WEBVTT
Kind: captions
Language: en
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welcome friends to my lecture on
solution of second order differential equations
00:00:26.830 --> 00:00:35.230
by changing independent variable now we
are going to discuss another method sometimes
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this method is found to be very useful and
this in this method we change the independent
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variable so far the differential equations
that we are considering they are of second
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order linear differential equation with
variable coefficients where x is the independent
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variable so in this method we will change
the independent variable from x to say another
00:00:56.040 --> 00:01:02.219
independent variable that is z so let us consider
a linear differential equation of second order
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that is y double dash plus p y dash plus q
y is equal to r where p q r are continuous
00:01:08.080 --> 00:01:14.430
functions of x on an on some interval i
we will be changing as i said we will be changing
00:01:14.430 --> 00:01:22.610
the independent variable from x to z the relationship
between x to z will be found from the
00:01:22.610 --> 00:01:28.840
fact that i the changed equation it becomes
readily integrable
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so once we have that aim in mind that
the changed equation has i mean is readily
00:01:36.510 --> 00:01:43.170
integrable from there we shall be able to
derive the relationship between x and z so
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let us see what we do since x is related
to z we can write d y by d x equal to d y
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by d z over into d z by d x and d square by
d x square if you find d square by d x square
00:01:57.509 --> 00:02:05.810
then d square by d x square will be equal
to d over d x of d y by d x which is equal
00:02:05.810 --> 00:02:18.040
to d over d x of d y by d z into d z by d
x so differentiating this product of functions
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of x we shall be having d over d x of d y
by d z into d z by d x plus d y by d z into
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d square z by d x square
now this can be further expressed as d over
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d z of because z is a function of x we can
write it as this into d z by d x into d z
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by d x plus d y by d z into d square z by
d x square so this will be equal to d square
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y over d z square into d z by d x whole square
plus d y by d z into d square z y d x square
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so the second derivative of y with respect
to x is equal to d two y over d z two into
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d z by d x whole square plus d y by d z
into d square z by d d x square
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now substituting these values of y dash and
y double dash in the equation one in this
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equation one we shall have this expression
d square y over d z square into d z by d x
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whole square plus d square x by d x square
plus p d z by d x into d y by d z plus q y
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equal to r ok so now let us denote
let us write this equation in the form
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d square y over d z in this standard form
d square y by d z square plus p one d y by
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d z plus q one by y equal to r one where
we write p one for this expression d square
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z by d x square plus p d z by d x over d z
by d x whole square and q one we write for
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d q over d z by d x whole square and r one
denotes r over d z by d x whole square so
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these p p one q one r one p one q one r
one are some functions of x and can
00:04:35.020 --> 00:04:40.810
they can be easily expressed as functions
of z by the relationship between z and x which
00:04:40.810 --> 00:04:45.310
we have yet to establish
so now what we do is let us discuss various
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possibilities which will give us a function
of z i mean z as a function of x and
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with those possibilities we shall be able
to integrate the equation two so the first
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possibility is that let us equate q over d
z by d x whole square q over d z by d x whole
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square is q one so let us equate q one to
a constant so if we do that so what do we
00:05:12.430 --> 00:05:24.979
have q one equal to q y d z by d x whole square
suppose it is some constant then if its so
00:05:24.979 --> 00:05:30.550
happens that p one also becomes a constant
if p one also becomes a constant then the
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equation two will be a linear differential
equation of second order in the independent
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variable z the dependent variable y with constant
coefficients and we have seen earlier how
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we solve a second order linear differential
equation with constant coefficients
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so if the function if z is a function of
x in such way that q over d z by d x whole
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square is a constant then if p one also
becomes a constant the equation two is at
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once integrable so this is one possibility
where we can find the general solution
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of equation two now since z is quite arbitrary
it can be chosen to satisfy any assignable
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condition s one condition that we put is
that q over d z by d x whole square is equal
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to a constant let us look at other possibilities
in the other possibility what we do is
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we choose z in such a way that the coefficient
of d y by d z in the equation two vanishes
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so coefficient of d y by z in the equation
two is p one so let us put one equal to zero
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so if p one equal to zero we get d square
z by d z square plus p d z by d x equal to
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zero ok so let us say suppose u is equal to
d z by d x then this a question the question
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can be expressed as d u by d x plus p into
u equal to zero we can separate the variables
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u and x are now let us integrate so integrating
both sides we have l n u equal to minus integral
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p d x ok are u equal to e to the power minus
integral p d x ok
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so thus d z by d x is equal to e to the
power minus integral p d x now integrating
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again with respect to x we get z as integral
e to the power minus integral p d x d x
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so so if we put the coefficient of d y
by d z that is p y equal to zero we get
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z equal to integral of e to the power minus
integral p d x d x and after that if q
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one with this choice of z as a function of
x if it so happens that q one becomes a constant
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then it is a second order linear differential
equation with constant coefficients it will
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be d square y over d z square plus a constant
times by equal to r one so it will be a linear
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differential equation of second order with
constant coefficients and therefore we can
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solve it we can find the general solution
of this equation
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now the other possibility is that suppose
q one is a constant divided by z square
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then z square times d square y by d x d
square plus a constant times y will be equal
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to r one times z square and so we will
have quasi euler equation and we know how
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to solve quasi euler equation so we can again
get the solution of this differential equation
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so if p one is taken equal to zero and
after that p one equal to zero we get z as
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a function of x with this choice of z as
a function of x if q one either becomes a
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constant or q one is a constant divided
by z square then we can find the general
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solution of equation two and hence we can
find the general solution of equation one
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so that is one more possibility and
so we can see here if the equation two will
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reduce to this form if the curve given comes
out to be a constant or a constant divide
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by z square the equation three becomes readily
integrable
00:09:48.160 --> 00:09:59.680
now let us see let us see when we put q
one as a constant ok say suppose q one we
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take as some constant a square ok we just
want z as a function of x so this constant
00:10:06.820 --> 00:10:13.260
can be chosen according to our requirement
so we we are we will be choosing q one equal
00:10:13.260 --> 00:10:26.300
to a square so then a q upon d z by d x whole
square this is equal to q one and q one is
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a square so we shall have d z by d x equal
to q upon a square r or root q divided by
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a ok or we can say z is a z is equal to one
upon a integral root q d x we will get z as
00:10:44.880 --> 00:10:51.160
a function of x
now so this is what we will get and the question
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will reduce to d square y over d z square
plus p one d y by d z plus a square y equal
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to r one if p one is a constant then the question
can be easily integrated ok because it will
00:11:00.740 --> 00:11:06.709
be a linear differential equation of second
order with constant coefficients ok now
00:11:06.709 --> 00:11:16.070
let us see some examples on this ok so
let us consider the question given in
00:11:16.070 --> 00:11:21.579
example six first we will write this equation
in the standard form so the coefficient of
00:11:21.579 --> 00:11:26.690
d square y by d x square you will get unity
which means that we will divide the equation
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by x to the power six
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so we shall have y double dash plus three
by x y dash plus a square by x to the power
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six into y equal to one by x to the power
eight now so when we compare this equation
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with the standard form we get p equal to three
by x q equal to a square by x to the power
00:12:02.880 --> 00:12:12.180
six and r equal to one by x to the power eight
so let us first explore the possibility when
00:12:12.180 --> 00:12:23.270
we put q as a constant ok so let us put
q as a constant so let q be equal to some
00:12:23.270 --> 00:12:29.790
constant say ok let us take it as one ok then
what we q by is q by d z by d x whole square
00:12:29.790 --> 00:12:40.029
we put as a constant so wait i think wait
a minute sorry so let us let q one be equal
00:12:40.029 --> 00:12:47.279
to q one is equal to q by d z by d x whole
square ok after we change the variable from
00:12:47.279 --> 00:12:53.060
x to z we get this q one ok so after
we change the variable from x to z we will
00:12:53.060 --> 00:12:57.990
get q one as this so this u let us put as
equal to this is equal to a square by x to
00:12:57.990 --> 00:13:06.769
the power six into one by d z by d x whole
square let us put it as a constant say one
00:13:06.769 --> 00:13:16.000
ok we can take any value of the constant if
you do not take here one you take some other
00:13:16.000 --> 00:13:22.320
constant then the relationship between x and
z will be changed accordingly so r d z by
00:13:22.320 --> 00:13:32.430
d x whole square equal to a square by x to
the power six which implies that d z by d
00:13:32.430 --> 00:13:42.330
x equal to a by x cube and this gives you
when we integrate with respect to z we get
00:13:42.330 --> 00:13:49.850
z equal to a x to the power minus two divided
by minus two ok
00:13:49.850 --> 00:13:59.080
so this is a upon minus a upon two x square
ok now let us find the value of p one p
00:13:59.080 --> 00:14:12.430
one equal to p d z by d x plus d z by d x
whole square d square z by d x square upon
00:14:12.430 --> 00:14:29.510
d z by d x whole square so p is equal to three
by x d z by d x is a by x cube plus d square
00:14:29.510 --> 00:14:39.519
z by d x square we can find from here so d
square z by d x square
00:14:39.519 --> 00:14:44.530
we differentiated once more see this is a
times x to the power minus three so we have
00:14:44.530 --> 00:14:51.480
minus three a x to the power minus four so
minus three a by x to the power four so we
00:14:51.480 --> 00:15:04.980
get minus three a over d z by d x whole square
which is a square by x to the power six
00:15:04.980 --> 00:15:08.699
now here we have three a by x to the power
four here minus three a by x to the power
00:15:08.699 --> 00:15:14.870
four so this is equal to zero so when q one
is assumed as a constant which is which
00:15:14.870 --> 00:15:20.630
we have taken here as one p one also comes
out to be a constant which is zero ok so we
00:15:20.630 --> 00:15:31.670
have the equation in in the independent
variable as d square y over d z square plus
00:15:31.670 --> 00:15:45.200
p one d y by d z plus q one into y equal to
r one ok this becomes d square y over d z
00:15:45.200 --> 00:15:54.589
square plus q one we have taken as one so
y and r one is r over d z by d x whole square
00:15:54.589 --> 00:16:02.290
so r one is one over x to the power eight
r one divided by d z by d x whole square
00:16:02.290 --> 00:16:12.829
so divided by a square over x t the power
six so we get here x to the power six over
00:16:12.829 --> 00:16:20.980
a square
so we get here d square y over d z square
00:16:20.980 --> 00:16:29.130
plus y equal to one by a square x square
now we have to change the variable independent
00:16:29.130 --> 00:16:38.870
from x to z and the relationship is z equal
to minus a by two x square so one by a
00:16:38.870 --> 00:16:50.300
one by x square is two z over minus a ok
so this is one by a square two z over minus
00:16:50.300 --> 00:17:02.210
a one by x square equal to two z over minus
a ok so this will be minus two z over a
00:17:02.210 --> 00:17:07.980
cube ok now this is a second order linear
differential equation with constant coefficients
00:17:07.980 --> 00:17:16.370
so we can or write the complimentary
function for this here the auxiliary equation
00:17:16.370 --> 00:17:21.040
will be m square plus one equal to zero so
the roots will be complex conjugate m equal
00:17:21.040 --> 00:17:31.250
to plus minus i so complementary function
will be c one cos z plus c two sin z and z
00:17:31.250 --> 00:17:40.950
we see is equal to minus a by two x square
so c one cos minus a by two x square plus
00:17:40.950 --> 00:17:51.230
c two sin minus a by two x square cos minus
theta is equal to cos theta sin minus theta
00:17:51.230 --> 00:17:59.730
is minus sin theta so we can write it as c
one cos a by two x square and here minus c
00:17:59.730 --> 00:18:09.900
two sin a by two x square minus c two can
be replaced by another constant say say
00:18:09.900 --> 00:18:15.420
c two dash and we can write c one cos a by
two x square plus c two dash sin a by two
00:18:15.420 --> 00:18:22.850
x square so that is ok
now a particular integral let us find so this
00:18:22.850 --> 00:18:30.710
y p x y p z let us find first find ok lets
find y p x y p x will be one over d
00:18:30.710 --> 00:18:42.310
square plus one one over d square plus
one acting on minus two z by a cube so this
00:18:42.310 --> 00:18:53.881
is equal to minus two by a cube one over d
square plus one acting on ok now this is a
00:18:53.881 --> 00:18:58.850
polynomial z is a polynomial of degree one
so one over z d square plus one we shall expand
00:18:58.850 --> 00:19:06.320
as in a in the form of binomial expansion
so minus two by a cube one plus d square raise
00:19:06.320 --> 00:19:12.480
to the power minus one operating on z when
we write binomial expansion for this we shall
00:19:12.480 --> 00:19:20.370
have one minus d square plus d four minus
d six and so on but the second derivative
00:19:20.370 --> 00:19:29.559
of z is zero so we will get minus two by a
cube one minus d square plus d four and so
00:19:29.559 --> 00:19:39.751
on operating on z which will give you minus
two by a cube minus two by a cube d square
00:19:39.751 --> 00:19:46.940
z d four z all are zero so we have minus two
z by a cube and z is equal to minus a by two
00:19:46.940 --> 00:19:55.340
x square
so this is
00:19:55.340 --> 00:20:08.309
minus a by two x square so this will give
you one by a square x square and thus the
00:20:08.309 --> 00:20:31.360
general solution is y equal to y c x plus
y p x this is y c x and this one is y p x
00:20:31.360 --> 00:20:37.090
so we take the sum of the two and it write
it equal to y so that is the general solution
00:20:37.090 --> 00:20:45.450
in this case now lets discuss one more problem
on this to make it more clear so let us take
00:20:45.450 --> 00:20:51.710
up another problem here in the example
two we are given the equation x d square y
00:20:51.710 --> 00:20:57.580
by d x square minus d y by d x minus four
x cube by equal to eight x cube sin x square
00:20:57.580 --> 00:21:03.150
so again we divide the equation by x to make
the coefficient of y double dash unity
00:21:03.150 --> 00:21:16.690
so we shall have y double dash minus one
over y dash minus four x square into y equal
00:21:16.690 --> 00:21:30.070
to eight x square sin x square so here p is
equal to minus one over x q equal to minus
00:21:30.070 --> 00:21:44.860
four x square and r is equal to eight
x square sin x square ok so again we try the
00:21:44.860 --> 00:21:52.370
method where we put q one as a q q one
q y q equal to a sorry q one equal to a constant
00:21:52.370 --> 00:21:59.419
so after we change the variable from x to
z we get the equation d square y by d z square
00:21:59.419 --> 00:22:16.780
plus p one d y by d z plus q one y equal to
r one so here q one is q y d z by d x whole
00:22:16.780 --> 00:22:24.960
square so let us put q one as a constant say
q one will be equal to minus four x
00:22:24.960 --> 00:22:34.110
square upon d z by d x whole square so q one
we can we can take as a constant let us take
00:22:34.110 --> 00:22:39.870
the constant as minus one to then it will
be simple so lets take q one equal to minus
00:22:39.870 --> 00:22:50.130
one then d z by d x whole square will be equal
to four x square so which will give us d z
00:22:50.130 --> 00:22:58.330
by d x equal to two x one can take d z by
d x equal to minus two x also that will give
00:22:58.330 --> 00:23:02.840
you other relation between x and z so the
equation will change accordingly so we are
00:23:02.840 --> 00:23:16.740
taking positive here sin here so d z by d
x equal to two x this gives you z as x square
00:23:16.740 --> 00:23:34.690
ok so does z has x x square let us find the
value of p one
00:23:34.690 --> 00:23:47.720
so d z by d x is two x p here is minus one
over x and d square z by d x square is two
00:23:47.720 --> 00:23:54.730
d z by d x whole square is four x square so
again this cancels with this minus two
00:23:54.730 --> 00:24:02.090
plus two is zero so we get p one equal to
zero and thus the equation one one reduces
00:24:02.090 --> 00:24:16.659
to d square y by d z square p one is
zero q is minus one so minus y equal to r
00:24:16.659 --> 00:24:25.460
one r one is r over d z by d x whole square
so a eight x square sin x square divided by
00:24:25.460 --> 00:24:35.210
d z by d x whole square which is four x square
so we shall have two sin x square ok so complementary
00:24:35.210 --> 00:24:40.240
function here is this is auxiliary equation
is i mean x minus one equal to zero so m is
00:24:40.240 --> 00:24:44.760
equal to plus minus one so the both the roots
of the auxiliary equation are real and distinct
00:24:44.760 --> 00:24:51.630
and therefore we will have c one e to the
power z plus c two e to the power minus z
00:24:51.630 --> 00:25:00.779
and z is equal to x square so c one e to
the power x square plus c two e to the power
00:25:00.779 --> 00:25:06.789
minus x square
now here this in order to find the particular
00:25:06.789 --> 00:25:12.360
integral the right hand side of this equation
which is linear differential equation with
00:25:12.360 --> 00:25:18.370
constant coefficients in the right hand side
x has to be changed to z so we have two sin
00:25:18.370 --> 00:25:26.650
z because z is equal to x square so particular
integral will be one by d square minus one
00:25:26.650 --> 00:25:34.260
d d here represents d over d z so one over
d square minus one operating on two sin z
00:25:34.260 --> 00:25:41.810
and this will be equal to two times one over
d square minus one operating on sin z so when
00:25:41.810 --> 00:25:49.260
one over d square plus alpha s square operates
on sin a at z we replace d square by minus
00:25:49.260 --> 00:25:56.929
a square so this will will be two times one
over minus one square minus one sin z this
00:25:56.929 --> 00:26:10.720
is minus sin z ok so so we have got the
particular solution as minus sin x square
00:26:10.720 --> 00:26:37.919
so thus the general solution is
00:26:37.919 --> 00:26:41.919
the other solution is y equal to c one e to
the power x square plus c two e to the power
00:26:41.919 --> 00:26:48.880
minus x square minus sin x square
so that is how we solve this equation
00:26:48.880 --> 00:26:57.559
given in example two so we have seen
in our in the past three lectures we have
00:26:57.559 --> 00:27:04.049
seen how we can solve a linear differential
equation of second order with variable coefficients
00:27:04.049 --> 00:27:10.720
when it is given in a when it when we are
given some special kinds of such equations
00:27:10.720 --> 00:27:16.179
as i said earlier such i mean a general solution
is not a well known for all second order
00:27:16.179 --> 00:27:20.290
linear differential equation with variable
coefficients only a special class of such
00:27:20.290 --> 00:27:27.750
equations can be solved so we have seen
those equations which can be solved
00:27:27.750 --> 00:27:34.370
by in our past three lectures now in
my next lecture we shall be discussing how
00:27:34.370 --> 00:27:40.300
to find the general solution of a higher
order homogenous linear differential equation
00:27:40.300 --> 00:27:45.679
with constant coefficients
thank you for your attention