WEBVTT
Kind: captions
Language: en
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hello friends welcome to my lecture on methods
for finding particular integral for second
00:00:23.610 --> 00:00:29.410
order linear differential equations with constant
coefficients on this topic there will be three
00:00:29.410 --> 00:00:36.450
lectures this is my first lecture on this
topic let us consider the second order
00:00:36.450 --> 00:00:42.640
differential equation y double dash plus a
y dash plus b y equal to r x here a and b
00:00:42.640 --> 00:00:47.750
are constants and r x is any function of x
so it is a second order linear differential
00:00:47.750 --> 00:00:52.930
equation with constant coefficients now we
have already discussed how to find the
00:00:52.930 --> 00:00:58.000
general solution of the associated homogenous
linear differential equation that is y
00:00:58.000 --> 00:01:04.560
double dash plus a y dash plus b y equal to
zero now we are going to consider we
00:01:04.560 --> 00:01:10.500
are going to find the particu a particular
solution of this non homogenous equation
00:01:10.500 --> 00:01:15.560
which is known as the particular integral
and we have already seen that once the
00:01:15.560 --> 00:01:21.080
general solution of the associated homogenous
linear differential equation is known and
00:01:21.080 --> 00:01:26.800
a particular solution of the non homogenous
equation is known then their sum gives
00:01:26.800 --> 00:01:31.840
us the general solution of the non homogenous
linear differential equation of second order
00:01:31.840 --> 00:01:38.580
so this equation of second order
can be written in a simple form by using the
00:01:38.580 --> 00:01:45.491
differential operator d d is the differential
operator d over d x so this what
00:01:45.491 --> 00:01:51.750
do we mean by d what do we mean by an operator
an operator t is a transformation from a space
00:01:51.750 --> 00:01:56.980
of functions v into the other space of function
w these are vector spaces that transforms
00:01:56.980 --> 00:02:02.780
a function f in v to another function t v
in w let us say v one is the set of all
00:02:02.780 --> 00:02:09.800
differentiable functions f on i i is any subset
i is any interval subset of r and d
00:02:09.800 --> 00:02:16.220
be the differential operator defined on v
one then d f at x d f at x means the derivative
00:02:16.220 --> 00:02:22.310
of f with respect to x v also denoted by f
dash for example if you take f x equal
00:02:22.310 --> 00:02:29.990
to x to the power n then d of x n that
is the derivative of x to the power n with
00:02:29.990 --> 00:02:35.890
respect to x is n times x to power n minus
one similarly derivative of cos x with
00:02:35.890 --> 00:02:42.910
respect to x is minus sin x etcetera since
d is a linear operator we can see that
00:02:42.910 --> 00:02:50.420
for example if you use you can see here
d is a d is an operator from b into
00:02:50.420 --> 00:02:56.100
w whenever you take any two function f and
g in b since b is a vector space and t
00:02:56.100 --> 00:03:04.110
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
00:03:04.110 --> 00:03:12.450
and a b r constants if f is a twice differentiable
function on i we write d d f as d of f dash
00:03:12.450 --> 00:03:19.060
or we can write it as d f dash over d x which
is same as f double dash s f double dash x
00:03:19.060 --> 00:03:30.430
the twice twice f double dash as d square
f over d xx square we write d d f s d square
00:03:30.430 --> 00:03:35.340
f also so this way we can define higher order
derivatives of a with respect to x
00:03:35.340 --> 00:03:51.840
now the given equation then y double dash
plus a times y dash x plus b times y x equal
00:03:51.840 --> 00:04:05.870
to r x can be written in a simpler form as
d square y plus a times d y plus b times
00:04:05.870 --> 00:04:22.620
y equal to r x r we can write it as d square
plus a d plus b operating on y equal to r
00:04:22.620 --> 00:04:31.580
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
00:04:31.580 --> 00:04:44.400
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
00:04:44.400 --> 00:04:54.400
a d by plus b y now let us see then the
particular integral which we also denote by
00:04:54.400 --> 00:05:01.410
p i particular integral be in short we also
write as p i and we have the notation
00:05:01.410 --> 00:05:11.470
that we have taken for the particular integral
is y p x so y p x is given y one over f d
00:05:11.470 --> 00:05:16.000
operating on r x
now let us see how we get this form of
00:05:16.000 --> 00:05:22.530
the particular integral y one over f d we
mean that it is the inverse of the operation
00:05:22.530 --> 00:05:29.570
that is defined by f d that is to say if
you operator by one over f d on this equation
00:05:29.570 --> 00:05:37.270
let us let me on this equation let us say
then um when we operate on the equation one
00:05:37.270 --> 00:05:49.310
by one over f d
00:05:49.310 --> 00:05:54.750
then since one over f d defines an operation
which is inverse to the operation defined
00:05:54.750 --> 00:06:05.860
by f d we get the left hand side equal to
y and which is equal to one over f d
00:06:05.860 --> 00:06:11.790
r x ok since y p solution of the sorry
y p is a particular solution of the differential
00:06:11.790 --> 00:06:17.930
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
00:06:17.930 --> 00:06:25.300
to one over f d r x we have assumed that y
p x is a particular solution of the non
00:06:25.300 --> 00:06:32.320
homogenous equation so f d y p f d y p will
be equal to r x and when we operate on
00:06:32.320 --> 00:06:38.550
that equation by the inverse operator one
over f d then what do we get is y p equal
00:06:38.550 --> 00:06:42.630
to one over f d r x
so y p x the expression for y p x is given
00:06:42.630 --> 00:06:51.530
by one over f d r x now we can see that this
f d f d can be broken into factors which
00:06:51.530 --> 00:06:57.080
are d minus m one into d minus m two and m
one m two are at the roots of the equation
00:06:57.080 --> 00:07:03.540
a m m square plus m plus b equal to zero see
in this equation let us write the auxiliary
00:07:03.540 --> 00:07:16.449
equation
00:07:16.449 --> 00:07:22.900
the auxiliary equation for the given differential
equation is m square plus a m plus b equal
00:07:22.900 --> 00:07:43.190
to zero let us say the roots of this equation
are let us say the roots of this equation
00:07:43.190 --> 00:07:59.560
be m one and m two then m square plus a m
square plus b is equal to m minus m one into
00:07:59.560 --> 00:08:09.200
m minus m two we can write it as m minus m
one and m minus m two now so then what we
00:08:09.200 --> 00:08:21.360
will have if you calculate this m square minus
m times m one plus m two and then plus m one
00:08:21.360 --> 00:08:32.500
m two so cutting the coefficients of the various
powers of m both sides we get m one plus m
00:08:32.500 --> 00:08:41.000
two equal minus a and m one m two equal to
b
00:08:41.000 --> 00:08:53.640
now d is f d is we are writing as d minus
m one into d minus m two
00:08:53.640 --> 00:09:04.510
so d minus m one into d minus m two we can
operate d minus m one on d that gives you
00:09:04.510 --> 00:09:13.300
m two is a m two is a constant we can write
it as m two times d minus m one so this
00:09:13.300 --> 00:09:25.750
gives you d square minus m one plus m two
into d plus m one m two now making use of
00:09:25.750 --> 00:09:39.410
m one plus m two equal to minus a and m one
m two equal to b what do we get is which is
00:09:39.410 --> 00:09:46.940
nothing but f d so f d can be written also
as d minus m one d minus m two we can we
00:09:46.940 --> 00:09:54.430
can write d minus m two into d minus
m one also because the that will also give
00:09:54.430 --> 00:10:00.250
the same thing so order of factors here is
immaterial we can write them in any order
00:10:00.250 --> 00:10:13.040
now so when we want to find the particular
integral y p x we can write one over f d s
00:10:13.040 --> 00:10:24.990
one over d minus m one
00:10:24.990 --> 00:10:31.500
now what we do is we first operate on r x
by the [op/operator] operator one over d minus
00:10:31.500 --> 00:10:37.660
m two and then whatever is the result of
that operation we operate on that result by
00:10:37.660 --> 00:10:43.860
the operator one over d minus m one as i said
in my as i just said this can also be
00:10:43.860 --> 00:10:56.310
written as the order of factors here is immaterial
so we can again we can also operate one
00:10:56.310 --> 00:11:01.230
over d minus m one on r x and then whatever
is the result of that operation we can operate
00:11:01.230 --> 00:11:08.330
on that by the operator one over d minus m
two to get the particular integral in what
00:11:08.330 --> 00:11:14.649
manner we will have to operate by this
operators on r x depends on the function r
00:11:14.649 --> 00:11:17.839
x which we shall see when we solve some problems
on this
00:11:17.839 --> 00:11:24.930
now so this is one way of finding particular
integral y p s where we operate by v operators
00:11:24.930 --> 00:11:30.490
one over d minus m one and one over d minus
m two now there is another method by which
00:11:30.490 --> 00:11:37.000
we can find the particular integral we
can resolve one over f d into partial fractions
00:11:37.000 --> 00:11:55.870
so one f d is if it is resolved into partial
fractions we have one over f d as one over
00:11:55.870 --> 00:12:10.180
f d is resolved into partial fractions as
n one over d minus m one and plus n two
00:12:10.180 --> 00:12:16.510
over d minus m two so that one over one over
f d operates on r x we have the operators
00:12:16.510 --> 00:12:22.750
n one over d minus m one operating on r x
plus n two over d minus m two operating on
00:12:22.750 --> 00:12:27.650
r x
now now we can see the following i mean see
00:12:27.650 --> 00:12:34.709
if you look at this expression to find one
over f d r x to find one over f d r x you
00:12:34.709 --> 00:12:39.730
have to operate on r x by the operator
one over d minus m one and then whatever is
00:12:39.730 --> 00:12:44.970
the result of that you have to multiply that
by the constant n one similarly here when
00:12:44.970 --> 00:12:50.880
you want to find this function of x
you have to operate on r x by the operator
00:12:50.880 --> 00:12:55.120
one over d minus m two and then whatever is
the result then you have to multiply that
00:12:55.120 --> 00:13:01.980
y and two so here the what we are doing is
we are operating on r x by an operator of
00:13:01.980 --> 00:13:07.600
the kind one over d minus alpha where alpha
is a constant in the previous method here
00:13:07.600 --> 00:13:12.840
also if you apply this method to find the
particular integral we are operating on r
00:13:12.840 --> 00:13:19.990
x y one over d minus m two r one over d minus
m one and then we operate by the other operator
00:13:19.990 --> 00:13:25.029
of the same kind on r x on the result of
the operation which we get by operating on
00:13:25.029 --> 00:13:30.460
r x by one of the operators so when operator
one over y r d minus m two on r x whatever
00:13:30.460 --> 00:13:36.140
function we get we operate on that by one
over d minus one and so here also we are operating
00:13:36.140 --> 00:13:40.160
on r x by an operator of the kind one over
d minus alpha
00:13:40.160 --> 00:13:44.279
so in both the methods what we are doing we
are doing in order to find the particular
00:13:44.279 --> 00:13:50.680
integral we are operating by an operator of
the kind one over d minus alpha on r x so
00:13:50.680 --> 00:13:58.740
let us see how we get the when
we operate on the r x by an operator of
00:13:58.740 --> 00:14:25.920
the kind one over d minus alpha what function
of x v get see let us see let us find this
00:14:25.920 --> 00:14:31.389
so let us say that let one over d minus alpha
when operates on r x you get a function of
00:14:31.389 --> 00:14:43.660
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
00:14:43.660 --> 00:14:50.930
u is a function of x now let us operate bo
bo on this equation by d minus alpha
00:14:50.930 --> 00:15:04.670
on both sides so d minus alpha
00:15:04.670 --> 00:15:09.200
when op when we operate by d minus alpha on
this equation since d minus alpha and one
00:15:09.200 --> 00:15:17.870
over d minus alpha are inverse to each other
ok we will get r x equal to d minus alpha
00:15:17.870 --> 00:15:27.450
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
00:15:27.450 --> 00:15:33.460
d u by d x minus alpha u equal to r x which
is a linear differential equation of the first
00:15:33.460 --> 00:15:38.660
order with constant coefficient alpha is a
constant and we know how to find the solution
00:15:38.660 --> 00:15:45.250
of this differential equation d u by d
alpha d u by d x minus alpha u equal to r
00:15:45.250 --> 00:15:58.760
x we find the par the integrating factor
of this equation so integrating factor is
00:15:58.760 --> 00:16:05.340
e to the power integral minus alpha d x which
is equal to e to the power minus [alp/alpha]
00:16:05.340 --> 00:16:13.050
alpha x then the then we know that we when
we multiply this equation by the integrating
00:16:13.050 --> 00:16:18.270
factor e to the power minus alpha x the left
hand side becomes an exact it becomes an
00:16:18.270 --> 00:16:27.829
exact equation so e to the power minus
alpha x into u d over d x of that is equal
00:16:27.829 --> 00:16:35.300
to e to the power minus alpha x into r x after
multiplying by the integrating factor the
00:16:35.300 --> 00:16:39.980
left hand side becomes d over d x of e to
the power minus alpha x into u which gives
00:16:39.980 --> 00:16:45.110
now let us integrate this equation with
respect to x so this gives you e to the power
00:16:45.110 --> 00:16:57.050
minus alpha x into u equal to integral e to
the power minus alpha x into r x d x
00:16:57.050 --> 00:17:11.870
e to the power minus alpha x is never zero
so we can write it further as i am not writing
00:17:11.870 --> 00:17:17.289
a constant of integration here because
the particular integral is free from arbitrary
00:17:17.289 --> 00:17:26.839
constants so one over d minus alpha when f
acts on r x the function u that we get is
00:17:26.839 --> 00:17:32.080
given by this formula e to the power alpha
x integral e to the power minus alpha x into
00:17:32.080 --> 00:17:40.239
r x d x now now let us see for example let
us see second order differential equation
00:17:40.239 --> 00:17:53.169
with constant coefficient and see how we can
find the particular integral ok so d square
00:17:53.169 --> 00:18:09.980
y over d x square plus n square by so or the
auxiliary equation is m square plus n square
00:18:09.980 --> 00:18:17.879
equal to zero which give us the complex roots
m equal to plus minus i n and so the complimentary
00:18:17.879 --> 00:18:31.610
function y c x is given by a cos n x plus
b sin n x now let us find the particular integral
00:18:31.610 --> 00:18:44.259
here so particular integral y p x this is
equal to one over d square plus n square because
00:18:44.259 --> 00:18:49.070
the differential equation given differential
equation can be written as d square plus
00:18:49.070 --> 00:18:55.269
n square operating on y equal to sec n
x so one over d square plus f d is d square
00:18:55.269 --> 00:19:01.559
plus n square so one over d square plus n
square sec n x ok
00:19:01.559 --> 00:19:09.529
now let us write the the function f d
in in terms of the its factors one over d
00:19:09.529 --> 00:19:24.869
minus i n d plus i n
now as i said we can then find the opera
00:19:24.869 --> 00:19:31.159
the result of the operation of one over
d plus i n on sec n x first are the result
00:19:31.159 --> 00:19:36.880
of the operation of operating by one over
d minus i n on sec n x [fro/first] first ok
00:19:36.880 --> 00:19:42.320
and then we can on the result of what
which which we get on that we can operate
00:19:42.320 --> 00:19:47.950
by the other thing now here we can also
break it into partial fractions let us suppose
00:19:47.950 --> 00:20:08.279
we are we wish to bracket into partial fractions
then we will write it as
00:20:08.279 --> 00:20:15.210
one over d minus i n minus one over d plus
i n divided by two i n that gives us one
00:20:15.210 --> 00:20:20.730
over one minus one over d minus i n into
one over into d plus i n now let us
00:20:20.730 --> 00:20:31.229
find let us apply this formula this is nothing
but one over d minus alpha operating on r
00:20:31.229 --> 00:20:41.330
x ok so let us make use of this formula so
let us find first we find
00:20:41.330 --> 00:20:53.470
one over d minus i n operating on r sec
n x so alpha is i n here so e to the power
00:20:53.470 --> 00:21:07.389
i n x integral e to the power minus i n x
into sec n x d x
00:21:07.389 --> 00:21:12.899
now this is nothing but e to the power i n
x integral let us apply the eulers formula
00:21:12.899 --> 00:21:24.840
here e to the power minus i theta is cos theta
minus i sin theta so cos n x
00:21:24.840 --> 00:21:36.879
minus i sin n x into sec n x d x now this
is further if you solve it this gives you
00:21:36.879 --> 00:21:48.340
cos n x into sec n x is one minus i tan n
x
00:21:48.340 --> 00:21:57.220
d x now we can easily integrate this this
is equal to e to the power i n x integral
00:21:57.220 --> 00:22:18.799
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
00:22:18.799 --> 00:22:24.099
sec n x when you differentiate with respect
to x what you get is one over sec n x then
00:22:24.099 --> 00:22:31.169
sec n x into tan n x into n so we get this
this is what we get and then we can write
00:22:31.169 --> 00:22:44.099
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
00:22:44.099 --> 00:22:51.710
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
00:22:51.710 --> 00:23:11.460
l n cos x cos n x ok so what we will get
let us multiply this
00:23:11.460 --> 00:23:32.470
this will be x cos n x
x cos n x plus i times x sin n x
00:23:32.470 --> 00:23:51.059
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
00:23:51.059 --> 00:24:07.799
n minus one over n sin n x l n cos n x so
this is what we get similarly we can find
00:24:07.799 --> 00:24:16.759
the value of one over d plus i n one over
d plus i n operating on sec n x if you obtain
00:24:16.759 --> 00:24:37.269
that in a similar we will get
so one over d plus i n sec n x if we get we
00:24:37.269 --> 00:24:58.229
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
00:24:58.229 --> 00:25:25.840
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
00:25:25.840 --> 00:25:32.609
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
00:25:32.609 --> 00:25:50.409
so that and divide two i n what do we get
so hence y p x is one over two i n now from
00:25:50.409 --> 00:25:58.609
this value from this value we subtract
this so x cos n x will cancel i x sin n x
00:25:58.609 --> 00:26:08.759
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
00:26:08.759 --> 00:26:22.739
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
00:26:22.739 --> 00:26:28.899
then this expression and this expression will
cancel each other so we have we divide by
00:26:28.899 --> 00:26:49.729
two i n so we get x sin n x divided by
n and then here we get cos n x
00:26:49.729 --> 00:26:56.479
so hence the general solution of the differential
equation d square over by d x square plus
00:26:56.479 --> 00:27:08.559
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
00:27:08.559 --> 00:27:28.190
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
00:27:28.190 --> 00:27:36.960
cos n x l n cos n x divided by n square so
that the general solution of this differential
00:27:36.960 --> 00:27:41.110
equation
and now we are going to study how to find
00:27:41.110 --> 00:27:49.200
the particular integral of one r x of
these some special forms say the first
00:27:49.200 --> 00:27:54.710
result that we are going to consider is the
result when r x is in exponential function
00:27:54.710 --> 00:28:00.469
so suppose r x is e to the power alpha x where
alpha e some real number alpha can be any
00:28:00.469 --> 00:28:06.820
real r complex constant so one over f d
when operates on e to the power alpha x let
00:28:06.820 --> 00:28:12.820
us find the particular integral when
r x is e to the power alpha x so one over
00:28:12.820 --> 00:28:18.019
f d e to power alpha x gives you one over
f alpha e to the power alpha x provided f
00:28:18.019 --> 00:28:22.349
alpha is not equal to zero we can make use
of this formula to determine the particular
00:28:22.349 --> 00:28:27.979
integral when r x is of the type e to the
power alpha x now to prove this formula we
00:28:27.979 --> 00:28:32.989
can see that when we differentiate e to the
power alpha x with respect to x what we get
00:28:32.989 --> 00:28:37.749
is alpha times e to the power alpha x if we
again differentiate this with respect to
00:28:37.749 --> 00:28:41.419
d that is we find the second derivative of
e to the power alpha x with respect to x
00:28:41.419 --> 00:28:45.969
which is given by d square e to the power
alpha x we get alpha x square e to the power
00:28:45.969 --> 00:28:50.109
alpha x
now so f d when operates on e to the power
00:28:50.109 --> 00:28:57.609
alpha x gives you f d is the d square plus
a d plus b so d square plus a d plus b when
00:28:57.609 --> 00:29:02.109
operated on e to the power alpha x we get
alpha x square e to the power alpha x then
00:29:02.109 --> 00:29:06.970
a times alpha e to the power alpha x then
b into e to the power alpha x so we can we
00:29:06.970 --> 00:29:12.039
get alpha a square plus a alpha plus b into
e to the power alpha x now this alpha a square
00:29:12.039 --> 00:29:18.229
plus alpha a alpha plus b can be written as
f alpha so we get f alpha into e to the power
00:29:18.229 --> 00:29:29.109
alpha x so f d when operates on e to the power
alpha x
00:29:29.109 --> 00:29:38.990
gives us now lets suppose that f alpha is
not equal to zero so when f alpha is not equal
00:29:38.990 --> 00:30:03.289
to zero let us first operate on both
sides by the operator one over f d
00:30:03.289 --> 00:30:10.320
since one over f d and f d are inverse
operators we have left hand side as e to the
00:30:10.320 --> 00:30:18.059
power alpha x and the right hand side is one
over f d operating on f alpha e to the power
00:30:18.059 --> 00:30:23.509
alpha x
now now f alpha is a non zero algebraic
00:30:23.509 --> 00:30:27.730
multiplier we have assumed alpha to f alpha
to be non zero it is a non zero algebraic
00:30:27.730 --> 00:30:35.080
multiplier so this is same as f alpha times
one over f d operating on e to the power alpha
00:30:35.080 --> 00:31:00.219
x now let us divide this equation by f alpha
so dividing
00:31:00.219 --> 00:31:04.330
we get one over f d operating on e to the
power alpha x equal to e to the power alpha
00:31:04.330 --> 00:31:13.039
x over f alpha so this is the proof and
this method fails when it happens
00:31:13.039 --> 00:31:18.349
that f alpha turns out to be zero let us apply
this method to find the particular integral
00:31:18.349 --> 00:31:24.139
in the case of the example d square y over
d x square plus y equal to e to the power
00:31:24.139 --> 00:31:31.259
x plus one whole square so this differential
equation can be written as d square plus one
00:31:31.259 --> 00:31:45.409
y equal to e to the power x plus one whole
square let us first first write the
00:31:45.409 --> 00:31:55.709
complimentary function so the auxiliary equation
is
00:31:55.709 --> 00:32:02.320
m square plus one equal to zero that is m
equal to plus minus i so complimentary function
00:32:02.320 --> 00:32:16.200
y c x equal to a cos x plus b sin x let us
find the particular integral y p x this is
00:32:16.200 --> 00:32:24.179
given by one over f d f d is d square plus
one operating on r x r x is e to the power
00:32:24.179 --> 00:32:31.859
x plus one whole square now this e to the
power r x is e to the power x plus one whole
00:32:31.859 --> 00:32:47.320
square so we have to square this and write
00:32:47.320 --> 00:32:53.659
now this is nothing but we operate by one
over d square plus one on each of the terms
00:32:53.659 --> 00:33:17.929
of this bracketed expression
ok now this is one over d square plus one
00:33:17.929 --> 00:33:22.119
e to the power two x can be found from the
formula one over f d e to the power alpha
00:33:22.119 --> 00:33:26.369
x equal to e to the power alpha x divided
by f alpha because here alpha is equal to
00:33:26.369 --> 00:33:32.450
two and when we replace d y in f d d square
plus one when we replace d y alpha there it
00:33:32.450 --> 00:33:40.479
is f alpha is not equal to zero so this is
equal to one over two square plus one e to
00:33:40.479 --> 00:33:46.740
the power two x now two is an algebraic
multiplier i can write it two times one over
00:33:46.740 --> 00:33:52.139
d d square plus one operating on e to the
power x so alpha is one here so i write one
00:33:52.139 --> 00:33:59.479
over one square plus one e to the power x
and here one can be regarded as e to the power
00:33:59.479 --> 00:34:09.000
zero x e to the power zero x so alpha can
be taken as zero and we can get the value
00:34:09.000 --> 00:34:17.530
of this so this is equal to further one over
five e to the power two x then two over two
00:34:17.530 --> 00:34:26.609
so e to the power x and then alpha is zero
here so one over zero square plus one so plus
00:34:26.609 --> 00:34:48.760
one so then general solution is y x equal
to y c x plus y p x
00:34:48.760 --> 00:34:59.490
now let us go to short methods of finding
one over f d sin c x and one over f d sin
00:34:59.490 --> 00:35:07.170
cos c x where c is any real number now
we are we are finding short methods of finding
00:35:07.170 --> 00:35:12.619
the result of operating y one over f
d on sin c x of one over f d operating on
00:35:12.619 --> 00:35:20.180
cos c x now if n is a positive integer notice
that d square to the power n sin c x is
00:35:20.180 --> 00:35:30.490
equal to minus c square to the power n and
sin c x let us prove this when we operate
00:35:30.490 --> 00:35:40.711
y c on sin c x that is we differentiate sin
c x with respect to x we get c times cos c
00:35:40.711 --> 00:35:53.250
x let us again operate by d on this c cos
c x so that is d square nine c x gives you
00:35:53.250 --> 00:36:05.299
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
00:36:05.299 --> 00:36:19.150
c x now d cube then sin c x will be minus
c square into c cos c x and d to the power
00:36:19.150 --> 00:36:31.299
four sin c x will be equal to minus c square
into minus c square into cos c x which can
00:36:31.299 --> 00:36:44.339
be written as minus c square is square cos
sin c x sorry sin c x so minus c square is
00:36:44.339 --> 00:36:55.779
square sin c x and thus we can say that
d square is square sin c x is nothing but
00:36:55.779 --> 00:37:03.630
minus c square is square sin c x it can be
by [math/mathematical] mathematical induction
00:37:03.630 --> 00:37:12.930
we can extend this to and so on d square
to the power n sin c x is equal to minus c
00:37:12.930 --> 00:37:27.559
square to the power n sin c x
similarly we can find d square to the power
00:37:27.559 --> 00:37:35.750
n cos c x this will come out to be minus c
square to the power n cos x now let us
00:37:35.750 --> 00:37:43.609
see suppose f d the expression of f d contains
only even powers of d and we denote it by
00:37:43.609 --> 00:37:50.480
phi d square then from here we can see
that phi d square one x on sin c x the
00:37:50.480 --> 00:37:56.809
effect of this operation of phi d square on
sin c x will be phi minus c square sin c x
00:37:56.809 --> 00:38:02.690
because here we see that whenever we operate
by d square on sin c x we get d square
00:38:02.690 --> 00:38:08.490
to power n on sin c x we get minus c square
to the power n sin c x so phi d square
00:38:08.490 --> 00:38:15.520
when updates on sin d x will get phi minus
c square sin c x now so what we get is
00:38:15.520 --> 00:38:26.760
when phi d square operates on sin c x we
get phi minus square sin c x
00:38:26.760 --> 00:38:32.310
now let us assume that phi minus a square
is non zero phi minus c square is non non
00:38:32.310 --> 00:38:54.039
zero then if we operate by one over phi
d square on both sides
00:38:54.039 --> 00:38:59.279
phi d square and one more phi d square are
inverse to each other will get sin c x equal
00:38:59.279 --> 00:39:10.890
to one over phi d square
phi minus c square is a algebraic multiplier
00:39:10.890 --> 00:39:19.290
non zero we divide it by minus phi c phi minus
c square and get one over phi d square operating
00:39:19.290 --> 00:39:30.269
on sin c x equal to sin c x divided by phi
minus c square so similarly we can show that
00:39:30.269 --> 00:39:35.331
one over phi d square one operates on cos
c x we get one over c minus c square cos c
00:39:35.331 --> 00:39:42.730
x provided phi minus c square is non zero
more generally if we have to operate by
00:39:42.730 --> 00:39:48.369
one over phi d square on sin c x plus d we
are c and d are com constants then we get
00:39:48.369 --> 00:39:52.519
one over phi minus c square sin c x plus d
b can similar the prove this in a similar
00:39:52.519 --> 00:39:57.039
manner
now let us take an example suppose we take
00:39:57.039 --> 00:40:10.030
the example one example one can be written
as d square minus four d plus one operating
00:40:10.030 --> 00:40:17.730
on y equal to two cos two x we know how to
find the complement function so i will discuss
00:40:17.730 --> 00:40:23.859
only particular integral so y p x equal to
one over f d which is d square minus four
00:40:23.859 --> 00:40:30.880
d plus one operating on two cos two x two
is an algebraic multiplier i can write it
00:40:30.880 --> 00:40:39.019
outside now a here c here is two so we replace
d square by minus c square so one over minus
00:40:39.019 --> 00:40:45.190
two square for d square we write and minus
four d plus one be leave just like that cos
00:40:45.190 --> 00:40:54.579
two x this is two times one over minus four
plus one so we get minus four d minus three
00:40:54.579 --> 00:41:02.309
cos two x and this also equal to minus two
times one over four d plus three operating
00:41:02.309 --> 00:41:09.890
on now what we do is to [ge/get] get the operation
of one four d plus three operating on cos
00:41:09.890 --> 00:41:16.029
two x let us operate by four d minus three
on cos two x and one over four d minus cos
00:41:16.029 --> 00:41:25.500
three so minus two times we we
write four d minus three and one over four
00:41:25.500 --> 00:41:34.950
d minus three they are inverse of each other
so
00:41:34.950 --> 00:41:44.380
we get this now this will be equal to minus
two
00:41:44.380 --> 00:41:54.190
in the denominator we have sixteen d square
minus nine cos two x again replace d square
00:41:54.190 --> 00:42:07.810
by minus a square we shall have so minus two
times d square by minus a square so sixteen
00:42:07.810 --> 00:42:18.769
times minus two square minus nine cos two
x so this is minus sixty four minus nine so
00:42:18.769 --> 00:42:24.930
that is minus seventy three so we get two
by seventy three and then four d minus three
00:42:24.930 --> 00:42:30.650
operating on cos two x
now we have to operate by the operator four
00:42:30.650 --> 00:42:37.309
d minus three on cos two x so we get two times
two over seventy three four d cos two x means
00:42:37.309 --> 00:42:44.450
derivative of cos two x with respect to x
so minus eight sin two x minus three cos two
00:42:44.450 --> 00:42:53.079
x so this is the particular integral in this
case ns we add to this the complementary function
00:42:53.079 --> 00:42:58.630
and write the general solution now in the
case of example two d square plus one y
00:42:58.630 --> 00:43:03.450
equal to cos square x by two what we do
is to find the particular integral we shall
00:43:03.450 --> 00:43:10.789
write now let us find the particular integral
in the case of example two so in the case
00:43:10.789 --> 00:43:27.060
of example two particular integral y p x is
equal to one over d square plus one cos square
00:43:27.060 --> 00:43:34.730
x by two
since this is cos square x by two we have
00:43:34.730 --> 00:43:41.520
to convert it to cosine x function to
apply the formula that we have just now
00:43:41.520 --> 00:43:50.980
proved so we shall write it as one over d
square plus one one plus cos x by two and
00:43:50.980 --> 00:44:03.660
this will then done by two one over d square
plus one operating on one plus cos x so this
00:44:03.660 --> 00:44:08.029
will be one by two one over d square plus
one operates on one one can be regarded as
00:44:08.029 --> 00:44:13.970
e to the power zero x so we will get one over
zero square plus one and then one over d square
00:44:13.970 --> 00:44:21.400
plus one operates on cos x we shall replace
d square one by d square by minus one square
00:44:21.400 --> 00:44:33.170
plus one so which is which is not defined
so how we shall deal with this in example
00:44:33.170 --> 00:44:38.380
we shall see in in in our lecture tomorrow
because here one over d square plus one becomes
00:44:38.380 --> 00:44:45.450
zero when d square is replaced by minus
a square so this example we cannot find a
00:44:45.450 --> 00:44:50.670
particular example like this this will be
found this will be tackled in my lecture
00:44:50.670 --> 00:44:55.230
which i give tomorrow
thanks