Introduction to Differential Equation Video Lecture - Computer Science Engineering (CSE)

welcome your lecture series on
differential equations for undergraduate
students
today's topic is introduction the first
question arises what do we mean by
differential equations before answering
this question let us try to remember one
thing which we have already done the
exponential function yes e to the power
X we know that this function has a
property that if we differentiate it
with respect to X we get the same
function that is dy over DX is e to the
power X thus we got a relationship that
dy over DX is equal to Y this
relationship is called
differential equation so now we are
ready to answer what do you is a
differential equation a differential
equation is a equation involving an
unknown function and its derivatives
thus dy over DX is equal to KY d 2y over
DX square plus 4y is equal to 0 1 plus
dy over DX square whole to the power 3
by 2 is equal to C times d2 y over DX
square x times dy over DX whole square
minus y times dy over DX plus X is equal
to 0 x times del u over del X plus y
times del u over del Y is equal to 2 u
del 2y over del T 2 is equal to C square
times del 2 y over del X 2 are all
examples of differential equations in
these examples we have seen that some
involves ordinary derivatives
while some
examples involved partial derivatives
thus we categorize our differential
equations into our energy differential
equations and partial differential
equations an ordinary differential
equation is an equation that contains
one or several derivatives of an unknown
function of one variable only thus in
our example the first example dy over DX
is equal to KY the second example d2 y
over DX 2 plus 4y is equal to 0 1 plus
dy over DX whole square complete power 3
by 2 is equal to C times d2 y over DX 2
x times dy over DX square minus y times
dy over DX plus X is equal to 0 or all
examples of ordinary differential
equations here we do have the unknown
function y and its derivatives and the
function y was of only one variable and
we had the ordinary derivative with
respect to that variable only now a
partial differential equation is an
equation involving an unknown function
of two or more variables and its
derivatives with respect to independent
variables thus in our examples x times
del u over del X plus y times del u over
del Y is equal to 2 u del 2y over del T
2 is equal to C square times del 2 y
over del X 2 are examples of partial
differential equations here in the first
example U is a function of two variables
x and y and the equation involved is
partial derivative with respect to x and
y while in the second equation y is the
unknown function and this function is of
two variables T and X and it involved a
second-order partial derivative with
respect to these independent variables
thus we have seen that all these
Asians are involving a function of one
or more variables the derivatives
ordinary are partial again the
derivatives of first-order are hired
again the derivatives may be having the
exponents thus we are now ready to give
some basic terminology of the
differential equations the first one is
order the order of a differential
equation is the order of the highest
derivative in the of the unknown
function involved in the equation thus
in our example let us see first example
dy over DX is equal to KY this involves
the first order derivative only thus the
order of this equation is one Wireless
the equation D 2 y over DX 2 plus 4y is
equal to 0 this involves the second
order derivative and that is the highest
derivative involved in this equation
thus order of this equation is 2
similarly in this equation plus dy over
DX square whole to the power 3 by 2 is
equal to C times D 2 y over DX 2 in this
equation it involves the first order
derivative as well as the second order
derivative so the highest derivative
involved is the second order derivative
thus this equation has ordered two now
we see in the last example that X dy
over DX square minus y times dy over DX
plus X is equal to 0 we see the
derivative involve are only of the 1st
order this order of this equation is 1
only so this equation is also being
called first order equation the second
term is degree the degree of a
differential equation is the exponent of
the highest derivative occurring in it
after the equation has been rationalized
with respect to derivatives let us
understand this
clearly with the help of examples so in
our first example we do have that the
derivative involved is only that highest
derivative is first order and it's
exponent is 1 thus the degree of this
equation is 1 in the second example the
derivative which is involved that a
second order derivative is the highest
derivative aren't its exponent is 1 does
this equation also has decree 1 now see
in this example we do have that
derivative involved is a first order
only but we say that gay y over DX has a
exponent 2 so the highest exponent is 2
thus degree of this equation will be 2
knowledge now let us say one more
example we had in our examples that is 1
plus dy over DX square whole to the
power 3 by 2 is equal to say times d 2 y
over DX 2 in this equation we do have a
power as 3 by 2 that is it is involving
a square root now this is not called a
rational equation because here the power
of dy over DX is having 1 would do
something so we will rationalize it how
will rationalize it we will square this
equation on both the sides and thus we
would get 1 plus dy over DX square whole
to the power 3 is equal to C square
times d2 y over DX 2 is equal to DX 2
here the highest derivative involved is
the second-order that is d 2 y over DX 2
and its exponent is 2
thus degree of this equation has to
now we can categorize the equations in
different manners so one more
categorization of differential equations
is coming as linear differential
equation a differential equation is
called linear if the function and its
derivatives are appearing in the
equation with degree 1 only and
separately if this is not happening then
it is called nonlinear let us understand
this more clearly with the help of
examples so first let us see the example
of linear differential equation see this
example X cube times D 3 y over DX 3
minus 3x squared times D 2 y over DX 2
plus 6 x times dy over DX minus 6y is
equal to 0 here we see that the
derivatives involve our first-order
second-order and third-order and all
these derivatives and the function y are
in degree one only none of them have any
exponent
moreover none of the places I do have
that the function and derivatives are
appearing simultaneously any component
thus this equation is linear in the
unknown function y and its derivative
this is called a linear differential
equation let's see one more example G 2
y over DX 2 minus y is equal to 0 here
we do see is that is the derivative is d
2i over DX 2 and y both are occurring in
degree 1 and separately thus this is
also an example of linear differential
equation let us see some example of
nonlinear equations see this first
example D 3 y over DX 3 plus 2 times D 2
y over DX 2 multiplied with dy over DX
plus X square times dy over DX whole to
the power 3
is equal to zero now we see here that
derivatives are coming with exponent as
the first derivative dy over DX has
degree 3 while is in the second
component D 2y over there so that is the
second order derivative and the first
order derivative are occurring
simultaneously thus this is a nonlinear
equation the second example D 2 y over
DX 2 minus M times 1 minus y square
times dy over DX plus y is equal to 0
here we see the derivatives are
occurring in the first degree only but
the function itself is occurring in the
second degree
moreover in one component we do have
that the function and it's derivative
are offering simultaneously thus this is
also an example of nonlinear
differential equations let us have
something more about linear differential
equations a linear differential equation
of order n is a differential equation
which is written in the form a + X times
D n y over DX n + n - 1 x times DN minus
1 Y over DX n minus 1 and so on
a 1 x times dy over DX plus a not x
times y is equal to F X we see here that
all the derivatives are occurring in
first degree only and their coefficients
are the function of X only none of them
involve the Y are its derivative this is
a linear equation now here the highest
occurring derivative is d + y over DX
and that is in eighth order thus order
of this equation will be in only if it's
coefficient a in X is not a zero
function thus this equation will be
called a linear equation of order n if a
in X is not a zero function y if NX is a
zero function the first term will not be
there
then the highest derivative would be
only DN minus 1 over DX n minus 1 that
is n minus 1 1/3 so this is a
differential equation of order n many
times this kind of equations are also
being written as n x times y n plus n
minus 1 times y n minus 1 and so on a 1
x times y 1 plus a naught X Y is equal
to F X what the change we have done here
if you see is that we have replaced this
Y n with DN y over D X n that is the nth
derivative we have used another notation
Y n similarly for n minus 1 F derivative
we have used the notation Y n minus 1
and so on for second derivative the y2
and for first derivative y1 these
notations are being used so that it is
convenient in writing so this is also
one way of writing the differential
equations and these notations are being
used for convenience now we have seen
that these differential equations are
functions and their derivatives aren't
there relationship thus a function will
satisfy these equations so any function
which is satisfying this equation will
call its solution so now we come to
another term solution a function is said
to be a solution of a differential
equation if the function and its
derivatives when substituted in the
differential equation satisfy the
equation identically let us try to see
it with the help of example let us take
this equation x times dy over DX is
equal to 2y we can check that Y is equal
to X square this is a solution how can
we check if see that is 4 y is equal to
X square what will be dy over DX that
will be 2
times X now this value of the function
that is X square and its derivative 2x
we can put into the given equation and
thus what we will get x times 2x is
equal to 2 X square and right hand side
is also 2 X square thus we get that this
equation is being satisfied identically
thus what we got that X square is a
solution of the differential equation x
times dy over DX is equal to 2y so we
have got a solution now let us see what
as soon as we talk about solution the
question comes existence and uniqueness
what do we mean by existence does a
differential equation has a solution the
answer to this question is called
existence then it comes uniqueness does
a differential equation has more than
one solution if yes can we find out a
particular solution that is if yes how
can we find out a solution which
satisfies the particular conditions
so now we have seen that a differential
equation may or may not have a solution
if it is having a solution it may have
more than one solutions so now we are
going to categorize the solutions also
the first categorization is general
solution the description of all
solutions to differential equation is
called the general solution let us
elaborate it little bit more we have
just seen this example that is x times
dy over DX is equal to 2y now let us see
a little bit more if I take this
function C times X square we can check
that this is also a solution of this
equation for all values of C we can
check it see Y is equal to CX square the
dy over DX will be 2 times C X now if I
incorporate these things into the
equation what I would get is that x
times 2 C times X is equal to 2 C X
square which is nothing but the right
hand side thus C X square is also
satisfying this equation for all values
of C what it says that is P are getting
not only one solution but many solutions
that is for each value of C I will get
one solution so we this CX square will
be called the general solution of this
differential equation now the second
question is uniqueness before answering
this let us try to define something more
over here initial value problem a
problem in which we are looking for the
unknown function of a differential
equation where the equation of unknown
function and its derivatives at some
points are known is called the initial
value problem this is also referred as
IVP insured if example
the equation d VAR x times dy over DX is
equal to 2y if I incorporate one extra
condition that the function value is 5
at X is equal to 1 this is called an
initial value problem a solution to the
initial value problem is known as
particular solution so now let us see
this with the help of example just now
we had seen that Y is equal to CX square
is the general solution of x times dy
over DX is equal to 2y now incorporate
this condition that is y at 1 is 5 that
means in C X square at X is equal to 1
we will put what we will get is that C
is equal to 5
hence we got that Y is equal to 5 X
square is a particular solution this
solution we have got when we have used
this initial condition that is the
function value is 5 at X is equal to 1
let us understand a little bit better
with the help of graph we see here is
that is this is the graph of all the Y
is equal to CX square here these graphs
I have made for different values of C
that is all these functions are solution
to the differential equation x times dy
over DX is equal to Y for different
values of C we are getting it now in our
condition what the initial condition was
given that at X is equal to 1 the value
of the function should be 5 we see is
that is this is that is at 1 if we see
here we are getting the value of file so
this is the graph which we are this is
the function which is coming as a
particular solution we see is at 1 no
other function has value 5 over here so
thus to arrive to a unique solution we
require a particular condition now we
have started this lecture with this
function that is exponential function
and we had seen that basis satisfying
the equation
over DX is equal to Y now let us take
another differential equation dy over DX
is equal to 2y this is closely related
to the previous one the only differences
of a constant Tori have used at Y but
now this e to the power X this does not
satisfies this equation
anymore now how to find out the answer
to this equation that is which function
will satisfy this equation ok let us
take this assumption that the function
should be of the same form that is
exponential one we have to do little bit
experiment with the same function so let
us assume that we take a function C
times e to the power K X where both this
C and K are constants if I differentiate
it we get dy over DX is equal to K C
times e to the power K X that is K times
y thus we have got that function y is
equal to C times e to the power K X
satisfies the equation dy over DX is
equal to k1 so we had started with dy
over DX is equal to 2y so now what I
will do I will replace this K by 2 so we
got that C times e to the power 2 X is
solution to the differential equation dy
over DX is equal to 2 1 now what we have
actually got let's summarize the result
we have actually got that C times e to
the power K X is the solution to the
differential equation dy over DX is
equal to K Y this is true for every
value of C
so that's what we have got actually we
have got a general solution to the
differential equation dy over DX is
equal to KY where this constant K is
determined by the equation as in our
example what we have chosen we have
chosen K is equal to 2 this K is also
referred as rate constant but this C is
not determined by the differential
equation that says we have actually find
out a whole family of possible solutions
and one for every possible value of this
constant is a solution let us see the
graph of this function this is the
family of exponential graphs so here I
have made this you see is that is here
we have different exponential curve all
these are e to the power 2 X functions
but with different values of C so all
these functions satisfy the equation
that dy over DX is equal to 2 y how do
we choose any particular solution that
says we require something more for that
one what that mode let us see that we
says that from where this function is
starting that is what is the initial
value so suppose I says that is my
initial value that is y at 0 that is
initially is 1 now if I get what is the
solution that C times e to the power K X
that will be actually 1 thus what we get
that C is equal to 1 if I replace C is
equal to 1 I will get dysfunction we see
that this function we are initially at 0
we are starting at 1 thus we are having
this as a function which is satisfying
this particular condition
so if we change our initial condition
say the function is starting at half way
we'll get that a different function that
is we will get here Y is equal to half
times e to the power 2 X this curve now
if I change my initial condition to
minus of that as the function is
starting at minus of we say that our
function has solution has been changed
so thus what we have got that we
summarize the result that we have got
the initial condition is picking up a
particular solution the differential
equation gives me a whole lot of general
solutions so now I give you some person
to think about find the solution to
differential equation dy over DX is
equal to minus 5y use the initial
condition that ad 0 the function is is
starting with chin C that is how the
graph of this function will look like
and what is the difference it has with
our example just now what I have done
ok moreover you can experiment little
bit more see that if you change the
initial value to 0 that as the function
is starting to 0 now find it out what
you are getting
does it drastically change your solution
think about it and then see
so here are some more questions for you
to think about find the general solution
to the differential equation dy over DX
is equal to 2x think about what should
be the solution of this differential
equation second question is dy over DX
is equal to three what will be solution
of this differential equation
moreover use the initial condition that
y 0 is equal to 2 in both these
questions and find out a particular
solution you can experiment with other
initial conditions also and see what you
are getting this would give you a little
bit insight that is what we mean by
differential equation and its solution
so we have learned what our differential
equations and what our solutions of
differential equations now the question
arises why do we study differential
equations are what we achieve by knowing
the solution of differential equations
in other words what is the application
of this differential equation well in
real life we'll apply this differential
equation so that we do study these
differential equations that says that is
what is application of differential
equation
the answer is modeling what is modeling
modeling is that actually transforming
the physical phenomena into mathematical
formulation by changing physical
phenomena into mathematical formulation
we do know more about that phenomenon
moreover we may predict the future
behavior of that phenomena
thus differential equations are very
important topic in mathematics science
and engineering in all those situations
where changes occur and predictions are
desirable the most of the systems which
come under study they observe certain
natural laws are observations these
physical changes when we transform into
the language of mathematics may turn out
to be differential equations thus we try
to see which kind of functions are which
kind of phenomena will translate into
differential equation a typical real
world application of differential
equations involve those phenomenons with
changes with the time and the change is
proportional to the
weight of the change are the
acceleration of the change
that is whenever these relationship we
can use as the differential equations
now
whenever a physical law involves a rate
of change of a function such as velocity
or acceleration etc we do get
differential equations thus in physics
and engineering application we do get
many differential equations an example
here I have taken here there is a
stationary object which is initially at
initial position a force is being
applied on it and this stationary object
starts moving so from where it starts
moving that point I am taking initial
position after time T 1 it reaches to a
position X and it is still continues to
move and at time T 2 it reaches to
position Y we have observed here that
this object is not covering equal
distance in equal time or in other words
the equal distance has not been covered
in equal time that says that the
movement of the object that is the
velocity of the object was changing and
why it is happening since the force has
been applied once only an object is
moving on certain surface that surface
is also giving some force so this
velocity is changing now this is an
example where we can say that we can use
the differential equation say for
simplicity if I use that the change
which is occurring is at constant rate
that is the distance it is covering with
respect to time that is changing with
constant rate so if I denote this
distance by is then DS by DT is a
constant say a and this thus we have got
that is the change in the distance is at
constant rate this is a differential
equation and by solving this weekend
that at time t3 where this object will
be this is what very simple application
here we do have so now in this
application what we have done we have
taken one assumption that the change is
at constant rate thus we have got that
everyday problem what scientist is
struggling is how do I translate a
physical phenomena into a set of
equations which can describe it in real
word actually it is not possible to find
out a system of equations are a set of
equations which describe this phenomena
completely and adequately so what we do
is we find out the set of equations
which can describe the system
approximately and when we say is
approximately of course we are making
certain assumptions of certain
observations and thus we come up with
some set of equations we try it we say
that how it is working we compare it
with real-world results and then again
we can improve upon our set of equations
this is what is the modeling we are
having and in all those situations where
changes are occurring and we want to
find out how these changes are happening
all these are coming modeling through
differential equations now let us try to
say this with a simple example the first
example is about the population growth
we have writing it is unlimited because
we had made an assumption that there is
no condition or don't control on the
growth the example is the number of
bacteria in a liquid culture is observed
to grow at a rate proportion to the
number of cells present
at the beginning of the experiment there
are 10,000 cells and after 3 hours there
are five leg cells now how many will be
there after one day of growth if this
unlimited growth continues what is the
doubling time of bacteria now let us
start with this example what we have
been given that is a liquid culture is
having certain number of cells and it's
growing with China now how do I
translate it you see here first we'll
assume that is from the starting of the
experiment the time elapsed that will
denote by thing and the time we will be
counting here in the hours second thing
we would be assuming is the unknown
function y that is the number of cells
at time T why we are calling it a
function because it is changing with
time T then what we have been given we
have been given that the starting of the
experiment the number of cells are
10,000 that is why at 0 is 10,000 here I
have used 1 another notation Y naught
and C that is this is also we are
sometime referring that is the value of
the function at that point
now what more we are being given in this
one that is the rate at which this
growth is taking place R that is the
number of cells are changing that is
called the growth rate so what will the
growth rate growth rate will be dy over
DT this is given that it is changing
according to the size of the population
at that time T that means it is given
that dy over DX is equal to K Y where K
is the constant of proportion no we do
remember this equation we have seen this
equation in our first
out of this lecture we do know that this
equation has a solution as YT is equal
to C times a to the power KT now we are
being given initial value that the
starting of the leg experiment we do
have Y at 10,000 cells now if I
incorporate this initial value we have
seen in our first example status there
should be C so C would be 10,000 thus
what we have got that this function
would be 10,000 times e to the power KT
now the question is what will be this
kill for that we have to see our
experiment a little bit more we are
being given that after three hours it
has been observed that the number of
cells are fine ler so now translate this
into the mathematical formulation we
have been given so at T is equal to
three my function would be 10,000 times
e to the power three K it is given that
number of cells at that time are 5000 so
if I create these two things what I do
get e to the power three K is equal to
50 now we can solve this to find out the
value of K how we will solve it will
just take the natural logarithmic on
both the sides of this one so we do get
log of e to the power three K is equal
to log of 50 which is giving us that 3 K
is equal to log 50 thus k is equal to
log 50 by 3 and now we will incorporate
the value of natural log of the amount
of 50 which is approximately 3 point 9 1
thus we are getting the value of K as 3
point 9 1 by 3 is equal to 1 point 3 so
what we have got that is the value of K
is 1.3 now this is the rate constant and
what weight we would have because we
have take
our tea as the time in our so it could
be that it is changing 1.3 per hour that
is rate of changes 1.3 per hour so now
what is our solution we have got that at
the time T the number of cells will be
10,000 times e to the power 1 point 3 T
that is this is we have got this
function as a solution R this is what we
are giving us the number of cells at the
time T now let us see this by the graph
we see that is this is the graph of the
function e to the power one point three
T and since initially we have started at
the value 10,000 we see here in the
graph that it is 10,000 is starting so
this is the solution of this one now we
are ready to answer the question that is
what will be the growth after one day
that is after 24 hours that is will
incorporate T is equal to 24 in our
solution what it will be 10,000 times e
to the power 1.3 and in place of T I
will put 24 which we will rinse all will
reach to 10,000 times e to the power 31
point 2 which will be giving me 3 point
5 into 10 to the power 17 that is these
are the number of cells and the end of
one day at the time of 24 hours
now let us try to solve the second part
of the problem that is what is the
doubling time of the bacteria so first
we should know what is the doubling time
doubling time is actually the amount of
time on takes for a given population to
double in size so let us explain it
let's save that is tau is the doubling
time then what we are being set is that
is the number of cells at time tau that
is why tau should be twice the number of
cells at the initial position that is
why not so now incorporate these
information in our solution because we
do know what is the white t why tau is
10,000 times e to the power K tau and
initially the value of the why not that
is the number of cells at the beginning
or 10,000 so twice of this would be
20,000 that's what we got that e to the
power K tau is equal to 2 now we do know
that in this solution that K is one
point three so we incorporate this K tau
is equal to log 2 after taking logarithm
and in copper in the value of K as 1
point 3 we do get that Chow is equal to
approximately point 6 9 divided by 1
point 3 that is approximately 0.5 3 3
what is this this is tau that is the
time and time how we are measuring in
hours so this is doubling time is
approximately 0.53 3 hours which if I
change into the minutes
I do get approximately 32 minutes so we
have here one problem where we are
interested to know the growth of
bacteria in a liquid culture we had
observed that experiment for certain
time we had initially started our
observations when there were 10,000
cells and after 3 hours we have observed
out there 5 lakhs
and one more assumption we have made
that the growth is to number of cells
present in the culture at that time with
the help of these three things we have
modeled our problem we said since the
rate is changing according to the size
of population we got the equation dy
over DT is equal to KY and we have solve
that equation with the help of other
observations which we have made the
first observation that initially we have
started at 10,000 cells then the fact
that after 3 hours we have observed it
has 5 lakh from there we have got the
rate constant that is scale and then we
could predict it that is after 24 hours
how many bacteria will be in that
culture moreover we can also answer the
question that is in how much time the
population will get doubled and these
answers we have made similarly let us
see one more example of the similar
nature but little bit different that is
radioactive decay
radioactive decay is a property
mischaracterizes the substances whose
atoms undergo spontaneous changes these
are being stored either in radioactive
isotopes are in step that our unstable
forms are in stable form on those
materials which decays these decays are
normally at a constant rate which when
they burst are being counted on a geezer
counter thus these bursts are being that
is atoms are being bused at a constant
rate on this isoh tibia that is the
radioactive material changes to the
stable material that says is that this
constant rate would be proportional to
the number of atoms present are the ISO
tip at that time so this decay is that
is the change in the mass of radioactive
material is proportional to the mass
present at that time
and this would be decreasing we can
understand this in another words as say
for example if there is half of the
material presenter the ISO tip then the
number of bursts in the atoms will be
half of the times whatever it is
previously does this change is in the
decreasing order and this is called
decay now this problem also we can study
as we have used this growth phenomena
what is the difference between the
growth phenomena on this decay phenomena
is the thing is that is there the things
were changing to they are growing up and
here the population is decreasing or
decaying one so we can use the same kind
of thing what we are being given that is
let us see we have been defining the
variable here first that is T is the
time and let us say empty is the
variable that is the mass of the ratio
radioactive isotope
it at time T it is given that this mass
is changing at a constant rate which is
proportional to the mass present at that
time
that is the rate of change of the mass
would be DM by DT this is proportional
to the mass given at a time thus we can
say that DM by DT is equal to minus km
now say here I have used minus sign
well whatever be the mass present or the
ISO tape that will always be a positive
number that will be always a positive
and it is decreasing K I have used as a
positive constant so I have to use the
sign minus because as the time is
changing the mass is decreasing so the
rate of change should be negative thus
we have used here minus K and this
solution of this differential equation
now we have already seen this kind of
differential equation the difference is
here that is rather than having plus K
we are having here minus K of course we
have changed the variable from Y to M so
the solution would be that is in the
solution itself also we would change it
to the minus K and the solution would be
that MT is equal to M naught times e to
the power minus KT here M naught I have
used as the constant R that is the the
constant in the general solution instead
of C Y M naught I have used because
there must be some initial value known
for this experiment also that is from
where we are starting observing this ISO
tape to decade that is that M naught is
initial mass present on the ISO tape now
we are ready to solve the problem let us
see that is what is this problem before
going to the problem let us see this
function a little bit more since this
function is negative exponential it is
having
see we have this function which is
decreasing
of course this function should be
decreasing because what we are having is
that solution must be a mass which is a
function which is decreasing and we say
as initially we are starting at
different values we do have different
solutions now try to solve one problem
let us see what the problem we are going
to solve here carbon-14 is a radioactive
isotope of carbon that has a high flat
half-life of 5600 years it is used
extensively in dating organic materials
that are very very good sometimes
thousands of years of so the question is
what fraction of original amount of
carbon-14 in a sample will be present
after 10000 of the ears now let us see
is that how we are going to model this
problem carbon-14 is a radioactive
isotope so it has a property that it
will decay and the decay would be at a
constant rate and that rate would be
proportion to the amount of the
radioactive material that is carbon-14
present at that time
thus the equation we will have is that
DM by DT s minus K we have already seen
that the solution of this equation is M
naught times e to the power minus K T
now for this problem we have to find out
what is M naught and what is K R to
answer this we have to know is that is
after 10,000 year what will be the
fraction of amount which will be left
that is we are not necessarily want to
know what is M naught whatever be M
naught after 10,000 years what fraction
of that amount will be there so we are
more interested in finding first what is
K how we are going to find it out we
will see in our example what other
conditions are given to us we have been
given this information that half
life now what is the half-life half-life
of a substance are a decaying material
is the amount of time it takes for 50%
of the original amount of the substance
to decay we have been given here that
half-life of this is 5,600 so let here
now be the half-life that is the Tau
will be the time it will take for
material to come as the half of that
percentage and let us say that M naught
is the amount of the substance present
initially then what it will give now
because we do know what is that solution
that is empty is equal to M naught times
e to the power minus KT we have been
given that at tau is equal to 5000 so at
tau I would get em tau s half times of M
naught now put tau is 5600 in our
solution and we do get that M naught
times is power minus K tau is 1/2 M
naught are e to the power minus K tau is
1/2 thus what we are getting is that 2
is equal to e to the power K tau and
thus tau is log 2 upon K so we are just
solving this equation without actually
incorporating the value of tau
we are given that tau is equal to 5,600
so we'll incorporate this and we get K
is equal to locked out log 2 upon tau
which is 0.6 9 divided by 5,600 which
gives approximately 1.2 into 10 to the
power minus 4 power here that is that
time we have incorporated here in the
terms of air because it is takes a long
time so we have got that K is equal to
this thus what we have got the solution
we have got the solution as MT is equal
to M naught times e to the power minus
1.2 times 10 to the power minus 4 T now
we are ready to answer of a question
question is that it after 10,000 years
the fraction of the original amount will
be M at 10,000 divided by M naught
because initially the amount is M naught
so now use these informations in our
solution M at 10,000 will be M naught
times e to the power minus 1 point 2
into 10 to the power minus 4 and at the
place of 8 I will use 10,000 which gives
me that it is e to the power minus 1.2
now we can calculate this value which is
coming up approximately as 0.3 what does
it says after 10,000 years the 30% of
the original sample will be lift on the
carbon-14 I so take that says that is
again we have seen here in this one that
if we are talking about a decade which
cannot be stopped by any outside source
so we are having the decay is happening
and that decays again at a constant rate
and that constant rate is proportional
to the population size are the material
present at that time that again we have
used the same kind of differential
equation and we have solved this problem
also so in both this
examples we have seen that we have used
the exponential function are its
differential equations that is why I had
use this example just because I have
started my lecture with exponential
function so this is not the only
application of differential equation
actually we can model many kind of
physical phenomena all those we know
knows where changes are taking place and
we are interested to know what will be
the future behavior of that phenomena so
wherever this change in changes are
occurring we do have the differential
equations
knowing the differential equations
knowing that how to solve those
differential equations we can know the
function that is we can know that
phenomena what this phenomena is the
function says is that if I say this is a
function it says is I do not know
completely about this thing so if any
phenomena I am saying is a function I
don't completely know the phenomena I
don't know the characteristic of that
phenomena I can predict the future
behavior of that phenomena are I may
actually calculate many things about
many characteristics of that particular
phenomena that is why this differential
equations are very very important here I
would conclude today's picture