two t you
00:29:01.860 --> 00:29:12.210
simply replace p by p plus two that means
the laplace transform of e k power minus
00:29:12.210 --> 00:29:23.190
two t sin three t will be nothing but three
upon p plus two whole square plus nine so
00:29:23.190 --> 00:29:28.879
this will be f f p here this will be f
p for this problem for the second problem
00:29:28.879 --> 00:29:36.519
and we have a division by t now for f t by
t we will apply the we will apply this
00:29:36.519 --> 00:29:43.450
this property so this will be nothing but
p to infinity f u d u so we have to integrate
00:29:43.450 --> 00:29:52.309
this up from p to infinity so that will give
the laplace transform of the second problem
00:29:52.309 --> 00:30:00.779
now let us try to prove this simple
problem so what is so from here we can
00:30:00.779 --> 00:30:09.700
see that laplace transform of f t is nothing
but zero to infinity f t by t e k power minus
00:30:09.700 --> 00:30:21.299
p t d t will be equals to p to infinity f
u d u this is from this expression now you
00:30:21.299 --> 00:30:29.440
take the limit p tend into zero plus both
the sides you take the limit p tend to zero
00:30:29.440 --> 00:30:37.109
plus both the side so what will obtained it
is zero to infinity f t by t d t because e
00:30:37.109 --> 00:30:47.200
k power zero is one and is equals to zero
to infinity f u d u so hence we got the first
00:30:47.200 --> 00:30:56.369
part of this problem now to prove that
zero to infinity sin t by t is integral of
00:30:56.369 --> 00:31:04.419
sin t by t is pi by two so what is f t for
for if we compare with this f t sin t so what
00:31:04.419 --> 00:31:14.490
is what is this problem now it is zero
to infinity sin t by t d t so it is equals
00:31:14.490 --> 00:31:21.230
to integral p to infinity
now f u is nothing but laplace of f t and
00:31:21.230 --> 00:31:30.749
laplace of sin t is nothing but one upon p
one upon u square plus one d u ok so it is
00:31:30.749 --> 00:31:40.799
tan invasive from p to infinity and ok from
zero to infinity sorry because we have zero
00:31:40.799 --> 00:31:48.529
to infinity in this result for zero to infinity
so at infinity is pi by two at zero is zero
00:31:48.529 --> 00:31:56.849
so we have this value that integral of zero
to infinity sin t by t nothing but pi by two
00:31:56.849 --> 00:32:01.649
so some integrals having limit lies to zero
to infinity or something can also be solved
00:32:01.649 --> 00:32:12.729
using laplace transforms ok
now now we can also solve these problems
00:32:12.729 --> 00:32:21.629
on based on this properties so i am just
illustrating on the how to solve the first
00:32:21.629 --> 00:32:30.369
part or the second part so it is in this
problem one minus cos t upon t square we have
00:32:30.369 --> 00:32:38.610
to integrate two times what what this integral
is basically it is one minus cos t first
00:32:38.610 --> 00:32:47.840
you will find laplace of this which is
nothing but integral p to infinity laplace
00:32:47.840 --> 00:33:03.109
of one minus cos t d u if if we are taking
as f f u ok now laplace of so this we can
00:33:03.109 --> 00:33:14.759
easily find this p to infinity one by
u minus it is u on u square plus one d
00:33:14.759 --> 00:33:27.849
u so it is it is l n u minus one by two
l n u square plus one from p to infinity
00:33:27.849 --> 00:33:37.159
so it is nothing but l n u upon under root
u square plus one from p to infinity at infinity
00:33:37.159 --> 00:33:45.940
is tend to zero at p it is sending to negative
of l n p upon under root p square plus so
00:33:45.940 --> 00:33:52.149
this is this is laplace transform of this
quantity for laplace transform of one minus
00:33:52.149 --> 00:33:59.610
cos t upon t square we again have to apply
the same property ok so what is laplace transform
00:33:59.610 --> 00:34:09.200
of one minus cos t upon t square it is nothing
but p to infinity laplace transform of one
00:34:09.200 --> 00:34:23.360
minus cos t upon t and d u ok because the
this property holds for f t by t so if you
00:34:23.360 --> 00:34:28.839
compare with this so what is the f t one minus
cos t by t so it is laplace transform of one
00:34:28.839 --> 00:34:36.809
minus cos t by t into d u and that we already
calculated so it is nothing but p to infinity
00:34:36.809 --> 00:34:44.159
so here this is the f so this is new f p u
suppose this is g p so it is g u so it is
00:34:44.159 --> 00:34:51.760
l n under root u square plus one upon u into
d u
00:34:51.760 --> 00:34:58.910
so this can easily be integrated if you take
u as suppose ten theta you substitute here
00:34:58.910 --> 00:35:05.660
u equal to ten theta proceed your integration
so this integral can the value of the integral
00:35:05.660 --> 00:35:13.539
can be find ok so how to find laplace inverse
of p upon p square minus nine whole square
00:35:13.539 --> 00:35:23.660
using this property so so from this we
got f t by t is nothing but laplace inverse
00:35:23.660 --> 00:35:42.010
of p to infinity f u d u here here what is
f u it is laplace of f t is a so so this f
00:35:42.010 --> 00:35:49.000
t is nothing but laplace inverse of this so
if you obtain f t that means we obtain
00:35:49.000 --> 00:35:55.230
laplace inverse of this expression
so take this as f p so laplace inverse of
00:35:55.230 --> 00:36:04.830
p to infinity it is u upon u square minus
nine whole square d u so this is because i
00:36:04.830 --> 00:36:12.200
am taking the expression given inside
the bracket as f p and laplace inverse of
00:36:12.200 --> 00:36:19.869
this f p to find out which is nothing but
this f t because we know this ok so now we
00:36:19.869 --> 00:36:31.049
divided multiply by two
00:36:31.049 --> 00:36:37.140
and one by two it is laplace inverse of
the [deri/derivatives] derivatives of denominator
00:36:37.140 --> 00:36:43.339
u square minus two is in the numerator so
it is a minus one upon u square minus nine
00:36:43.339 --> 00:36:51.359
from p to infinity and it is one by two [lap/laplace]
laplace inverse of at infinity it is tending
00:36:51.359 --> 00:36:58.089
to zero and at p it is nothing but one by
p square minus nine
00:36:58.089 --> 00:37:04.460
now you divide by a divide multiply by three
so it is one by six laplace inverse of three
00:37:04.460 --> 00:37:15.670
upon p square minus nine and this is nothing
but one by six it is hyperbolic three t so
00:37:15.670 --> 00:37:26.520
so what will be f t f t will be f t will be
t by six sin hyperbolic three t so and f t
00:37:26.520 --> 00:37:32.090
is nothing but laplace inverse of f p that
is this quantity so laplace inverse of this
00:37:32.090 --> 00:37:38.400
quantity is nothing but this expression so
hence we can find laplace inverse also
00:37:38.400 --> 00:37:41.849
using this property so so
thank you