welcome to the series of lecture series on mathematical methods and its applications
so in the lectures we have seen about laplace transforms their property inverse laplace
transform various functions how to find the laplace inverse and laplace transform etcetera
so now we will see applications of laplace transforms what are the various applications
of laplace transforms so laplace transform can be used in solving various ordinary
differential equations which we faced in various engineering problems like in engineering
like electrical engineering problems or in mechanical engineering problems we face
some ordinary differential equations or partial differential equations which can be solved
using laplace transforms some integral equations some involving discontinuous functions
which can be solved using laplace transforms so let us see the application laplace transform
[pa/part] part one so the first problem is the differential equation for a current i
in an electrical circuit containing an inductance l and a resistance r in series and acted on
by an electromotive force given by e sin omega t satisfies the differential equation this
so what is the differential equations it is the d l d i by d t plus r i is equal to e
sin omega t e sin omega t is the e m f electromagnetic force ok so we have to find out the value
of the current at any time t if initially there is no current in the circuit that is
it is given to us at i at zero is zero that is i at t equal to zero is zero it is
given to us so how to solve so we can solve this problem using our differential equations
techniques also but we can solve the same problem using laplace transforms also so how
can we solve this using laplace transform let us see
so first you take laplace both the sides ok so this l is inductance so lap l into laplace
transform of d i by d t plus r is constant laplace transform of i is equals to e is constant
laplace transform of sin omega t so it is l ok laplace transform of d i by d t we already
know that laplace transform of f dash t is what f dash t is nothing but p f p minus f
zero where f p is laplace transform of f t ok now here f t is here it is i dash t
d i by d t it is i dash derivative of i respect to t so what what the laplace transform
of this using that expression is will be it is p f p minus f zero f zero means i zero
plus r into suppose laplace transform of i is f p ok which is equals to e and sin
omega t is omega upon p square plus omega square so and assuming laplace transform of
i s f p you can call it some other function also f p so it is p into laplace transform
of i which are call which i am calling as f p minus i zero or f zero plus r into laplace
transform of i which are which i am calling as f p equal to e into laplace transform of
sin omega t it is sin omega upon p square plus omega square now i zero is zero is given
to us ok you can collect f p so f p will be nothing but l p plus r it is zero and it is
e omega upon p square plus omega square so from here f p is nothing but e omega upon
p square plus omega square into l p plus r so so what will be this is f p and f p
is nothing but laplace transform of i so what will be i basically i is nothing but function
of t ok i is equal to i of t i is nothing but function of t here so this implies i t
will be nothing but laplace inverse of e omega upon p square omega square into l p plus r
so now find out the solution of this differential equation i t we have to find out the laplace
inverse of this expression this is the only thing left now ok so how can you find out
laplace inverse of this so let us focus on the laplace inverse now laplace inverse of
this expression so this f p is what is e omega e omega is constant free from p it
is p square omega square into l p plus r so one way is you can make partial fractions
or you can use theorem so we can make partial fraction as well so it is a p plus b upon
p square plus omega square and plus c upon l p plus r
so e omega will be nothing but a p plus b we can find out a b and c now [co/compare]
compare the coefficients both the sides p [co/coefficients] coefficients of p square
here is zero and here it is a l plus c coefficient of p square now coefficient of p suppose coefficient
of p will be nothing but it is a r plus b l and here it is nothing so it is zero and
constant quantity is b r plus c omega square which is e omega so using these three expressions
we can easily find out the value of a b c sub where you substitute these values over
here and take the laplace inverse both the sides so we can find out what will be i t
i t can we find out ok so what will be i t i t will be nothing but laplace inverse of
this expression means this expression so this laplace inverse this will be nothing but a
into cos omega t plus b upon omega sin omega t plus c upon l we can take out common which
is nothing but e to the power minus r by l times t ok so these a b c can be find out
using these three expressions using these three equations so we can substitute the
values of a b c from these three equations over here which will give us the value of
i t function i t ok now see the next problem again it is a electrical
circuit problem in a electrical circuit with e m f e t resistance r and inductance l the
current i builds up at the rate given by this equation ok now here if the if the switches
is connected at t equal to zero and disconnected t equal to a find the current i at any time
t ok so how can we proceed for this problem so what is a what is the governing equation
here i d i by d t plus r i equals to e t ok the same equation it is l c r circuit equation
i mean e m f resistance and [inductan/inductance ] inductance this is the equation given by
differential equation given by this expression now it is given that in a switches is connected
at equal to zero and disconnected at two a what was it mean that electromotive force
which is e t it is some constant quantity say e when t varying from zero to a and disconnected
at t equal to a means it is zero when t is greater than or equal to so this is how electromotive
force works for this particular problem ok now to solve to find i at n is ten t we will
use laplace transforms so it is take laplace transform both the side it is l it is again
laplace transform of d i by d t which is given by p into f p minus i zero plus r into f p
is equals to is equal to e t e t nothing but this expression i mean given by this
hm expression so it is zero to infinity because you have to find out laplace transform
of e t so laplace transform of e t is nothing but zero to infinity e t into e power minus
p t d t and it is nothing but zero to a e e to the power minus p t d t otherwise it
is zero so which is equal to e e to the power minus p t upon minus p zero to a and which
is equals to minus e by p e to the power minus a p minus one so these are this is the right
hand side and what is f p f p is nothing but laplace transform of i t this i am assuming
here ok now you assumed that a t equal to zero initially there is no current in the
circuit so i zero is zero ok so i zero is zero so this implies basically f p p l plus
r is equals to e upon p one minus e to the power minus a p because i zero is zero initially
i am assuming that there is no conductance circuit it is zero f p is l p plus r and right
hand side is e by p one minus e to the power minus a p
now the only thing is find i t which is nothing but laplace inverse of f p and f p
from here is nothing but laplace inverse of e by a p into p l plus r one minus e to the
power minus a p so now now we have to find out the laplace inverse of this expression
this is the only problem left now so how to find laplace inverse of this now now again
what is f p f p is e upon e upon p into p l plus r minus e into e to the power minus
a p upon p into p l plus r this is this is f p and we have to find out laplace inverse
of this f p ok so first let us find the partial fraction of this expression so this is
partial fraction can be find out when you put is equal to zero so this is r p plus when
you put equal to this so it is minus l by r into p l plus r so you can check when you
take the l c m you can get the right hand side so this is the partial fraction of this
expression so what will be f p now it is e upon r p minus
minus e into e to the power minus a p upon r p minus minus plus e into e to the power
minus a p now the laplace inverse that is i t will be
laplace inverse of f p so this will be the laplace transform of this expression so we
can further divided into two when t is more than a so it is one so it is one it
is one so that will give the i t when t is greater than a and when t is less than
a and greater than zero it is zero it is zero so that we give gives the value of i t
when t is lying between zero and a zero and a ok so this will be the i t for this expression
so this is how we can solve some problems using laplace transforms
now we will see some more problems which can be solved using laplace transforms ok these
are simple problems just just to illustrate how can we use laplace transform to solve
ordinary differential equations so the problem is simple it is y double dash minus six y
dash plus five y is equal to e to the power two t and y zero is one and y dash zero is
also one it is given ok it is one minus one so it is one y zero is one and y dash zero
is minus one ok so now how to solve this problem is a laplace transforms so first take
laplace transform both the sides ok so it will be laplace transform of y double dash
minus six times laplace transform of y dash plus five times laplace transform of y is
equals to laplace transform of e to the power two t now from the laplace transform derivatives
we know that the laplace transform of y double dash is nothing but p square f p minus p f
zero minus f dash t now you call it f zero or you can call it y zero there is no problem
now because here instead of f we are having y ok so minus six and laplace of y dash will
be nothing but p f p minus y zero plus five times laplace of ok laplace of y is f p and
it is nothing but one upon p minus two laplace of y i am calling as f p so y zero
is one and y dash zero is minus one it is given so f p will be that is p square minus
six p plus five and y zero is one that is minus p minus minus plus one and it is
minus minus plus six into one is six is equals to one upon p minus two ok it is it is one
it is minus one so it is p it is plus one minus minus plus six into one is one so f
p will be nothing but one upon t minus two into t square minus six p plus five and it
is this will go to that side so it is plus p minus five of seven ok six plus one is seven
so it is p minus seven upon p square minus six p plus five so it is nothing but one upon
p minus two into it is nothing but five and one so that is p minus one into p minus five
plus it is p minus seven upon p minus one into p minus five now what is what will be
y t y t will be nothing but laplace inverse of f p so laplace inverse of this can be find
out using partial fractions ok so you can use partial fractions to find out the laplace
of this
so hm that will give the final answer of final expression for y t ok
now the next now we will see the next problem involving t also in the y y double dash or
y let us solve this problem so this this problem can be completed ok you use you take the partial
fraction make the partial fractions and then take the laplace inverse of these expressions
now next problem t y double dash plus two y dash plus t y equal to zero and it is give
to you that t y at zero is one and y at pi is some value zero ok this is given to you
now how to solve this problem again you will use laplace transforms you take laplace
transform both the sides so laplace transform of t y double dash plus two times laplace
transform of y dash plus laplace transform of t y dash t y equal to zero now what is
laplace transform y double dash i mean f double dash ok f double dash t is nothing but p square
f p minus p f zero minus f dash zero and laplace transform of t f double dash would be what
will be nothing but minus one to the power one d by d p of laplace transform f double
dash t by by the property of multiplication with t ok so now laplace transform of t into
y double dash would be what will be minus one to the power one d by d p of laplace transform
of y double dash so laplace transform of y double dash is p square f p minus p into y
zero minus y dash zero plus two into laplace transform of y dash laplace transform of y
dash is nothing but p f p minus y zero plus laplace transform of t into y dash again is
nothing but minus one to the power one d by d p of laplace transform of y which is f p
it is equal to zero because i am assuming the laplace transform of y t as f p
now let us simplify this it is minus derivative with respect to p so it is p square into f
dash p plus f p into two p minus derivative of with respect to p is y zero y zero is
one and derivative this is zero plus two p f p and y zero is oneth minus two minus f
dash p is equal to zero ok it is two f p two f p y zero is one so two into one is two ok
now it is nothing but this implies f dash p into minus p square minus one with f p what
will be having now it is minus this and plus this so both will cancel out ok it is minus
minus plus one minus two so it is minus one equal to zero so what will be f dash p f dash
p will be nothing but one upon negative p square plus one so what will be f p you integrate
both sides f p will be nothing but minus tan inverse p plus c now now you find laplace
inverse both the sides to find out y t ok so so y t will be nothing but this implies
so it is f dash p minus p square minus one so it is ok it is minus minus plus one ok
it is minus minus plus one ok plus one minus two is minus one ok and it is one upon p square
plus one and p square plus one is tan inverse p plus c ok
now to calculate this c we will use corollary of existence theorem for laplace transforms
we know that as p tend to infinity f p must tend to zero so take p tend to infinity both
the sides so p tend to infinity so at p tend to infinity limit p tend to infinity
f p must be zero so this is zero equal to minus tan inverse pi by two tan must infinity
ok plus c so this implies c so implies c is pi by two so f p is nothing but minus tan
inverse p plus pi by two and pi by two minus tan inverse p is nothing but cot inverse p
so now we have to find out laplace inverse of this f p ok so take the derivative both
the sides or we can take laplace inverse here itself if we take laplace inverse here so
it is nothing but minus t f t and it is nothing but minus of sin t so f t will
be nothing but sin t by t if we take the laplace inverse here itself we can we can
proceed from here also from here to take the if you want to find out the laplace inverse
of f p again we have to differentiate it so we will go back to a same expression ok
then the laplace inverse of this will be nothing but sin t upon t so this is nothing
but y t now y at pi must be zero which is satisfied y at pi is zero so which is satisfied
so hence this is the expression their solution for this problem
now similarly this problem can be solved using laplace transforms whenever we involve
with delta function right hand side so such problems can also be solved using
laplace transforms ok you take laplace transform both the sides and then take laplace inverse
that will give you the final answer of the same problems so
thank you