hello friends welcome to my lecture on methods for finding particular integral for second
order linear differential equations with constant coefficients on this topic there will be three
lectures this is my first lecture on this topic let us consider the second order
differential equation y double dash plus a y dash plus b y equal to r x here a and b
are constants and r x is any function of x so it is a second order linear differential
equation with constant coefficients now we have already discussed how to find the
general solution of the associated homogenous linear differential equation that is y
double dash plus a y dash plus b y equal to zero now we are going to consider we
are going to find the particu a particular solution of this non homogenous equation
which is known as the particular integral and we have already seen that once the
general solution of the associated homogenous linear differential equation is known and
a particular solution of the non homogenous equation is known then their sum gives
us the general solution of the non homogenous linear differential equation of second order
so this equation of second order can be written in a simple form by using the
differential operator d d is the differential operator d over d x so this what
do we mean by d what do we mean by an operator an operator t is a transformation from a space
of functions v into the other space of function w these are vector spaces that transforms
a function f in v to another function t v in w let us say v one is the set of all
differentiable functions f on i i is any subset i is any interval subset of r and d
be the differential operator defined on v one then d f at x d f at x means the derivative
of f with respect to x v also denoted by f dash for example if you take f x equal
to x to the power n then d of x n that is the derivative of x to the power n with
respect to x is n times x to power n minus one similarly derivative of cos x with
respect to x is minus sin x etcetera since d is a linear operator we can see that
for example if you use you can see here d is a d is an operator from b into
w whenever you take any two function f and g in b since b is a vector space and t
is a linear operator t d a f plus b g becomes a d of plus b d g where f g belongs to b one
and a b r constants if f is a twice differentiable function on i we write d d f as d of f dash
or we can write it as d f dash over d x which is same as f double dash s f double dash x
the twice twice f double dash as d square f over d xx square we write d d f s d square
f also so this way we can define higher order derivatives of a with respect to x
now the given equation then y double dash plus a times y dash x plus b times y x equal
to r x can be written in a simpler form as d square y plus a times d y plus b times
y equal to r x r we can write it as d square plus a d plus b operating on y equal to r
x now let us say let f d if you take f d equal to d square plus a d plus b then you can write
it as f d y equal to so f d is a is an operator which when x on y gives you d square by plus
a d by plus b y now let us see then the particular integral which we also denote by
p i particular integral be in short we also write as p i and we have the notation
that we have taken for the particular integral is y p x so y p x is given y one over f d
operating on r x now let us see how we get this form of
the particular integral y one over f d we mean that it is the inverse of the operation
that is defined by f d that is to say if you operator by one over f d on this equation
let us let me on this equation let us say then um when we operate on the equation one
by one over f d
then since one over f d defines an operation which is inverse to the operation defined
by f d we get the left hand side equal to y and which is equal to one over f d
r x ok since y p solution of the sorry y p is a particular solution of the differential
equation f d y p i can write it f d y p ok so f d y p equal to this and so y p equal
to one over f d r x we have assumed that y p x is a particular solution of the non
homogenous equation so f d y p f d y p will be equal to r x and when we operate on
that equation by the inverse operator one over f d then what do we get is y p equal
to one over f d r x so y p x the expression for y p x is given
by one over f d r x now we can see that this f d f d can be broken into factors which
are d minus m one into d minus m two and m one m two are at the roots of the equation
a m m square plus m plus b equal to zero see in this equation let us write the auxiliary
equation
the auxiliary equation for the given differential equation is m square plus a m plus b equal
to zero let us say the roots of this equation are let us say the roots of this equation
be m one and m two then m square plus a m square plus b is equal to m minus m one into
m minus m two we can write it as m minus m one and m minus m two now so then what we
will have if you calculate this m square minus m times m one plus m two and then plus m one
m two so cutting the coefficients of the various powers of m both sides we get m one plus m
two equal minus a and m one m two equal to b
now d is f d is we are writing as d minus m one into d minus m two
so d minus m one into d minus m two we can operate d minus m one on d that gives you
m two is a m two is a constant we can write it as m two times d minus m one so this
gives you d square minus m one plus m two into d plus m one m two now making use of
m one plus m two equal to minus a and m one m two equal to b what do we get is which is
nothing but f d so f d can be written also as d minus m one d minus m two we can we
can write d minus m two into d minus m one also because the that will also give
the same thing so order of factors here is immaterial we can write them in any order
now so when we want to find the particular integral y p x we can write one over f d s
one over d minus m one
now what we do is we first operate on r x by the [op/operator] operator one over d minus
m two and then whatever is the result of that operation we operate on that result by
the operator one over d minus m one as i said in my as i just said this can also be
written as the order of factors here is immaterial so we can again we can also operate one
over d minus m one on r x and then whatever is the result of that operation we can operate
on that by the operator one over d minus m two to get the particular integral in what
manner we will have to operate by this operators on r x depends on the function r
x which we shall see when we solve some problems on this
now so this is one way of finding particular integral y p s where we operate by v operators
one over d minus m one and one over d minus m two now there is another method by which
we can find the particular integral we can resolve one over f d into partial fractions
so one f d is if it is resolved into partial fractions we have one over f d as one over
f d is resolved into partial fractions as n one over d minus m one and plus n two
over d minus m two so that one over one over f d operates on r x we have the operators
n one over d minus m one operating on r x plus n two over d minus m two operating on
r x now now we can see the following i mean see
if you look at this expression to find one over f d r x to find one over f d r x you
have to operate on r x by the operator one over d minus m one and then whatever is
the result of that you have to multiply that by the constant n one similarly here when
you want to find this function of x you have to operate on r x by the operator
one over d minus m two and then whatever is the result then you have to multiply that
y and two so here the what we are doing is we are operating on r x by an operator of
the kind one over d minus alpha where alpha is a constant in the previous method here
also if you apply this method to find the particular integral we are operating on r
x y one over d minus m two r one over d minus m one and then we operate by the other operator
of the same kind on r x on the result of the operation which we get by operating on
r x by one of the operators so when operator one over y r d minus m two on r x whatever
function we get we operate on that by one over d minus one and so here also we are operating
on r x by an operator of the kind one over d minus alpha
so in both the methods what we are doing we are doing in order to find the particular
integral we are operating by an operator of the kind one over d minus alpha on r x so
let us see how we get the when we operate on the r x by an operator of
the kind one over d minus alpha what function of x v get see let us see let us find this
so let us say that let one over d minus alpha when operates on r x you get a function of
x say u x so u x equal to one over d minus alpha r x we get a function say u x all right
u is a function of x now let us operate bo bo on this equation by d minus alpha
on both sides so d minus alpha
when op when we operate by d minus alpha on this equation since d minus alpha and one
over d minus alpha are inverse to each other ok we will get r x equal to d minus alpha
u now since d is d over d x so we get d u by d x minus alpha u so what we get is
d u by d x minus alpha u equal to r x which is a linear differential equation of the first
order with constant coefficient alpha is a constant and we know how to find the solution
of this differential equation d u by d alpha d u by d x minus alpha u equal to r
x we find the par the integrating factor of this equation so integrating factor is
e to the power integral minus alpha d x which is equal to e to the power minus [alp/alpha]
alpha x then the then we know that we when we multiply this equation by the integrating
factor e to the power minus alpha x the left hand side becomes an exact it becomes an
exact equation so e to the power minus alpha x into u d over d x of that is equal
to e to the power minus alpha x into r x after multiplying by the integrating factor the
left hand side becomes d over d x of e to the power minus alpha x into u which gives
now let us integrate this equation with respect to x so this gives you e to the power
minus alpha x into u equal to integral e to the power minus alpha x into r x d x
e to the power minus alpha x is never zero so we can write it further as i am not writing
a constant of integration here because the particular integral is free from arbitrary
constants so one over d minus alpha when f acts on r x the function u that we get is
given by this formula e to the power alpha x integral e to the power minus alpha x into
r x d x now now let us see for example let us see second order differential equation
with constant coefficient and see how we can find the particular integral ok so d square
y over d x square plus n square by so or the auxiliary equation is m square plus n square
equal to zero which give us the complex roots m equal to plus minus i n and so the complimentary
function y c x is given by a cos n x plus b sin n x now let us find the particular integral
here so particular integral y p x this is equal to one over d square plus n square because
the differential equation given differential equation can be written as d square plus
n square operating on y equal to sec n x so one over d square plus f d is d square
plus n square so one over d square plus n square sec n x ok
now let us write the the function f d in in terms of the its factors one over d
minus i n d plus i n now as i said we can then find the opera
the result of the operation of one over d plus i n on sec n x first are the result
of the operation of operating by one over d minus i n on sec n x [fro/first] first ok
and then we can on the result of what which which we get on that we can operate
by the other thing now here we can also break it into partial fractions let us suppose
we are we wish to bracket into partial fractions then we will write it as
one over d minus i n minus one over d plus i n divided by two i n that gives us one
over one minus one over d minus i n into one over into d plus i n now let us
find let us apply this formula this is nothing but one over d minus alpha operating on r
x ok so let us make use of this formula so let us find first we find
one over d minus i n operating on r sec n x so alpha is i n here so e to the power
i n x integral e to the power minus i n x into sec n x d x
now this is nothing but e to the power i n x integral let us apply the eulers formula
here e to the power minus i theta is cos theta minus i sin theta so cos n x
minus i sin n x into sec n x d x now this is further if you solve it this gives you
cos n x into sec n x is one minus i tan n x
d x now we can easily integrate this this is equal to e to the power i n x integral
d x is x minus i times the integral of tan n x is one over n l n sec n x because l n
sec n x when you differentiate with respect to x what you get is one over sec n x then
sec n x into tan n x into n so we get this this is what we get and then we can write
further this as e to the power i n x is cos n x plus i sin n x and this is to be multiplied
by so if i write l n one over cos x then it will be minus l n cos x so x plus i y n
l n cos x cos n x ok so what we will get let us multiply this
this will be x cos n x x cos n x plus i times x sin n x
and then i y m cos n x l n cos n x and then mi i square is minus one so minus one over
n minus one over n sin n x l n cos n x so this is what we get similarly we can find
the value of one over d plus i n one over d plus i n operating on sec n x if you obtain
that in a similar we will get so one over d plus i n sec n x if we get we
simply have x cos n x minus i x sin n x and then minus i by one n cos n x l n cos x n
x and then plus no sorry minus one by n sin n x l n cos n x ok so then we we we one
ob one over d minus i n sec n x we have to subtract one over d plus i n sec n x if we
so that and divide two i n what do we get so hence y p x is one over two i n now from
this value from this value we subtract this so x cos n x will cancel i x sin n x
and i x sin x will add up to i x sin n x we will get and then i y n cos n x l n cos n
x and i y n cos n x l n cos n will add up so two i y n cos n x l n cos n x we get and
then this expression and this expression will cancel each other so we have we divide by
two i n so we get x sin n x divided by n and then here we get cos n x
so hence the general solution of the differential equation d square over by d x square plus
n square by a equal to sec n x we can write as y x equal to y c x plus y p x which is
equal to a cos n x plus b sin n x plus x sin n x by n plus x sin n x by n plus
cos n x l n cos n x divided by n square so that the general solution of this differential
equation and now we are going to study how to find
the particular integral of one r x of these some special forms say the first
result that we are going to consider is the result when r x is in exponential function
so suppose r x is e to the power alpha x where alpha e some real number alpha can be any
real r complex constant so one over f d when operates on e to the power alpha x let
us find the particular integral when r x is e to the power alpha x so one over
f d e to power alpha x gives you one over f alpha e to the power alpha x provided f
alpha is not equal to zero we can make use of this formula to determine the particular
integral when r x is of the type e to the power alpha x now to prove this formula we
can see that when we differentiate e to the power alpha x with respect to x what we get
is alpha times e to the power alpha x if we again differentiate this with respect to
d that is we find the second derivative of e to the power alpha x with respect to x
which is given by d square e to the power alpha x we get alpha x square e to the power
alpha x now so f d when operates on e to the power
alpha x gives you f d is the d square plus a d plus b so d square plus a d plus b when
operated on e to the power alpha x we get alpha x square e to the power alpha x then
a times alpha e to the power alpha x then b into e to the power alpha x so we can we
get alpha a square plus a alpha plus b into e to the power alpha x now this alpha a square
plus alpha a alpha plus b can be written as f alpha so we get f alpha into e to the power
alpha x so f d when operates on e to the power alpha x
gives us now lets suppose that f alpha is not equal to zero so when f alpha is not equal
to zero let us first operate on both sides by the operator one over f d
since one over f d and f d are inverse operators we have left hand side as e to the
power alpha x and the right hand side is one over f d operating on f alpha e to the power
alpha x now now f alpha is a non zero algebraic
multiplier we have assumed alpha to f alpha to be non zero it is a non zero algebraic
multiplier so this is same as f alpha times one over f d operating on e to the power alpha
x now let us divide this equation by f alpha so dividing
we get one over f d operating on e to the power alpha x equal to e to the power alpha
x over f alpha so this is the proof and this method fails when it happens
that f alpha turns out to be zero let us apply this method to find the particular integral
in the case of the example d square y over d x square plus y equal to e to the power
x plus one whole square so this differential equation can be written as d square plus one
y equal to e to the power x plus one whole square let us first first write the
complimentary function so the auxiliary equation is
m square plus one equal to zero that is m equal to plus minus i so complimentary function
y c x equal to a cos x plus b sin x let us find the particular integral y p x this is
given by one over f d f d is d square plus one operating on r x r x is e to the power
x plus one whole square now this e to the power r x is e to the power x plus one whole
square so we have to square this and write
now this is nothing but we operate by one over d square plus one on each of the terms
of this bracketed expression ok now this is one over d square plus one
e to the power two x can be found from the formula one over f d e to the power alpha
x equal to e to the power alpha x divided by f alpha because here alpha is equal to
two and when we replace d y in f d d square plus one when we replace d y alpha there it
is f alpha is not equal to zero so this is equal to one over two square plus one e to
the power two x now two is an algebraic multiplier i can write it two times one over
d d square plus one operating on e to the power x so alpha is one here so i write one
over one square plus one e to the power x and here one can be regarded as e to the power
zero x e to the power zero x so alpha can be taken as zero and we can get the value
of this so this is equal to further one over five e to the power two x then two over two
so e to the power x and then alpha is zero here so one over zero square plus one so plus
one so then general solution is y x equal to y c x plus y p x
now let us go to short methods of finding one over f d sin c x and one over f d sin
cos c x where c is any real number now we are we are finding short methods of finding
the result of operating y one over f d on sin c x of one over f d operating on
cos c x now if n is a positive integer notice that d square to the power n sin c x is
equal to minus c square to the power n and sin c x let us prove this when we operate
y c on sin c x that is we differentiate sin c x with respect to x we get c times cos c
x let us again operate by d on this c cos c x so that is d square nine c x gives you
d of c cos c x which is equal to minus c square sin c x i can write it as minus c square times
c x now d cube then sin c x will be minus c square into c cos c x and d to the power
four sin c x will be equal to minus c square into minus c square into cos c x which can
be written as minus c square is square cos sin c x sorry sin c x so minus c square is
square sin c x and thus we can say that d square is square sin c x is nothing but
minus c square is square sin c x it can be by [math/mathematical] mathematical induction
we can extend this to and so on d square to the power n sin c x is equal to minus c
square to the power n sin c x similarly we can find d square to the power
n cos c x this will come out to be minus c square to the power n cos x now let us
see suppose f d the expression of f d contains only even powers of d and we denote it by
phi d square then from here we can see that phi d square one x on sin c x the
effect of this operation of phi d square on sin c x will be phi minus c square sin c x
because here we see that whenever we operate by d square on sin c x we get d square
to power n on sin c x we get minus c square to the power n sin c x so phi d square
when updates on sin d x will get phi minus c square sin c x now so what we get is
when phi d square operates on sin c x we get phi minus square sin c x
now let us assume that phi minus a square is non zero phi minus c square is non non
zero then if we operate by one over phi d square on both sides
phi d square and one more phi d square are inverse to each other will get sin c x equal
to one over phi d square phi minus c square is a algebraic multiplier
non zero we divide it by minus phi c phi minus c square and get one over phi d square operating
on sin c x equal to sin c x divided by phi minus c square so similarly we can show that
one over phi d square one operates on cos c x we get one over c minus c square cos c
x provided phi minus c square is non zero more generally if we have to operate by
one over phi d square on sin c x plus d we are c and d are com constants then we get
one over phi minus c square sin c x plus d b can similar the prove this in a similar
manner now let us take an example suppose we take
the example one example one can be written as d square minus four d plus one operating
on y equal to two cos two x we know how to find the complement function so i will discuss
only particular integral so y p x equal to one over f d which is d square minus four
d plus one operating on two cos two x two is an algebraic multiplier i can write it
outside now a here c here is two so we replace d square by minus c square so one over minus
two square for d square we write and minus four d plus one be leave just like that cos
two x this is two times one over minus four plus one so we get minus four d minus three
cos two x and this also equal to minus two times one over four d plus three operating
on now what we do is to [ge/get] get the operation of one four d plus three operating on cos
two x let us operate by four d minus three on cos two x and one over four d minus cos
three so minus two times we we write four d minus three and one over four
d minus three they are inverse of each other so
we get this now this will be equal to minus two
in the denominator we have sixteen d square minus nine cos two x again replace d square
by minus a square we shall have so minus two times d square by minus a square so sixteen
times minus two square minus nine cos two x so this is minus sixty four minus nine so
that is minus seventy three so we get two by seventy three and then four d minus three
operating on cos two x now we have to operate by the operator four
d minus three on cos two x so we get two times two over seventy three four d cos two x means
derivative of cos two x with respect to x so minus eight sin two x minus three cos two
x so this is the particular integral in this case ns we add to this the complementary function
and write the general solution now in the case of example two d square plus one y
equal to cos square x by two what we do is to find the particular integral we shall
write now let us find the particular integral in the case of example two so in the case
of example two particular integral y p x is equal to one over d square plus one cos square
x by two since this is cos square x by two we have
to convert it to cosine x function to apply the formula that we have just now
proved so we shall write it as one over d square plus one one plus cos x by two and
this will then done by two one over d square plus one operating on one plus cos x so this
will be one by two one over d square plus one operates on one one can be regarded as
e to the power zero x so we will get one over zero square plus one and then one over d square
plus one operates on cos x we shall replace d square one by d square by minus one square
plus one so which is which is not defined so how we shall deal with this in example
we shall see in in in our lecture tomorrow because here one over d square plus one becomes
zero when d square is replaced by minus a square so this example we cannot find a
particular example like this this will be found this will be tackled in my lecture
which i give tomorrow thanks