WEBVTT
Kind: captions
Language: en
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hello friends welcome to my second lecture
on solution of lagrange's equation it is in
00:00:23.590 --> 00:00:30.310
continuation of my previous lecture on
this topic here first we discuss the how
00:00:30.310 --> 00:00:36.500
geometrical interpretation of the lagrange's
linear equation which is p into p plus q into
00:00:36.500 --> 00:00:41.650
q equals r where small p and small q represent
the partial derivative subject with respect
00:00:41.650 --> 00:00:49.870
to x and y now we may write this equation
pp plus q q equal to rs pp plus qq plus
00:00:49.870 --> 00:00:53.910
r times minus one equal to zero now we know
that the direction cosines of the normal
00:00:53.910 --> 00:01:01.440
to a surface z is equal to f x y are given
by delta z over delta x delta z over
00:01:01.440 --> 00:01:06.640
delta y and minus one so direction cosines
of the normal to the surface z equal to f
00:01:06.640 --> 00:01:09.970
x y are proportional to p q and minus one
+
00:01:09.970 --> 00:01:14.340
now if the surfaces f x y z equal to zero
then the direction cosines of the normal to
00:01:14.340 --> 00:01:20.190
the surface are proportional to its
partial derivatives of f with respect to x
00:01:20.190 --> 00:01:27.700
by z because we know that del f is a
vector normal to the surface f x y z equal
00:01:27.700 --> 00:01:41.970
to zero and del f is we know that del f is
equal to i del f by del x plus j del f by
00:01:41.970 --> 00:01:50.690
del y plus k del f by del z so it is known
that if the surface is given by f x by
00:01:50.690 --> 00:01:56.829
z equal to zero then del f is a vector normal
to the surface so the direction cosines of
00:01:56.829 --> 00:02:02.409
the normal to the surface are proportional
to partial derivatives of f with respect
00:02:02.409 --> 00:02:06.770
to x y z that is [dual f/delta f] over delta
f over delta x delta f over delta y delta
00:02:06.770 --> 00:02:07.770
f over delta z
00:02:07.770 --> 00:02:13.650
now or we can also say that this is
there are proportional to minus delta f over
00:02:13.650 --> 00:02:19.200
delta x over delta f over delta z minus delta
f over delta y over delta f over delta z and
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minus one or delta z over delta x delta z
over delta y and minus one because if f x
00:02:25.700 --> 00:02:47.260
y z equal to zero then when we differentiate
it with respect to x what we get is so which
00:02:47.260 --> 00:03:04.569
implies that
and similarly
00:03:04.569 --> 00:03:30.450
when we differentiate with respect to y the
equation f x y z equal to zero we get
00:03:30.450 --> 00:03:36.810
ok so the ratio is minus delta f by delta
x over delta f by delta z becomes delta z
00:03:36.810 --> 00:03:42.060
by delta x minus delta f over delta y over
delta f by delta z becomes delta z over delta
00:03:42.060 --> 00:03:47.340
y and so we can say that these ratios are
same as the partial derivatives of z with
00:03:47.340 --> 00:03:48.340
respect to x and y
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so delta z over delta x delta z over delta
y and minus one that is p q and minus one
00:03:54.569 --> 00:04:01.819
now hence the geometrical interpretation of
the equation one is that let us look at this
00:04:01.819 --> 00:04:08.170
p into p plus q into q plus r into minus one
equal to zero now we are seeing that the direction
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cosines of the normal to the surface are proportional
to p q and minus one and therefore we can
00:04:17.200 --> 00:04:21.799
geometrically interpret this equation one
as the normal at a point to a certain surface
00:04:21.799 --> 00:04:27.050
is perpendicular to a line whose direction
direction cosines are in the ratio p q and
00:04:27.050 --> 00:04:31.690
r because we know that if the two lines
are perpendicular to each other then the dot
00:04:31.690 --> 00:04:38.640
product of their direction ratio is equal
to zero so we have normal to a point
00:04:38.640 --> 00:04:42.979
at a point to a certain surface is perpendicular
to join whose directional ratios are in the
00:04:42.979 --> 00:04:44.320
ratio p q and r
00:04:44.320 --> 00:04:51.230
now let us look at the simultaneous equations
d x by p equal to d y by q equal to d z by
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r it represents the family of curves such
that the tangent at any point has the direction
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ratios proportional to p q and r ok now if
u is equal to a and v is equal to b are two
00:05:01.870 --> 00:05:07.390
particular integrals of these equations then
phi u v equal to zero represents a surface
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through these curves let us take a point
p dash on the surface represented by phi
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u v equal to zero then through this point
passes a curve of the family which lies entirely
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on the surface the normal to the surface at
the point p dash must therefore be at right
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angles to the tangent at this point to the
curve that is it is perpendicular to a line
00:05:30.640 --> 00:05:37.380
whose direction cosines are proportional to
p q and r and this is exactly what we require
00:05:37.380 --> 00:05:39.830
in the partial differential equation one
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so one and two one and two this is two one
and two define the same set of surfaces and
00:05:47.040 --> 00:05:55.630
are therefore equivalent so if the solutions
the solutions u and v of the equations
00:05:55.630 --> 00:06:01.880
subsidiary equations d x over p equal to d
y over q equal to d z over r can be obtained
00:06:01.880 --> 00:06:07.060
easily when the variables are separable if
the variables are not separable then we shall
00:06:07.060 --> 00:06:13.010
express it in the form d x over p equal to
d y over q equal to d z over r equal to lambda
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d x plus mu d y plus delta d z over lambda
p plus mu q plus delta r where lamda mu delta
00:06:20.380 --> 00:06:26.550
are certain functions of x y z such that lambda
p plus mu q plus delta r equal to zero
00:06:26.550 --> 00:06:32.560
so we search the functions lambda mu and
delta in such a way that lambda into p plus
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mu q plus delta r equal to zero now if this
equation one lambda p plus mu q plus delta
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r equal to zero we will have lambda d x plus
mu d y plus delta z equal to ze[ro] d z
00:06:47.250 --> 00:06:54.510
equal to zero now you on choosing lambda mu
new and lambda mu and delta in such a way
00:06:54.510 --> 00:07:01.029
that lambda p plus mu q plus delta r equal
to zero from here it turns out that lambda
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d x plus mu d y plus delta d z equal to zero
00:07:04.459 --> 00:07:10.190
so now if the equation lambda d x plus mu
d y plus delta d z equal to zero is integrable
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we will obtain a solution of the subsidiary
equations then we can make another choice
00:07:14.590 --> 00:07:20.520
of lambda mu delta and obtain another solution
are from this solution using this solution
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we can obtain another independent solution
of the subsidiary equations so how we will
00:07:24.880 --> 00:07:29.980
arrive at the general integral of a partial
differential equation of first odd and first
00:07:29.980 --> 00:07:35.159
degree using this lagrange's method let it
is illustrated in the examples below so let
00:07:35.159 --> 00:07:41.650
us first take the example of y square z into
p minus x square z into q equal to x square
00:07:41.650 --> 00:07:42.650
y
00:07:42.650 --> 00:07:54.140
so y square z into p minus x square z into
q equal to x square y so the subsidiary equations
00:07:54.140 --> 00:08:14.750
are d x over p p is y square z equal to d
y over minus x square z that which is q and
00:08:14.750 --> 00:08:21.880
then r is x square y so d z upon x square
y we will find two independent solutions u
00:08:21.880 --> 00:08:34.969
and v from the subsidy equations so let
us first consider d x over y square z equal
00:08:34.969 --> 00:08:42.640
to d y over minus x square z
now here so this z and z can be cancelled
00:08:42.640 --> 00:08:51.390
out and then we see that x square d x plus
y square d y equal to zero so the variables
00:08:51.390 --> 00:08:57.300
x and y can are separated here and we can
integrate this integrating this we get x cube
00:08:57.300 --> 00:09:09.060
by three plus y cube by three equal to a constant
or we can say x cube and y cube equal to a
00:09:09.060 --> 00:09:16.020
where a is an arbitrary constant now we
can find another solution of this is u
00:09:16.020 --> 00:09:22.970
x y z equal to a we can find v x y z equal
to v from the other equations d y over minus
00:09:22.970 --> 00:09:32.870
x square z equal to d x upon x square y now
here we can cancel x square and we get y d
00:09:32.870 --> 00:09:41.640
y plus z d z equal to zero so again z d
z equal to zero so again the variables are
00:09:41.640 --> 00:09:46.780
separable and therefore here we we can integrate
and we will get y square by two plus z square
00:09:46.780 --> 00:09:48.780
by two equal to a constant
00:09:48.780 --> 00:09:58.080
so we can write y square plus z square as
some constant let us say b ok so this is b
00:09:58.080 --> 00:10:12.870
x y z equal to b and thus the general integral
is phi [pi/pi] is an arbitrary function phi
00:10:12.870 --> 00:10:24.560
u v equal to zero so x cube plus y cube y
square plus z square equal to zero so this
00:10:24.560 --> 00:10:32.300
is how we will find the general solution of
the lagrange's equation given in example
00:10:32.300 --> 00:10:40.490
one here we have seen that the it is easy
to solve this subsidiary equation because
00:10:40.490 --> 00:10:46.760
it is the case of case where the variables
are separable we can easily separate the variables
00:10:46.760 --> 00:10:53.810
x and y so it was easy to solve now let
us take the example two
00:10:53.810 --> 00:11:09.200
so in example two z minus y into p plus x
minus z into q equal to x by minus x ok so
00:11:09.200 --> 00:11:17.510
we will have subsidiary equations as d x over
z minus y equal to d y over x minus z equal
00:11:17.510 --> 00:11:27.070
to d z over y minus x ok now we can see that
if you can see that d x over z minus y equal
00:11:27.070 --> 00:11:33.790
to d y over x minus z the variables x y z
are not separable so it is a case where the
00:11:33.790 --> 00:11:42.080
variables are not separable so we shall consider
this equal to lambda d x plus mu d y plus
00:11:42.080 --> 00:11:55.530
delta d z divided by lambda times x minus
y plus mu times x minus z plus delta times
00:11:55.530 --> 00:11:56.970
y minus x
00:11:56.970 --> 00:12:02.680
now let us make a choice of lambda mu and
delta in such a way that the denominator here
00:12:02.680 --> 00:12:12.760
is zero ok so let us choose lamda mu and delta
all to be equal to one ok let us choose lambda
00:12:12.760 --> 00:12:29.200
mu delta equal to one then we see that x minus
y plus x minus z z minus y z minus y this
00:12:29.200 --> 00:12:45.550
is z minus y z minus y x minus z ok this is
z minus y plus x minus z plus y minus x
00:12:45.550 --> 00:12:50.810
this cancels with this y cancels with y and
z cancels with z so this is equal to zero
00:12:50.810 --> 00:13:06.600
ok and hence d x plus d y plus d z equal
to zero from here and so x integrating we
00:13:06.600 --> 00:13:13.540
get x plus y plus z equal to a constant a
so we have got one solution u x y z equal
00:13:13.540 --> 00:13:16.660
to a where u x y z is x plus y plus z
00:13:16.660 --> 00:13:27.470
now let us make another choice of lambda
mu delta now let us choose
00:13:27.470 --> 00:13:41.810
lambda equal to x mu equal to y and delta
equal to z then we notice that lambda times
00:13:41.810 --> 00:13:55.470
z minus y plus mu times x minus z plus delta
times y minus x is again zero because this
00:13:55.470 --> 00:14:07.770
is x times z minus y plus y times x minus
z plus z times y minus x so we have x y which
00:14:07.770 --> 00:14:14.900
cancels with x y here we have x z which
cancels with x z here then we have y z y z
00:14:14.900 --> 00:14:21.000
cancels with this is equal to zero
so hence we have another equation hence
00:14:21.000 --> 00:14:31.460
x d x plus y d y plus z d z equal to zero
and on integration we get x square by two
00:14:31.460 --> 00:14:41.290
plus y square by two plus z square by two
equal to a constant which can be written as
00:14:41.290 --> 00:14:48.861
x square plus y square plus z square equal
to a constant b ok so this is v x y z equal
00:14:48.861 --> 00:15:10.710
to b and the general solution is phi x plus
y plus z x square plus y square plus z square
00:15:10.710 --> 00:15:11.710
equal to zero
00:15:11.710 --> 00:15:16.960
so now we have seen an example of a lagrange's
linear equation were the variables are
00:15:16.960 --> 00:15:22.840
not separable and we have to make a choice
of lambda mu and delta such that lambda p
00:15:22.840 --> 00:15:29.720
plus mu q plus delta r equal to zero now by
choosing two by making two different choices
00:15:29.720 --> 00:15:35.260
of lambda mu and delta we arrive at lambda
p plus mu q plus delta r equal to zero and
00:15:35.260 --> 00:15:41.420
then integrating the two for cases on integration
we get the two independent integrals u and
00:15:41.420 --> 00:15:47.160
b u equal to a and b equal to the b of the
subsidiary equations and so we can we are
00:15:47.160 --> 00:15:49.310
obtain we obtain the general solution
00:15:49.310 --> 00:15:56.050
now let's take one more case another case
where we will see that we obtain one solution
00:15:56.050 --> 00:16:02.060
of the auxiliary equation and using
that solution we shall be getting the second
00:16:02.060 --> 00:16:09.360
solution so in example three it is a it is
a different type of example here we have
00:16:09.360 --> 00:16:27.750
d x by one d y by three equal to d z upon
ten by five z plus ten by minus three x so
00:16:27.750 --> 00:16:41.740
this the subsidiary these are subsidiary equations
now solving the equation d x over one equal
00:16:41.740 --> 00:16:56.600
to d y over three we obtain
d y minus three d x equal to zero are y minus
00:16:56.600 --> 00:17:00.350
three x equal to a where a is some constant
00:17:00.350 --> 00:17:12.890
now in order to find the second solution what
we do is let us consider next
00:17:12.890 --> 00:17:33.220
let us take d y over three are ok we can take
d x over one equal to d z over five z plus
00:17:33.220 --> 00:17:43.559
ten by minus three x ok now we shall in order
to solve this question we shall be using
00:17:43.559 --> 00:17:49.130
the solution y minus three x equal to
a so y minus three x equal to a we can put
00:17:49.130 --> 00:18:02.380
here so the which implies that d x by
one equal to d z upon five z plus tan a now
00:18:02.380 --> 00:18:08.340
we can this is the case of separation separation
of variables on one side we have x on the
00:18:08.340 --> 00:18:20.750
other side we have z so we can integrate so
we get x equal to one by five l n five z plus
00:18:20.750 --> 00:18:43.530
tan a plus some constant a constant ok or
we can say five x minus l n five z plus
00:18:43.530 --> 00:18:52.730
tan a equal to some constant equal to a constant
00:18:52.730 --> 00:19:06.090
now let us replace back the value of a so
five x minus l n five z plus ten by minus
00:19:06.090 --> 00:19:17.730
three x equal to a constant let us say b
ok so thus we have then thus the general solution
00:19:17.730 --> 00:19:45.790
is phi y minus three x and then five x
minus l n five z plus tan y minus three x
00:19:45.790 --> 00:19:54.510
equal to zero so the the general integral
in this case here we have seen that by
00:19:54.510 --> 00:20:01.760
using one solution we can obtain the second
independent integral of the subsidiary
00:20:01.760 --> 00:20:14.350
equations now lets take one more example of
a similar type so z minus x p minus y q equal
00:20:14.350 --> 00:20:23.929
to a under root x square plus y square plus
z square where a is some constant it is given
00:20:23.929 --> 00:20:33.799
now we can put it in this standard form or
x p plus y q equal to z minus a under root
00:20:33.799 --> 00:20:40.400
x square plus y square plus z square
00:20:40.400 --> 00:20:47.309
now this is in the form of the lagrange's
equation p p plus q q equal to r so the subsidiary
00:20:47.309 --> 00:21:08.500
equations are dx over x equal to d y over
y equal to d z over z minus a under root x
00:21:08.500 --> 00:21:23.211
square plus y square plus z square now d x
over x equal to d y over y gives the solution
00:21:23.211 --> 00:21:35.360
very easily so this is l n x equal to l n
y plus a constant
00:21:35.360 --> 00:21:44.350
that this can be written as y equal to
c into x i can cut the constant as minus l
00:21:44.350 --> 00:21:50.670
n c so then we will get l n y equal to l n
x plus l n c are by equal to c x where c one
00:21:50.670 --> 00:21:59.850
x let us say where c one is some constant
00:21:59.850 --> 00:22:12.960
now let us try to find the second solution
so we have ok so d x over x equal to d
00:22:12.960 --> 00:22:23.410
y over y equal to d z upon z minus a under
root x square plus y square plus z square
00:22:23.410 --> 00:22:35.910
is also equal to x d x plus y d y let us multiply
y x y and z plus z d z divided by x square
00:22:35.910 --> 00:22:50.860
plus y square plus z square minus a z under
root x square plus y square plus z square
00:22:50.860 --> 00:23:01.080
so let us now take let
u square b equal to x square plus y square
00:23:01.080 --> 00:23:18.360
plus z square ok then two u d u is equal to
two x plus two y two x d x plus two y d y
00:23:18.360 --> 00:23:34.680
plus two z d z are u d u equal to x dx plus
y d y plus z d z so this is equal to u d u
00:23:34.680 --> 00:23:46.080
divided u square minus a z into u on putting
where u square is equal to x square plus y
00:23:46.080 --> 00:24:10.280
square plus z square and this is same as d
u over u minus a z d u over u minus a z we
00:24:10.280 --> 00:24:13.419
get
00:24:13.419 --> 00:24:25.520
now what we do is so we have d z upon z
minus a u thus d z upon z minus a u is equal
00:24:25.520 --> 00:24:39.820
to this is z minus a u z minus a u equal to
d u over u minus a z ok which is equal to
00:24:39.820 --> 00:25:07.600
d z minus d u divided by z minus u minus
z z minus a u plus a z so this is equal to
00:25:07.600 --> 00:25:23.410
d z by d u divided by one plus a times
z minus u so d z upon z minus a u is equal
00:25:23.410 --> 00:25:31.610
to d u over u minus a z which is equal to
d z minus d u divided by z minus u minus a
00:25:31.610 --> 00:25:38.990
u plus a z so we get z minus u times
one plus a ok now we can inte[grate]- easily
00:25:38.990 --> 00:25:50.330
integrate this so what we will do is ok
so this is equal to this so thus d x by x
00:25:50.330 --> 00:25:58.110
ok d x by x equal to d y by y equal to d z
by z minus a u equal to d u by u minus a z
00:25:58.110 --> 00:26:12.610
equal to d z by d u minus upon one plus
a into z minus u so we get this
00:26:12.610 --> 00:26:25.380
and this will give you l n x equal to one
plus a one over one plus a l n z minus
00:26:25.380 --> 00:26:30.830
u can put z minus u s t then do z minus d
u equal to d t so d t over one plus a into
00:26:30.830 --> 00:26:36.960
t and when integrate when we integrate we
get one over one plus a ln z minus u so this
00:26:36.960 --> 00:26:49.580
is we can put it as one plus some constant
we can simplify this now you can multiply
00:26:49.580 --> 00:27:03.820
by one plus a so one plus a l n x equal to
l n z minus u plus some constant
00:27:03.820 --> 00:27:20.400
and this can then be written as l n x to the
power one plus a
00:27:20.400 --> 00:27:31.070
so we can write it as z minus u equal to some
constant c two times x to the power one plus
00:27:31.070 --> 00:27:39.330
a this can be written as minus l n c two so
we will get z minus u equal to c two
00:27:39.330 --> 00:27:48.490
x to the power of one plus a now r z minus
under root x square plus y square plus z square
00:27:48.490 --> 00:27:58.840
equal to c two times x to the power one plus
a and so thus we get the following solutions
00:27:58.840 --> 00:28:01.880
general so integral we get as follows
00:28:01.880 --> 00:28:08.650
so one solution was y equal to c one x
the the other solution is i can say x to
00:28:08.650 --> 00:28:24.130
the power minus one plus a into z minus under
root x square plus y square plus z square
00:28:24.130 --> 00:28:38.419
equal to c two so the general integral is
phi y over x and x to the power minus one
00:28:38.419 --> 00:28:51.650
plus a z minus under root x square plus y
square plus z square equal to zero
00:28:51.650 --> 00:28:58.809
so this the general integral in this case
so we have discussed four different
00:28:58.809 --> 00:29:04.809
types of examples in the first case we have
the case of a suppression of variables and
00:29:04.809 --> 00:29:09.860
the second case we had the example where
we had to make two different choices of lambda
00:29:09.860 --> 00:29:16.309
mu and delta so that lambda p plus mu q plus
delta r equal to zero in the third case we
00:29:16.309 --> 00:29:22.370
had an example where we use the one
solution y minus three x equal to a to find
00:29:22.370 --> 00:29:29.150
the second solution independent solution in
the fourth case in the fourth case it was
00:29:29.150 --> 00:29:34.679
a typical example where we one the solution
bar very bar very easy to find which was y
00:29:34.679 --> 00:29:43.809
equal to c minus x but for the second solution
we had to do i mean several
00:29:43.809 --> 00:29:49.669
this thing substitutions and all so
the second solution was not so easy to obtain
00:29:49.669 --> 00:29:54.170
but that is how we get the second solution
and the general solution is therefore given
00:29:54.170 --> 00:30:01.180
by phi u v equal to zero so with that
i would will like to conclude my lecture and
00:30:01.180 --> 00:30:02.180
i
00:30:02.180 --> 00:30:03.650
thank you all for attention