Linear Differential Equations and their Solution using Integrating Factor Video Lecture | Mathematics (Maths) Class 12 - JEE

everything will work out nicely in this
video I'm going to talk about the idea
and a strategy for how to solve a linear
differential equation but before I do
this with you guys let's talk about an
integral first let me ask you guys this
what's the integral of sine x over X and
I know I did not put down with the X by
was for good purpose anyways what if the
answer to this and you should know that
this right here has no answer right this
is actually a non elementary integral so
no matter what you do use substitution
integration by parts Y or whatever the
answer idea is just not going to be nice
right but what if I tell you that maybe
I can add something right here let me
ask us this following two instead what
if I'm going to ask you guys what's the
integral sine x over X plus Ln X times
cosine X and now suppose this is my
actual question I want to ask you guys
I'll put parentheses around this and a
DX so I can be legitimate what's the
answer to this integral now okay maybe
you see it maybe not
but let me tell you that this right here
he actually has a much better answer and
the answer to this integral just going
to be sy x times Ln X and of course we
are done I'll put on a plus C at the end
well how come I can just get this for
the answer well does look at this
backwards okay if you just kind of pay
attention to the form first of all you
see that you have a addition in between
of two things right and then here we
have sex over X and then here we have ln
x times cos x and guess what if you put
this down you see signs x ln x that's
the function part and if you just go
ahead and differentiate this instead
this is how to entire derivative right
you put on the answer and you want to
know if this is the right answer well
now you just have to take that the root
of this if you can get back that would
be the antiderivative
I will be the integral heart you have to
use the product rule because this is a
product of two functions and the product
one to show you this well I will keep
the first function which is sine X and
then we multiply by the derivative
second which is going to be one over X
and then we add you see this is why I
know that I will use the product to
function because earlier this is a
adding situation right we add the second
function which is Ln X and that's a
peahen DC and we multiply by the
derivative of the first function the
relative sex is positive cosine X you
see we get back to the original
integrand and of course this will be the
antiderivative for that and that would
be the answer earlier so that will be a
way to do integral right you just kind
of look at the form and then going to
just notice that this right here he
actually came from a product rule
situation for the derivative okay now
why do I bring this up let's talk about
the linear differential equation if you
look at this right here we have this is
the phone now you want to begin with
dy/dx which is the derivative plus
another function times y and y it's a
function of X and then you put
everything else on the right-hand side
and you see this is linear because the
derivative okay it's just to a first
power it's nine the cosine it's not
being square whatsoever likewise for the
Y it's only to the first power if you
have Y squared if you have school of Y
this wouldn't work okay so I'm going to
show you guys the strategy of how to
deal with this kind of situation look at
this again we have the derivative and
the original hmm and I'm going to use
this idea again let me write it down for
you guys here is the product rule I just
want to put this down and I would like
to write it down as if you want to take
the derivative of two functions and I'll
just use F times G F is the first and G
is the second of course and the floor I
want to write it down for you guys is
this we will keep the first function and
we multiply by the derivative second and
we
add the second function times the
derivative of the first this is the
product rule right and now I'm going to
differentiate this for you guys you see
we have the derivative and its original
we have the derivative or in this case
with the G Prime in front of match like
right here and the original which is
right here very similar right
however over there we have an issue
because there was nothing in front and
of course it's not all the case that you
have a one in front I guess a function
wise anyways I want to just kind of
match the phone for you guys today I
will to differentiate something times y
and I'm not sure what this is yet
because I didn't have anything right
here so that's right just leave it blank
okay all right I want to differentiate
something times y you can imagine this
equity function okay all right using the
product rule we'll have something times
the derivative the second function which
I will just put down dy/dx so I can
match what's that from over there and
then we add well this right here I don't
know what it is yet
and I'll just kind of like no deeper
blank but then the second part is we
will keep the second function which is y
times y over this was and now you see
this is why I put down for the linear
differential equation I'm looking at the
product rule once again I didn't have
anything the front so that's why I leave
a space right here and then for this
right here is the one I have to
differentiate all right now the issue is
seriously we didn't have anything in
front and this may not be a good phone
that one to use so this is the idea and
the strategy we're going to use
something that we can force the left
hand side here to be the derivative that
came from the product of two functions
and what we are going to do right here
is that we'll just say we are going to
use and the thing is that we will
multiply what we mean by the inter
grading factor okay so we can force it
the left hand side to be the product of
two functions and we have to use the
product rule to match that well we are
going to use another function and that's
called the integrating factor and our
label uses mu of X okay so I'm going to
kind of just draw a line between this
and this is what we're going to do we
will multiply everything by mu of X so
from here we will end up mu of x times
dy/dx plus mu x times P of x times y and
then on the right hand side we have to
just multiply this in that and I'll just
have mu of x times Q of X just like that
all right now you see I'm just going to
put on mu of X right that's a Newseum to
have otherwise 0 see I just have blanks
that's no fun right
so the blank will be the mu of X and
right here let me just write down mu
instead of MU effects for simplicity
purposes hopefully gets don't mind today
if we differentiate mu which is the
function of x times y which is also the
function of X we use the product rule we
keep the first function which is mu and
then for the second part right here see
we will have to differentiate the mu and
then put it here and now you see we can
mix and match the form mu of x times
dy/dx which is this right here and then
we add the words we have the Y right
here that's matched right we are in
order for us to come over to formula
from U of X what must happen is you'll
see me up x times P of X which P of X
was finally original
you must match moon Prime so the key is
we are going to set this to be Moon
Prime and they mean ready stand right
here for you guys
mu of X times P of X we want this to be
so let's set this to be mu prime of X
because this way you will see the
left-hand side will be the derivative of
a product of two functions right and
found this right here will be able to
come up with a formula to get mu of X
and we will multiply everything by the
mu of X the actual function and you'll
see everything will work out nicely of
course I'll show you guys an example
with actual functions so be sure to
continue to watch this video and also I
would suggest you guys to watch another
video it's called fake prada rule you
can click the link in the description
it's a lot of fun and a strategy right
here it's very similar okay
now let's continue what I want to do
next is I would like to divide both
sides by mu of X okay so that they can
so and let me write this down first
right here for you guys we will have mu
prime of X over mu of X and this is
equal to P of X and now let's look at
the left hand side we have the original
function in the denominator and the
derivative of the numerator this it's a
form that came from the derivative of a
logarithm right in another word if I
integrate this and the reason that were
integrating is toda at a somehow get rid
of the derivative right anyways let me
integrate this of course we'll have to
integrate both sides this right here
it's going to be Ln X the value of MU of
X well double check today if we
differentiate this the derivative of Ln
of X it's going to be 1 over the inside
1 over mu of X and then because of the
chain rule we multiply by the derivative
the inside that's how we get the new
prime of X ok so the integral of this is
equal to Ln of this
if you differentiate that you get that
don't worry about the plus D and on the
right hand side I will just write D
dance how this integral of P of X yet
and once again don't worry about the
plus C because we don't care about the
constant we just care about the function
part we just care about that function
part so we can multiply the original
equation by that function part every
single will come next
at the end we see that we have milk eggs
inside the Ln what can we do to clear
off the iron we do e to that e to the
right so that E and now can so finally
this is the formula and let me write
this down right here for you guys to get
milk bags and you don't even need to
worry about the absolute value because
once again you can have positive and one
negative function both of them will work
you just need to have one function that
you need to make things work so let me
just write down mu of X this is equal to
e okay and the power is an integral you
have to integrate P of X DX this is how
you are going to find out what that
integrating factor is and now let me
show you guys an example so every single
make much more sense so let me show you
guys how can solve this linear
differential equation with the strategy
that we just talked about it and if
you're doing this on your own you should
pay attention to the following we see
that we have the Y DX the first
derivative right and here we have the Y
to the first power
there's no otherwise there's no other
derivative this is indeed the first
water linear differential equation so
you can use the integrating factor the
one that just showed you guys earlier
right now this right here it's not in
the correct folder because we have this
cosine of X in the front but it's okay
because we can just go ahead and divide
everything by cosine of X right this way
we'll be able to get into the correct
form you want
we will have dy/dx plus sine of X over
cosine of X we can just write that down
tangent of X and you see right here I
put a parentheses around the X because
we are to emphasize X is the only thing
that insert the tangent the Y it's being
multiplied it with tangent of X and this
is equal to 1 over cosine of X which is
just secant of X right let's put it down
from here we are in business
because we can see that the P of X this
right here is positive tangent of X from
here we will be able to work out a
formula to find ourselves an integrating
factor and yes you have to know that
formula so right here let me write it
down
mooom of X for that integrating factor
it's going to be e where's the base and
the power is an integral right we have
to integrate P of X and let me just
write this down for you guys we have to
integrate tangent of X right here in
this situation and let's just go ahead
and work things out first work out the
integral the integral of tangent of X is
Ln absolute value of C and X right
so we still have the e for the base but
this is going to give us Ln absolute
value of secant of X
don't worry about putting down classy
right here we're not because we just
need one integrated factor with just in
that function part the plus see later on
yes you will be found the process of
solving the differential equation but
for the integrating factor don't worry
about it and in fact what I'm going to
do next is I will just try to cancel the
e and ELN and I'm just going to say mute
legs instead of saying absolute value of
secant X I'm just gonna use the positive
version secant of X this is it okay and
the reason for that is we just want to
get we just need to have one function
you don't want to multiply with absolute
value of secant X you don't multiply by
negative nobody that's negative use the
positive version pass the secant of X
this right here is a function that we
are going to multiply throughout with
our equation everything will work out
nicely and I know I didn't put on the
parentheses right here right here but
they shouldn't bother you too much right
anyways let's just multiply by secant of
eggs throughout this equation and name
it ready stand right here for you guys
you want to apply everything by secant
of X and of course we want to be sure to
distribute distribute distribute and let
me put on the result right here first
we'll have secant of x times dy/dx and
let me write it down right here
next we add your words secant of x times
tangent X right so let me put this down
secant of x times tangent of X and then
we still have this Y right here and on
the right hand side secant of x times
secant of X we have secant square of X
all right here is the punch line on the
left hand side this you see we have
secant of x times the derivative Y plus
y times the derivative of the secant X
right
this right here is nothing but just the
derivative of secant x times y you see
if you go back to the phone earlier this
was the move X and the y and once again
you can always write down the derivative
and you just differentiate it to
double-check the relative secant x times
y is you keep the second egg and times
the derivative of Y and then you add you
keep the white right here and you
multiply by the derivative second X
which is secant tangent X so nice and
you see this right here is just that
will force you to be the derivative of
the product is a nap the right hand side
is still the secant squared of X okay
and the left hand side we have a
derivative how can we get rid of this
derivative well we can just integrate
right and of course we will have to
integrate both sides on the death inside
this and that will cancel and we just
have this right so that's right down
secant of x times y I know this looks
sexy but this is secant of x times y and
then this is equal to the integral
secant squared X this is just tangent of
X and then right here I will put down
the plus C and this is because you know
we are solving that differential
equation this is the constant that we
need right now at the end of course we
want to isolate the Y it's Napper to do
right but let me just write this as 1
over cosine of X first
and this is x y and this is equal to
tangent of X let me just ready S sub X
over cosine of X and we still have this
plus C because if I do care the equation
this for now we can just multiply
everything by cosine of X this and that
cancel out and we will just have the
wire left on the left hand side and then
be sure you have to distribute right
multiply everything by cosine of X
cosine of X times this will just have
sine of X and then cos of X times plus C
we will just put down plus C times
cosine of X and this right here is the
general solution to that differential
equation and of course suppose you have
initial value let's say if you have this
equation here along with an initial
value that say y of pi is equal to 2
well now we can just plug in these
values into this phone and so for C this
means X is equal to PI and Y will be 2
now that's just plug in 2 into this Y
will have 2 this is equal to sine of pi
right because X is PI so let's write
that down
sine of PI plus C we don't know yet and
we multiply by cosine of X which is the
same as constant pi and just work this
out real quick sine of pi is 0 cosine of
PI is negative 1 so this means we have
the 2 right here and this is just 0 and
we have just C times negative 1 that's
negative C and of course we can justify
what set by negative 1 C will give us
maybe 2 at the end if you have this
initial value the solution will be Y
it's equal to sine of X
and here we have the plus C but you know
C is negative two so we have minus two
times cosine of X this right here will
be the answer with this initial
condition so hopefully you guys like
this