yeah hello everyone so in the last lecture we were studying about properties of laplace
transform some of the properties we have seen like shifting property first shifting or
first translation or shifting property then we see second translation or shifting property
some problems based on this that we have seen in the last lecture so now we will see some
more prosperities of laplace transforms in this lecture so the next property is change
of scale property now what does states it states that if laplace transform of some f
t is f t then laplace transform of f a t is nothing but one by a f p by a of course a
should not equal to zero so this is called change of scale property
now how we will how we it we will prove it so the proof is very simple see simply
based on the definition of laplace transforms so what is laplace transform of f a t it is
nothing but zero to infinity e key power minus p t f a t now you can substitute a t equal
to some new variable z so a d t will be equals to d z so where you substitute this thing
over here so when t is zero z is zero so it will be zero when t tend into infinity z will
tend to infinity so it is infinity e key power now t is nothing but z by a so it is minus
p into z upon a into f of z and d t is nothing but d z upon a so that is one upon a times
d z so this quantity is nothing but one by a zero to infinity e key power minus p z upon
a f z into d z or we can write it like this one upon a integer zero to infinity e key
power minus some zeta times p into f z d z where zeta is nothing but p upon a
now when you compare this with laplace transform of f t so laplace transform of f t is is given
by a zero to infinity e key power minus p t f t d t and this we are calling as f p this
we are calling as f p because p is a parameter two is a function of f p now here the same
thing is same thing what we have in with this expression so this can also be written
as this expression can also be written as zero to infinity one upon a times e key power
minus zeta t p into f t d t so that is nothing but one upon a f zeta because instead
of ok it is it is z because variable is z so it is z so now z is here so instead of
it is t now t is here so instead of p we have zhy here so it is will be f zhy so what
is zhy it is one by a f of p upon a so that is a proof of change of scale property
that is laplace of f a t is nothing but one by a f of p by a
now we will see a [pro/some] some problems based on this suppose it is given to us
that laplace transform of f t is nothing but one by p p square plus one and we have to
find laplace transform of e key power minus two t into f of three t now there are two
ways to solve this problem the one way is you find the laplace inverse of this f p this
is f p you find the laplace inverse of this f p that will be nothing but f t find f of
three t and then f of three t will be find and then using shifting property you can find
laplace transform of e key power minus three t f three t and second way is you apply change
of scale property and then the shifting property to directly find out the laplace transform
of this expression now how you can find a laplace inverse of
this again using partial fractions this is nothing but some a upon p plus b p plus c
upon p square plus one so one is equals to a into p square plus one plus b p plus c into
p so when you take p equal to zero this implies one is equals to a so a is nothing but
one now you compare the coefficient of p square here p square is zero here p square is a plus
b so this implies b is minus a and hence is equals to minus one because a is one now for
c you can compare the coefficient of p here p coefficient is only this which is c and
here there is no coefficient involving p so zero equal to c so c must be zero so what
are the partial fraction of this expression one by p into p square plus one the partial
fraction will be nothing but a is one so it is one by p plus b is minus one and c is zero
so it is minus p upon p square plus one so this ultra partial fraction of this f p so
what will be laplace inverse of this so by the linearity property of laplace inverse
this is nothing but laplace inverse of one by p minus laplace inverse of p upon p square
plus one and this is nothing but one minus cos t so this is f t laplace inverse of f
p is nothing but f t now we have to find out first f three t so what will be f three t
it is nothing but one minus cos three t and laplace of e key power minus two t into f
three t will be nothing but first we will find out laplace transform of f three t
that will be new f p and in that new f p we will apply shifting property
so what is laplace transform of f three t which is nothing but one by p minus three
upon p minus p upon sorry p upon p square plus nine where cos a t is nothing but laplace
of cos a t is p upon p square plus a square so laplace of so this is this we are taking
as suppose g p function of p suppose it is g p so laplace of f three t is g p then laplace
of this expression will be nothing but g you replace p by a p minus a here a is minus
two so it is nothing but p plus two so that is nothing but one upon p plus two minus p
plus two upon p plus two whole square plus nine so if you take l c m and simplify it
is clearly [vis/visible] visible that it is nothing but p plus two into p plus two
whole square plus nine so that will be the laplace of e key power minus three t into
f three t e key power minus two t into f three t
so this is the one way now finding laplace inverse of given f p this f p is very
simple so we have taken the partial fraction and simply computed first computed laplace
inverse of this f p let it suppose it is f t and then using shifting property we have
find the laplace transform of e key power minus two t into f three t now if you have
complicated expression so finding laplace inverse is a difficult job i mean is
a time consuming job so for this particular type of problems we can solve directly also
using first change of scale property and then using shifting property how now let us
see so this this is suppose f p ok this is suppose now what is laplace of f three t it
is nothing but one by a f a here is three so one by three f of p by three so it is nothing
but one by three one upon p by three into p square by nine plus one so that is nothing
but three three cancels out ok this is p by three and this is one by three ok this is
one by three so this is one by three into so this is
three into nine upon p into p square plus nine so this three three cancels out now
what is laplace of e key power minus two t f three t so this this suppose this is some
g p function of p suppose this is g p so laplace of e key power a t or in some f t will be
nothing but by a shifting property it is f of p minus a here laplace of this function
is g p so we have to apply shifting property for this function so this will be nothing
but by the shifting property it is p plus two because here a is minus two so it is nothing
but nine upon p plus two into p plus two whole square plus nine so that is how using change
of scale property and shifting property we can find the laplace transform of this particular
problem very easily ok now let us find out laplace transforms of some special functions
special functions like bessel function so i am not going into much detail of special
functions simply assume that j naught x is given by this series and error function
all is given by this particular expression so how to find laplace transform of j naught
t which we call as bassel function and how to find laplace transform of error function
so let us try to find out this thing and then once we find laplace transform of bessels
function which is j naught t or error function we can solve those two problems using shifting
property and change of scale property so let us find first j naught t so j naught t is
nothing but one minus t square by two square plus t key power four upon two square into
four square minus t key power six upon two square four square into six square and so
on as given in the problem ok so what is laplace transform of j naught t
now laplace transform one is one by p laplace transform of t square is factorial two upon
p square into two square laplace transform p q it is because laplace transform of t key
power n is what is nothing but factorial n upon p key power n plus one if n is one two
three and so on for integer for positive integers so again it is factorial four upon p key power
five two square into four square minus factorial six upon p key power seven two square four
square six square and so on so it is nothing but one by p take one by p common so it is
one minus one two two cancels out so it is one by two p square then it is nothing but
four into three into two into one upon p key power four into two square into four square
minus it is nothing but six into five into four into three into two into one upon p key
power seven p key power six because one p is common
so and it is two square four square and six square and so on now one two two cancel out
one four four cancel out so we have observed that numerator always contain odd terms and
denominator always contain even terms that that is it is nothing but one by p minus one
minus one by two p square plus one into three upon two into four into one by p square ka
whole square it is nothing but minus one into three into five upon two into four into six
into one by p square ka whole cube and so on
ok so this is nothing but one by p one by p into meh this is one by p into one by p
into now this is one minus this is nothing but you can say plus minus one by two minus
one by two into p one by p square now this is nothing but minus one by two into two minus
one by two minus one upon factorial two into one by p square when you simplify this expression
so what we get it is minus one into minus three when you taken l c m so one into three
minus minus plus and two cube so two cube in the denominator
ok next is it is plus minus one by two minus one by two minus one into minus one
by two minus two upon factorial three into one by p square ka whole cube so again it
is nothing but when you simplify so you will get back the same expression it is one into
three into five upon two cube into factorial three which is nothing but two into four into
six so what this expression basically this is nothing but one minus one by p sq[uare]
one plus p square key power minus half this is this is by the binomial theorem one plus
m x plus m minus upon factorial to into x square and so on so now this is nothing but
when you simplify it this is one by p into p square plus one upon p square whole power
minus half which is nothing but one upon p square plus one ka under root so this is the
laplace transform of j naught t which we are calling as bessel function ok so you simply
open the series this is the series is given to you you take the laplace transform both
the side simplify so this will be reduce to some binomial expansion
of one plus one by p square whole key power minus half and when you simplify we will get
one upon under root p square plus one at the laplace of laplace transform of j naught
t now now we can solve first problem since now we know laplace transform of j naught
t so what will be the laplace transform of j naught two t first so this is using
change of scale property because if laplace transform f t is some f p then laplace transform
f a t is nothing but is nothing but one by a f p by a so here here here this is f
p and this is this is laplace transform of j naught t so what will be laplace transform
of this here it will be one by two f p by two so this nothing but one by two f of one
by under root p by two ka whole square plus one which is nothing but two upon two under
root p square plus one p square plus four so it is nothing but two two cancel out
so it is one by under root p square plus four so what should be laplace transform of e key
power minus three t of this j naught two t this will be equal to now let us suppose this
is g p the laplace transform of j naught two t there a suppose it is g p so laplace transform
of e key power minus three t into lap[lace] into this is nothing but by using shifting
property it is g of p plus three because in the laplace transform this j naught two t
you replace p by p plus three so laplace transform of this is g p so you replace p by p minus
a and a is minus three so it is nothing but one upon under root p plus three whole square
plus four so this will be the final answer you can simplify and that should be the laplace
transform of e key power minus three t j naught two t now let us find out laplace transform
of error function under root t so and then using change of scale property
we can solve that problem so what is laplace transform of error function let us see so
error function what is error function of under root t how you define it it is nothing but
zero to under root t two upon pi times zero to under root t e u square e key power minus
u square d u so now it is equal to two upon under root pi zero to under root t now
we know the expression of e key power minus x we can use this expression here it is one
minus u square upon factorial one mi[nus] plus u square ka whole square upon factorial
two minus u square ka whole square upon factorial three and so on into d u so this is nothing
but two upon under root pi so integral of one will be u and when you apply a upper limit
upper limit is sorry it is under root t ok so when you apply upper limit minus lower
limit so this will be nothing but under root t minus
now integration of u square is u cube upon three so u cube and when you apply upper limit
minus lower limit is nothing but t key power three by two upon three into factorial one
plus u key power four ka integral is u key power five upon five and when you apply the
limit is t key power five by two upon five factorial two similarly here it will be
u key power six that is t key power seven by two upon seven factorial three and so on
so this would be the expression of error function in the series from now when you take
laplace transform of e key error function of under root t so this will be nothing but
two upon under root pi you take common and this is nothing but gamma three by two upon
three by two upon p key power three by two minus it is gamma five by two upon three into
factorial one into p key power five by two ok plus gamma seven by two upon five into
two factorial into p key power seven by two and so on this is nothing
but again two upon under root pi take gamma three by two common and p key power three
by two common from the entire expression from the entire series now gamma three by two is
nothing but one by two under root pi this we know as gamma and plus one is n gamma n
so this is one by two under root pi this i have taken common and also i have taken p
key power three by two as common this nothing but one minus now gamma five by two is
three by two gamma three by two gamma three by two already common so it is three by two
into three into factorial one and p key power three by two is common so it is p ok plus
now gamma seven by two is five by two gamma five by two and gamma five by two can be [ag/again]
again written as three by two gamma three by two
so it is nothing but five by two into three by two gamma three by two which is common
and upon five into factorial two p key power two and so on now again let us simplify this
expression two two cancels out this is cancels out one upon p key power three by two and
it is one minus this three three cancel out so it is one minus one by two p plus this
five five cancels out it is nothing but one into three upon two into four upon one by
p ka whole square and so on so it is something the same expression which we arrived in
the derivation of jacobian derivation of sorry bessel function j naught t so it
is something like that so we can also write this as one upon p key power three by two
one remain one one minus this so this can be written as minus one by two minus one by
two minus one upon factorial two into one by p square and so on
similarly the other expression can be write in the same way so it is nothing but it is
nothing but one upon p key power three by two one min[us] one plus one by p whole
key power minus half in that expression we have a binomial series of one plus one
by p square key power minus half here we have a expre[ssion] binomial series of one plus
one by p key power minus half so when you simplify it so it is nothing but one upon
p key power three by two into into p key power half upon under root t plus one so it
is nothing but one upon p under root p plus one so this is the laplace transform of error
function of under root t this is how we can find out laplace transform of error function
of under root t ok now we can solve that problem very easily
because we know the laplace transform under root t which is given by this f p now laplace
transform of two under root t we can use change of scale property it will be one by
two f p by two hm if it is f p if it is f p so laplace transform of error function of
two under root t will be nothing but one by two f p by two this is by the scale property
change of scale property so we can use this so it is one by two p by two under root p
by two plus one so this will be the laplace transform of that expression
now the next problem let us first find laplace transform l n x what is l n x l n
x is luxurious polynomial ok so we are not going into the detail of this [bus/but]
l n x is given by this particular expression this expression this expression is called
as l n x let us try to find out the laplace transform of this and again using shifting
property or scaling property we can find out the laplace transform of this expression
so let us try to find out laplace transform of l n x
then the rest of the part you can easily solve yourself that how can you find out laplace
transform of e key power two minus two t l n three t using shifting or scaling
property so what is l n x l n x is given by it is e
key power x upon factorial n n th derivative of n th derivative of x key power n e key
power minus x so laplace transform of l n x by the definition of laplace transform is
nothing but zero to infinity e key power minus p t e key power minus p t into e key power
x upon factorial n n x derivative of this into x key power n e key power minus x
so we have to use one symbol because here t is involved so we have to define this in
terms of t l n t because laplace transform f t we are defining
so we have to express l n t and terms of t so this is the expression for l n x so we
can reduce in terms of t replace x by t so how can you find out laplace transform of
this function so this is nothing but one by factorial n can be taken out so it is zero
to infinity e key power minus p minus one times t into n th derivative respect to t
t key power n e key power minus t into d t now you can make use of integration by
paths so let us suppose this first function and the second function so first as it is
integration of second it is n minus one derivative now
integration zero to infinity of this expression minus integration derivative of first integration
of second because integration is anti derivative so derivative powers [will] reduce now
when t tends to infinity because this have negative powers and we are assuming p greater
than one we can assume so this will tends to zero [to/so] and hence entire term will
tend to zero and when t tend to zero this is this this is one and since it has a in
the n th n minus one derivative of this function all the terms involved at least one power
of t because here the highest power of t is t key power n and when we differentiate n
minus one times of this aporta function so all the terms will involve at least t key
power one so when you take t equal to zero that value will be zero so this is one by
factorial n into zero minus now minus minus plus so it is p minus one times integer zero
to infinity e key power minus p minus one times t n minus one derivative of this expression
into t key power n e key power minus t into d t
now now t key yeah it is d t sorry it is d t because we are differencing with respect
to t now so it is d t so so we have observed that in in the integration of this p minus
one comes out when we integrate first time one times and n minus one derivative will
come here so if we repeat this process n times so then we will be having if it if integrate
again for second time this is the first function this is the second function again we will
take one one power of p minus one one more power of p minus one outside the bracket
so that is p minus one whole square will be here now when we when we integrate this expression
n times so hm this will be nothing but power n and it will be involve zero to infinity
e key power minus p minus one times t and there will be no derivative of this expression
[so/to] t key power e key power minus t into d t
so this will be nothing but is equals to one by factorial n this will be nothing but
now now i write here so this expression will be nothing but one by factorial n p minus
n key power n into now e key power t and t cancels out so this will reduce to laplace
transform of t key power n and [so/though] this is equals to one by factorial n p minus
one key power n and laplace on this is nothing but factorial n upon p key power n plus one
because n is a integer so these two cancels out so hence we have shown that laplace transforms
l n x is t minus n key power n upon p key power n plus one now using scale property
and shifting property we can solve the remaining part now laplace same property that
is change of scale property we can read in terms of inverse also if laplace
involves of f p of f t then laplace inverse of p f p by a is nothing but a f a t if a
is not equal to zero now this [pro/problems] problems we can easily solve using this
property suppose first problem we want to solve so laplace inverse of f p is given to
us is is e key power t sin under root t and you want to find out laplace inverse of
f two p so when you compare with this a is nothing but one by two a is nothing but one
by two so you simply replace a by one by two so that means laplace inverse of f two p will
be nothing but one by one by two f two by two because a is half and this expression
is f t so what should be the laplace inverse of this expression so laplace inverse of this
expression will be nothing but nothing but here a is one by two so it is one by two
f t by two so it is one by two f t is this so it is e key power t by two and sin under
root t by two so using the same property we can solve the other two problems also its
very easy you take nine common or four common and then you can use scaling property to
solve these two problems also so thank you very much