WEBVTT
Kind: captions
Language: en
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hello friends welcome to my lecture on
solution of second order linear differential
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equation by changing the dependent variable
so we begin with a differential equation
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linear differential equation of second
order in the standard form that is y double
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dash plus p y dash plus q y equal to r where
p q r are continuous functions of x on an
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interval say j ok in this method what we do
is when in my previous lecture we studied
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how to find the solution of this second
order linear differential equation with variable
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coefficients when one integral included
in the complimentary function is known
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now there are cases of second order differential
equations where one integral included in
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the complimentary function is not easy to
find so if an integral included in the complimentary
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function is not obvious by instruction or
we cannot determine it easily then what we
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will do is we reduce the given differential
equation of second order to the normal form
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by normal we means we will reduce it to a
form where the first order derivative term
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is missing so how we do it let us see in
the second slide let us put y equal to
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u v in the equation one
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so we will put y equal to u in equation one
then what do we get so we have y double dash
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plus p y dash plus q y equal to r let us put
y equal to u v then y dash is equal to u dash
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v plus u v dash y double dash will be equal
to y double dash v plus u dash v dash n plus
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u dash v dash plus u v double dash let us
replace these values in the differential equations
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one
so y double dash plus p y dash plus q y equal
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to r will then become u double dash v plus
two u dash v dash plus u v double dash plus
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p into y dash so u dash v plus u v dash plus
q into u v equal to r ok now let us write
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it as a second order differential equation
in v so we have u v double dash first we write
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this term u v double dash then we write v
dash times so v dash times what we get two
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u dash plus p u and then we write the coefficient
of v the coefficient of v is u double dash
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plus p u dash plus q u equal to r ok
let us divide it by u ok the coefficient of
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v double dash so r we can write v double dash
plus v dash times two u dash by u plus
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p plus v times u double dash by u plus
p times u dash by u plus q equal to r by u
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ok
now we are going to reduce this second order
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differential equation in v to the normal form
normal form means where the first order derivative
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term is missing so let us put two u dash by
u plus p equal to zero ok so putting
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we get u dash by u equal to minus half p when
we integrate this with respect to x what we
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get is l n u equal to minus half integral
p d x ok so this will give you u equal to
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e to the power minus half integral p d x ok
now let us say suppose so this term is
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now zero ok with this choice of u now let
us put this as say i ok so if i call i let
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i be equal to u double dash by u plus p times
u dash by u plus q then let us simplify this
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expression for i we can see here that u
dash by u equal to we know that u dash by
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u equal to minus half p
so if we differentiate with respect to x what
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do we get u double dash into u minus u dash
into u dash divided by u square equal to minus
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half d p by d x [or/r] if you divide by u
square you get u double dash by u minus
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u dash by u whole square equal to minus half
d p by d x now u dash by u equal minus half
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p so let us put here so u double dash by u
is equal to minus one by four p square equal
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to minus half d p by d x so let us replace
the value of u double dash by u here and the
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value of u dash by u and let us find find
a simplified expression for i ok
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so i will be equal to one by four p square
minus half d p by d x ok p times u dash by
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u is minus half p plus q so this is minus
half p square that is one by four p square
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so we get minus one by four p square so this
q minus half d p by d x minus one by four
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p square so we get a simplified expression
for i and let us denote this r by u by
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x ok so let us say further let r by u equal
to s so then we shall have
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d square p by d x square plus i into v equal
to s where i is again i write i equal to
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q minus half d p by d x minus one by four
p square and s is equal to r by u and u is
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e to the power minus half integral p d x
so r times e to the power half integral p
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d x now let us note here that this equation
this equation is called as the normal form
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let us say let me call it as equation two
the equation two
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is called the normal form
now we can integrate this equation or
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we can find a solution of this general solution
of this equation provided r is a constant
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if r is a constant then this will be linear
differential equation of second order with
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constant coefficients r i is a some constant
divided by x square which if you assume
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that i is some constant divided by x square
then multiplying the equation by x square
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it will reduce to a quasi euler equation
and you know how and we know how to solve
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a quasi euler equation so we have a catch
here so let us note that
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the equation two can can be solved if either
i is a constant or i is a constant divided
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by x square
so yes just as we said in the beginning that
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only a special class of second order
linear differential equation can be solved
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with variable coefficients so there is
a catch here such equations where
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i is either a constant or a constant divided
by x square can be solved by reducing the
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given equation to the normal form so
let us see how this is these to the
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this this the normal form we are given the
value of or we have the value of i and s and
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this is what we have noted just now if the
value of i is a constant or a constant divided
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by x square then the constant becomes readily
integrable this form is said to be the normal
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form
now let us take an example and see how we
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will use this method so ok so let us say
d square y over d x over that is y double
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dash we have y double dash minus four x y
dash plus four x square minus one into
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y equal to minus three e to the power x square
sin two x so when we compare it with the standard
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form we have the value of p equal to minus
four x q equal to four x square minus one
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and r equal to minus three e to the power
x square sin two x
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so let us find first when we reduce this
equation to the normal form the value of u
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ok u is equal to e to the power minus half
integral p d x so e to the power minus half
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integral minus four x d x so this is e to
the power integral two x d x and which gives
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you e to the power x square
so we know the value of u now let us find
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i i is equal to q minus half d p by d x minus
one by four p square so q is four x square
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minus one minus half d p by d x when you find
d p by d x here you get minus four and then
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minus one by four p square so you get minus
four x square whole square so what we get
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is four x square minus one and here we
get plus two and here we get minus four
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x square
so this cancels with this two minus one is
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one so we get i equal to a constant and therefore
we can see from this article that this
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given equation is readily integrable the normal
form is now
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d square v over d x square plus i into v equal
to x so i is equal to one and s is r over
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u ok so s is r over u and r over u is minus
three e to the power x square sin two x divided
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by u u is e to the power x square
so e to the power x square gets cancelled
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and we get minus three sin two x so we get
d square v by d x square i is one so plus
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v equal to minus three sin two x now this
is a linear differential equation of second
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order with constant coefficients so we
can write auxiliary equation auxiliary equation
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is m square plus one equal to zero so we
get m equal to plus minus one plus minus i
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sorry so complimentary function y c
x is equal to a cos x plus v sin x where a
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and v are arbitrary constants and particular
integral y p x equal to one over d square
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plus one where d represents d over d x operating
on minus three sin two x ok so this is minus
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three times one over d square plus one operating
on sin two x when we replace d square by
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minus two square the denominator does not
become zero so this is minus three one over
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minus two square that is minus four plus one
sin two x so this is minus three minus three
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cancel and you get sin two x
so this way we get the value of v s so hence
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v is equal to y c x plus y p x so a cos
x plus v sin x plus sin two x now general
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solution is y equal to u into v that is u
is e to the power x square so e to the
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power x square into a cos x plus v sin x plus
sin two x so this is the general solution
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in the case of example one
let us take one more example to make the
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things clear so we have y double dash in example
two minus two tan x plus y into y dash
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plus five by equal to e to the power x into
sec x so here p is equal to minus two tan
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x q equal to five and r equal to e to the
power x into sec x now u is equal to y over
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formula e equal to u equal to e to the power
minus half integral p d x so this is e to
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the power minus half integral minus two tan
x d x which is e to the power integral
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now integral of tan x is log sec x so we have
e to the power l n sec x ok so this is equal
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to sec x so we have found the value of u let
us now determine i i equal to q minus half
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d p by d x minus one by four p square so q
is five minus half d p by d x will be minus
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two sec square x the derivative what tan x
is sec square x minus one by four p square
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p square is four tan square x ok so this will
cancel with this and we have five this
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will cancel with this we have five plus sec
square x minus ten square x sec square x equal
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to one plus ten square x so this is one sec
square x minus ten square x is one so this
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is six so i is again a constant and therefore
we can easily integrate the normal form
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that is d
so thus we have the normal form as
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i into v that is six into v equal to s
s is r by u so s is equal to r by u which
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is equal to or e to the power x sec x divided
by sec x so we cancel this and now this is
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easy to integrate auxiliary equation is m
square plus six equal to zero it is our auxiliary
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equation so m equal to plus minus i root
six and therefore complimentary function is
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c one cos root six x plus c two sin root six
x and particular integral y p x is equal to
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one over d square plus six operating on e
to the power x
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now we can apply one over f d e to the power
a x formula because here when d is replaced
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by a a is one here f d does not become zero
so one over one square plus six e to the
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power x so we have e to the power x by seven
so we have thus v equal to c f so c one cos
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root six into x plus c two sin root six into
x plus one by seven e to the power x this
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is the and we multiply it by u to get the
general solution u is sec x so hence the general
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solution is y equal to u into v equal to sec
x times c one cos root six x plus c two
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sin root six x plus one by seven e to the
power x
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so this is the general solution of equation
given in the case of example two in my
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next lecture i will discuss how to solve
second order linear differential equation
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with variable coefficients by changing changing
the independent variable in this article we
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have changed the dependent variable from y
to v in that next lecture we shall be changing
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the independent variable from x to z now how
how we get the relationship between x and
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z that we shall see in the net lecture ok
thank you for your attention