so we were studying some of the properties of laplace transforms we have already seen
that what is shifting property and how can we solve the questions related to shifting
property and change of scale property also we have seen in the last lecture and some
problems based on change of scale property also we have seen
now let us see some more properties of laplace transforms so this is laplace transform of
derivatives so if function f t is continuous for all t greater than and equal to zero and
is of exponential order say alpha and f dash t is piecewise continuous on zero to infinity
then laplace of f dash t exist for p greater than alpha and laplace of f dash t which is
a derivative of f respect to t is given by p f p minus f zero we are f p is nothing but
laplace of f t so how can we proof this so laplace of f dash t so what is laplace of
f dash t it is zero to infinity e k power minus p t f dash t d t
now now you can apply integration by a parts you can use integration by a parts
to find out the integral of this so first this second this let us suppose so first
as it is integral of second which is f t from zero to infinity minus integral derivative
of this integral of second d t and it is from zero to infinity now when t is tending to
infinity this value tends to zero why because f t is of exponential order ok since f t is
an exponential order so there will exists some k and alpha such that mot of f
t will be less than or equals to k e k power n alpha t so mode of e k power minus
p t into f t will be less than equals to k into e k power alpha t into e k power minus
p t because it is a non negative quantity we can
multiply both the sides if will not change the inequality so this is equal to k into
e k power minus p minus alpha times t and as as t tending to infinity so since p is
greater than alpha we are assuming so this value will tend to zero so hence the limit
hence this quantity will tends to zero as t tends to infinity and this is because
of the property of exponential order so this will tends to zero so as t tend to infinity
this will tends to zero and as t tends to zero this is nothing but minus f zero because
e k power zero is one so this is minus minus plus p times zero to infinity e k power minus
p t f t d t and what is what is the integral the integral is nothing but laplace of f t
so it is minus f zero plus p into laplace of f t and laplace of f t we are calling as
f p though it is minus f zero plus p into f p
so this is how we can obtain this result that laplace of f dash d t will be nothing but
p f p minus f zero now on the same lines if f t and f dash t continuous and are of exponential
order and f double dash t is piecewise continuous then laplace of f double dash t is given by
this expression so this also we can prove it is very easy to prove this once we obtain
for f dash t now what is laplace of f double dash t so the first expression which we have
just obtained you simply replace f by f dash in that expression you simply replace f by
f dash so what we will obtained this is p into laplace of f dash t minus f dash zero
because you see that this result the first result holds for every p i mean every f
so this will hold for f dash also so we replace f by f dash in the first expression so we
will we get this result this expression now this is equals to p into what is laplace of
f dash t now you again apply the result it is this minus f zero minus f dash zero
so this is nothing but p square f p minus p f zero minus f dash zero so hence we obtain
the next result of laplace of f double dash t
now we can generalize this result for n th order derivative for laplace of f of i mean
n th or derivative of f respect to t will be given by this expression this can be
proved very easily because in order to prove this we can use mathematical induction
when n equal to one that is laplace of f dash t we have just proved that laplace of f dash
t is p f p minus f zero when you substitute n equal to one in this expression so we will
get by the same expression that is p f p minus f dash zero p f p minus f zero sorry ok
now let us assume it to be true for n equal to k so assume it to be true for n equal to
k so what we have assumed we have assumed that laplace of f k th derivative is k th
derivative with respect to t is nothing but you can simply replace n by k
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
so on minus k minus one derivative of at t equal to zero now we have to show we have
to show the same result for n equal to k plus one this is mathematical induction we first
prove it for n equal to one assume it to be true for n equal to k and try to prove the
result for n equal to k plus one if you obtain the result for k equal to n equal to k
plus one this means the result hold for all n this is the concept of mathematical
induction now for n equal to k plus one laplace of f k plus one t is nothing but
it is zero to infinity it is zero to infinity e k power minus p t k plus one th derivative
respect to t into d t again by part first second so it is nothing
but first as it is integral of second zero to infinity minus integral derivative of first
it is minus p e k power minus p t integral of second it is k th derivative t d t so again
since all the derivatives up to n minus one that is up to k here so are of exponential
order some exponential order so this will tends to zero as t tends to infinity this
can be proved in the same lines and when t equal to zero it is nothing but minus k th
derivative at t equal to zero minus minus plus p and it is nothing but laplace of k
th derivative of t this so this will be the expression now the laplace of this we have
already assumed this equal to this expression now this is minus f k th derivative at t equal
to zero plus p into laplace of k th derivative of t we have assumed equal to this so it is
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
k minus one derivative at t equal to zero so this is nothing but p k plus one from this
expression f p minus p k f zero and so on minus f k th derivative at zero and this is
at infinity it is zero and at zero it is minus k and when you multiply by p it is
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
f k minus one and minus f k derivative at zero so what we have obtained that if we
replace in this expression and by k plus one so the result holds this we have
shown that if we replace in this expression and by k plus one so we hold this the result
hold hence the results holds for all n so hence the hence proved this result
now the next is how can we solve some problems using derivatives using this concept so
just for illustration let us try to find out laplace of this simple problems using derivatives
laplace of derivatives so we already know that laplace of f dash t is nothing but p
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
have to find laplace of laplace of t e k power two t so so here f t is t e k power
two t so we if we apply this result though laplace of f dash t so first we find f dash
t what is derivative of f respect to t it is first as it is derivatives of second plus
second as it is derivative of first so laplace of use this result laplace of f
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
is laplace of f t e t e k power two t f p is nothing but laplace of f t minus f zero
so what is f zero it is zero so it is nothing but two times laplace of t e k power two t
plus laplace of e k power two t equal to p laplace of p e k power two t so it is nothing
but this implies two minus p laplace of t e k power two t is equals to minus of laplace
of e k power two t which is nothing but minus of one upon p minus two so laplace of p e
k power two t will be nothing but this will go in the denominator and negative if we adjust
though this is nothing but one upon p minus two square ok so this will be the laplace
of this expression so this we can directly find of course because
we know the laplace of t that is one upon p square and using shifting property if we
replace p by p minus two though we we get back of this expression but it is just
for illustration that how can we solve some problems using laplace of derivatives also
so similarly we can solved the second problem also it is also based on laplace transform
of derivatives that also we can find out otherwise also we can find out the laplace on sin square
by writing sin square as one minus cos two t whole divided by two and then use linearity
property of laplace transform we can find out laplace transform of sin square t
now let us try to solve few problems again of laplace inverse based on laplace
transform derivatives so we get back the same result again now this is laplace transform
of derivatives so what is what does it imply this implies f dash t is nothing but laplace
inverse of p f p minus f zero and here f p is nothing but f p is nothing but laplace
of f t so this expression we can derive from laplace transform of derivatives now it
is given towards at laplace inverse of laplace inverse of p upon p square plus one whole
square is equal to one by two t sin t it is given to us in the problem
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
sin t so if i use this expression so f dash t is one by two t cos t plus sin t will be
equal to laplace inverse of p into f p f p is laplace transform of f t and that is nothing
but that is nothing but p into p upon p square plus on whole square may if laplace inverse
of this expression is f t so laplace of f t will be p upon p square plus on whole square
and minus f zero f zero from here is nothing but zero so so hence we get back to laplace
inverse of p square upon p square plus one whole square which is nothing but one
by two t cos t plus sin t so in this way we have solved first part of this problem
now the second part can be solved now what is laplace inverse of p square upon p square
plus one whole square we have just obtained that it is equals to one by two t cos t plus
sin t it is it is this expression we have just obtained half whole expression now
this we have obtained in the first part in order to obtain the second part laplace
inverse of p square upon p square plus one whole square can be rewritten as laplace inverse
of p square plus one minus one upon p square plus on whole square where does subtract to
one in the numerator this is nothing but laplace inverse of one upon p square plus one minus
one upon p square plus one whole square now this is nothing but laplace inverse of one
upon p square plus one minus laplace inverse of one upon p square plus one whole square
and this is nothing but sin t minus laplace inverse of one upon p square plus one on whole
square but laplace inverse of this is equal to this
this we have obtained in the first part of this problem so this is equal to one by two
t cos t plus sin t so what should be the laplace inverse of one upon p square plus one whole
square so this implies laplace inverse of one upon p square plus one whole square is
nothing but you can put this in the left and right hand side and this here this will be
nothing but half of sin t minus one by t two cos t so that is nothing but one by
two sin t minus t cos t so this will be the laplace inverse of one upon p square plus
one whole square so so hence we can say that the laplace
transform derivatives can also be used to find out the laplace inverse of some problems
now the next property if laplace transform f t is f p then laplace transform of f t by
t is nothing but integral p to infinity f u d u where this is f u is nothing but again
laplace transform of f t so provided this limit exist because function must be piecewise
continuous for existence of laplace transform so f t by t will be will be piecewise continuous
if laplace if limit t tend in to zero plus f t by t exist so that must that is required
in this problem so what is how to find this so take this side integral p to infinity f
u d u the take the right hand side and we will obtain the left hand side so what is
the what is f u what is capital f u it is nothing but laplace transform of f t so it
is p to infinity e k power minus u t instead of p we have u f t d t of course integral
from zero to infinity it is d t and whole multiplied by d u so this we are having in
this problem now we will this is the double integral so
now we will make use of change of order of integration how can we proceed for changing
th order of integration suppose this is line of t this is u now t is varying from zero
to infinity this is the limit for t t is varying from zero to infinity so this is first and
forth (( )) ok t is varying from zero to infinity ok and u is varying from p to infinity
so suppose this is u equal to p line and u is varying from p to infinity that is from
this to this side so this is the region this is the required region this is the required
region of integration ok now we have to change the order of integration that means it is
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
you want the limit for u t so we have to make a strip parallel to u axis because you want
d u first so we make a strip parallel to u axis
so u axis this way the strip here now here u is p and goes up to infinity so u is varying
from p to infinity and t is varying from zero to infinity so e is varying from zero to infinity
so that is how we can change this order of integration ok so now this is equals to t
equal to zero to infinity now we have the u term only in this expression so we can integrate
this respect to u so it is nothing but e k power minus u t upon minus t because their
integration respect to u ok so this is this p to infinity f t d t and when u tending
to infinity it is tending to zero and when u tending to p it is nothing but zero to infinity
e k power minus p t upon t f t d t because when t tend to infinity tend into zero i i
mean sorry it is limit for u so u tend to infinity tend to zero but when u is p so it
is e k power minus p t upon t so it is nothing but laplace transform of f t by t so it is
nothing but laplace transform of f t by t so hence we obtain this result that laplace
transform f t by t is nothing but p to infinity f u d u now how this result can
be use to find out laplace transform of some function of laplace inverse let us see by
solving few problems based on this so let us find out laplace transform of sin hyperbolic
t upon t very simple problem so we know this is result ok laplace transform of f t by t
is something p to infinity f u d u we are we are laplace transform of f t is f p so
we know this result so using this result we will try to solve this problems so laplace
transform of sin hyperbolic upon t so first first we have to find out laplace transform
of numerator quantity here numerator is here f t is sin hyperbolic t so what a laplace
transform of sin hyperbolic t this you already know this is nothing but one upon t square
minus one sin hyperbolic t a upon a square a upon p square minus a square and division
by t is here for division by t we have integral of p to infinity f u d u it is integral p
to infinity so here this is f p so what will be f u one
by u square minus one into d u so now we can integrate this using partial fraction
or so it is one one u minus one into u plus one d u the difference of these two difference
of this and this and this two which is a quotient quantity though you we can multiply a two
and divide by two in the numerator so one by two p to infinity and two can written as
u plus one minus u minus one so this is nothing but one by two integral
p to infinity one upon u minus one minus one upon u plus one d u
this is the partial fraction of this so this is equals to one by two now integral of this
is l n u minus one and integral of this is l n u plus one from p to infinity so this
is one by two l n u minus one upon u plus one from p to infinity at infinity it is tending
to one log one is zero and at p it is tending to l n p minus one upon p plus one so because
of negative sin integral will change this log u plus one upon u minus p plus one
upon p minus one will come because we have upper limit minus lower limit so because of
that negative these two will interchange so it is p plus one upon p minus one so this
will be the laplace transform of the first problem the second problem can be solved
on the same lines first you find laplace transform of the numerator you see numerator what
is f t here in the numerator it is e k power minus two t sin three t what a laplace transform
of sin t sin three t it is three upon p square plus nine and for e k power minus two t you
simply replace p by p plus two that means the laplace transform of e k power minus
two t sin three t will be nothing but three upon p plus two whole square plus nine so
this will be f f p here this will be f p for this problem for the second problem
and we have a division by t now for f t by t we will apply the we will apply this
this property so this will be nothing but p to infinity f u d u so we have to integrate
this up from p to infinity so that will give the laplace transform of the second problem
now let us try to prove this simple problem so what is so from here we can
see that laplace transform of f t is nothing but zero to infinity f t by t e k power minus
p t d t will be equals to p to infinity f u d u this is from this expression now you
take the limit p tend into zero plus both the sides you take the limit p tend to zero
plus both the side so what will obtained it is zero to infinity f t by t d t because e
k power zero is one and is equals to zero to infinity f u d u so hence we got the first
part of this problem now to prove that zero to infinity sin t by t is integral of
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
is what is this problem now it is zero to infinity sin t by t d t so it is equals
to integral p to infinity now f u is nothing but laplace of f t and
laplace of sin t is nothing but one upon p one upon u square plus one d u ok so it is
tan invasive from p to infinity and ok from zero to infinity sorry because we have zero
to infinity in this result for zero to infinity so at infinity is pi by two at zero is zero
so we have this value that integral of zero to infinity sin t by t nothing but pi by two
so some integrals having limit lies to zero to infinity or something can also be solved
using laplace transforms ok now now we can also solve these problems
on based on this properties so i am just illustrating on the how to solve the first
part or the second part so it is in this problem one minus cos t upon t square we have
to integrate two times what what this integral is basically it is one minus cos t first
you will find laplace of this which is nothing but integral p to infinity laplace
of one minus cos t d u if if we are taking as f f u ok now laplace of so this we can
easily find this p to infinity one by u minus it is u on u square plus one d
u so it is it is l n u minus one by two l n u square plus one from p to infinity
so it is nothing but l n u upon under root u square plus one from p to infinity at infinity
is tend to zero at p it is sending to negative of l n p upon under root p square plus so
this is this is laplace transform of this quantity for laplace transform of one minus
cos t upon t square we again have to apply the same property ok so what is laplace transform
of one minus cos t upon t square it is nothing but p to infinity laplace transform of one
minus cos t upon t and d u ok because the this property holds for f t by t so if you
compare with this so what is the f t one minus cos t by t so it is laplace transform of one
minus cos t by t into d u and that we already calculated so it is nothing but p to infinity
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
l n under root u square plus one upon u into d u
so this can easily be integrated if you take u as suppose ten theta you substitute here
u equal to ten theta proceed your integration so this integral can the value of the integral
can be find ok so how to find laplace inverse of p upon p square minus nine whole square
using this property so so from this we got f t by t is nothing but laplace inverse
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
t is nothing but laplace inverse of this so if you obtain f t that means we obtain
laplace inverse of this expression so take this as f p so laplace inverse of
p to infinity it is u upon u square minus nine whole square d u so this is because i
am taking the expression given inside the bracket as f p and laplace inverse of
this f p to find out which is nothing but this f t because we know this ok so now we
divided multiply by two
and one by two it is laplace inverse of the [deri/derivatives] derivatives of denominator
u square minus two is in the numerator so it is a minus one upon u square minus nine
from p to infinity and it is one by two [lap/laplace] laplace inverse of at infinity it is tending
to zero and at p it is nothing but one by p square minus nine
now you divide by a divide multiply by three so it is one by six laplace inverse of three
upon p square minus nine and this is nothing but one by six it is hyperbolic three t so
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
is nothing but laplace inverse of f p that is this quantity so laplace inverse of this
quantity is nothing but this expression so hence we can find laplace inverse also
using this property so so thank you