WEBVTT
Kind: captions
Language: en
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hello friends welcome to my third lecture
on solution of first order nonlinear equations
00:00:25.950 --> 00:00:32.379
in previous two lectures we have discussed
first order nonlinear equations of type
00:00:32.379 --> 00:00:39.559
one two and three in this lecture we shall
discuss the forth type of standard type
00:00:39.559 --> 00:00:44.710
of first order nonlinear equations and
then we shall discuss the general method of
00:00:44.710 --> 00:00:50.760
solving such nonlinear equations first
of first order which cannot be reduced
00:00:50.760 --> 00:00:57.399
to any those four standard forms so let us
begin with fourth order of first order
00:00:57.399 --> 00:01:03.690
nonlinear equations of type four
so in the in this category we consider
00:01:03.690 --> 00:01:11.310
those kind of equations which are of the form
z equal to p x plus q y plus f p q and as
00:01:11.310 --> 00:01:18.340
you remember p and q denotes the partial derivatives
of z with respect to x and y so if z
00:01:18.340 --> 00:01:26.110
is p x plus q y plus f p q then this equation
you can see analogous to clairauts ordinary
00:01:26.110 --> 00:01:31.810
differential equation the clalelout clairauts
clairauts ordinary differential equation
00:01:31.810 --> 00:01:41.750
is y plus p x plus f p where p is d z by dx
in the ordinary differential equation clairauts
00:01:41.750 --> 00:01:57.000
form is p x plus f p where p is dy by d x
so so the partial differential equation
00:01:57.000 --> 00:02:05.440
z equal to p x plus q y plus f p q is analogous
to the clairauts ordinary differential
00:02:05.440 --> 00:02:12.360
equation y equal to p x plus f p the complete
integral of equation one is then given
00:02:12.360 --> 00:02:18.950
by z equal to a x plus b y plus f a b because
we can you can see that when z equal to
00:02:18.950 --> 00:02:29.480
a x plus b y plus f a b then your partial
derivative with of z with respect to x which
00:02:29.480 --> 00:02:37.560
is which we we have denoted by p it will be
equal to a and q which is the partial derivative
00:02:37.560 --> 00:02:45.750
of z with respect to y it will be equal to
b and therefore y equal therefore z
00:02:45.750 --> 00:02:53.530
equal to a x plus b y plus f a b this is
a solution of equation one because when
00:02:53.530 --> 00:03:00.280
you put z equal to a x plus b y plus f a b
into one it is satisfied due to the fact that
00:03:00.280 --> 00:03:03.590
the value of p is a and the value of q is
b
00:03:03.590 --> 00:03:11.590
now so this complete solution you can see
z equal to p a x plus b y plus f a b this
00:03:11.590 --> 00:03:20.660
solution it represents plane and
it is at two para[meter]- plane family
00:03:20.660 --> 00:03:26.500
of two two parameter family of planes
because there are two parameters a and b so
00:03:26.500 --> 00:03:31.319
it is at two parameters family of planes now
inorder to obtain the general integral as
00:03:31.319 --> 00:03:37.391
we have earlier discussed let us put b equal
to phi a in this solution and where
00:03:37.391 --> 00:03:42.871
phi denotes an arbitrary function then z will
be equal to a z equal to a x plus y into phi
00:03:42.871 --> 00:03:49.720
a plus y into phi a plus f a phi a we differentiate
this equation with respect to a to get z d
00:03:49.720 --> 00:03:57.240
zero equal to x plus y phi dash a plus d over
d a of f a phi a and then eliminate a between
00:03:57.240 --> 00:04:04.220
equation three this equation and the equation
four to get the general integral
00:04:04.220 --> 00:04:09.180
now inorder to obtain the singular integral
we differentiate equation two this one
00:04:09.180 --> 00:04:14.070
z equal to a x plus b y plus f a b which is
the complete integral with respect v we differentiate
00:04:14.070 --> 00:04:20.340
this with respect to a so we will get zero
equal to x plus d over delta over delta
00:04:20.340 --> 00:04:26.850
a of f a b zero equal to x x plus delta f
over delta a and the second one will be when
00:04:26.850 --> 00:04:30.900
we differentiate with respect to b we will
get zero equal to y plus delta f over delta
00:04:30.900 --> 00:04:37.120
v so we have three equations now equation
two which is the complete integral this one
00:04:37.120 --> 00:04:42.930
and then the equations three sorry f[ive]-
five z zero equal to x plus delta f over delta
00:04:42.930 --> 00:04:49.310
a and six that is zero equal to y plus delta
f over delta v so we eliminate a and b between
00:04:49.310 --> 00:04:53.000
these three equations two reach the singular
integral
00:04:53.000 --> 00:04:59.000
let us see how we apply this method to
some differential equations let us begin with
00:04:59.000 --> 00:05:07.600
the first example z equal to p x plus q y
plus f p q so let us consider z equal to p
00:05:07.600 --> 00:05:23.100
x plus q y plus f p q and we can see of
the clairauts form z equal to p x plus sorry
00:05:23.100 --> 00:05:31.560
here it is p q not f p q so let us consider
z equal to p x plus q y plus p q is here f
00:05:31.560 --> 00:05:40.160
pq is equal to p q so it is a equation of
type four now let us so the complete integral
00:05:40.160 --> 00:06:06.690
is z equal to a x plus b y plus a b so this
is the complete integral to find the singular
00:06:06.690 --> 00:06:24.889
integral let us differentiate one with
respect to a differentiating
00:06:24.889 --> 00:06:42.960
with respect to a and
b we obtain zero equal to x plus b and
00:06:42.960 --> 00:06:47.600
when we when we differentiate with respect
to v we will get zero equal to x plus b and
00:06:47.600 --> 00:06:55.830
when we differentiate with respect to b we
will get zero plus y plus a so which imply
00:06:55.830 --> 00:07:01.389
x is equal to minus b b equal to minus
x
00:07:01.389 --> 00:07:10.960
lets write like this b equal to minus x anda
equal to minus y so now eliminating
00:07:10.960 --> 00:07:15.660
eliminating a and b between equation one
and these two equations we have eliminating
00:07:15.660 --> 00:07:37.340
a and b we obtain z equal to a is minus
y so minus x y b is minus x so minus x y plus
00:07:37.340 --> 00:07:54.600
minus x into minus y so we get minus x y see
this is the which is the singular integral
00:07:54.600 --> 00:08:17.669
in this case to determine the general
integral let us put
00:08:17.669 --> 00:08:34.300
b equal to phi a where phi is an arbitrary
function
00:08:34.300 --> 00:08:54.579
ok so then z will be equal to a x plus y into
five a plus a into phi a now we differentiate
00:08:54.579 --> 00:09:09.459
with respect to a
so differentiating
00:09:09.459 --> 00:09:29.829
we get zero equal to x plus y times phi
dash a plus phi a into a phi dash a let
00:09:29.829 --> 00:09:37.449
let me call it as equation number two and
this as equation number three so then we eliminate
00:09:37.449 --> 00:09:55.369
a between equations two and three to determine
the general integral so eliminating
00:09:55.369 --> 00:10:14.550
a between two and three we ob[tain]- obtain
the desired result so we will get the general
00:10:14.550 --> 00:10:20.569
integral so that is how we solve the problem
given in example one now lets take up the
00:10:20.569 --> 00:10:34.029
problem number two so we have z equal to
p x plus q y plus under root one plus p square
00:10:34.029 --> 00:10:41.800
plus q square
now we can see that here one plus f p q is
00:10:41.800 --> 00:11:04.360
equal to under root one plus p square plus
q square so it is of the clairauts form
00:11:04.360 --> 00:11:20.079
so this the complete integral is therefore
a z equal to a x plus b y plus under root
00:11:20.079 --> 00:11:30.439
one plus a square plus b square to determine
the complete integral to determine the this
00:11:30.439 --> 00:11:42.899
is the complete detail to obtain the singular
integral we differentiate with respect to
00:11:42.899 --> 00:12:13.480
a and b differentiating we get zero equal
to x plus a upon under root one plus a square
00:12:13.480 --> 00:12:22.529
plus b square and when we differentiate
with respect to b we get zero equal to y plus
00:12:22.529 --> 00:12:28.989
b upon under root one plus a square plus b
square
00:12:28.989 --> 00:12:35.399
now from these two equations we get x equal
minus a upon under root one plus a square
00:12:35.399 --> 00:12:39.389
plus b square and y equal to minus b upon
under root a square plus plus one plus a square
00:12:39.389 --> 00:12:47.019
plus b square so x square plus y square is
equal to a square plus b square divided by
00:12:47.019 --> 00:12:57.170
one plus a square plus b square which can
be written as one minus one upon one plus
00:12:57.170 --> 00:13:10.699
a square plus b square and so one upon
one plus a square plus b square is equal to
00:13:10.699 --> 00:13:25.299
one minus x square minus y square or we
can say one plus a square plus b square
00:13:25.299 --> 00:13:31.499
is equal to one over one minus x square minus
y square
00:13:31.499 --> 00:13:37.889
now x is equal to minus a upon under root
one plus a square plus b sq[uare]- we see
00:13:37.889 --> 00:13:52.421
so x is equal to minus a upon under root one
plus a square plus b square we will
00:13:52.421 --> 00:14:02.429
get a as a equal to minus x into under root
one plus a square plus b square so minus
00:14:02.429 --> 00:14:15.779
x times
00:14:15.779 --> 00:14:20.189
ok so a is equal to minus x into under root
one plus a sq[uare]- which is this equal to
00:14:20.189 --> 00:14:28.429
this and similarly b is equal to b is equal
to minus y minus y upon under root one
00:14:28.429 --> 00:14:37.549
minus x square minus b square yeah so now
eliminating a and b between equation one
00:14:37.549 --> 00:15:08.339
and ok so eliminating
a and b between equations one two three we
00:15:08.339 --> 00:15:21.949
get the singular integral we get the singular
integral and for general integral as we
00:15:21.949 --> 00:15:32.999
have discussed earlier
so for the general integral
00:15:32.999 --> 00:15:50.859
let us put b equal to
phi a then one becomes z equal to a x plus
00:15:50.859 --> 00:16:06.309
b phi plus by phi a plus under root one plus
a square plus phi a whole square we differentiate
00:16:06.309 --> 00:16:13.449
this equation with respect to a and get zero
equal to this implies zero equal to x plus
00:16:13.449 --> 00:16:27.499
y phi dash a plus one over two times under
root one plus a square plus phi a whole square
00:16:27.499 --> 00:16:41.809
and then we have two a plus two times phi
a into phi dash a ok this two can be cancelled
00:16:41.809 --> 00:16:49.949
so we have this equation so eliminating
a between this equation and this equation
00:16:49.949 --> 00:16:56.230
this we call it as this as four and this as
five so eliminating a between four and five
00:16:56.230 --> 00:17:12.679
we get the
00:17:12.679 --> 00:17:28.140
desired inte[gral]- desired integral ok
this is how we solve these two equations now
00:17:28.140 --> 00:17:34.720
we go to the general method of solving
first order nonlinear equations which we
00:17:34.720 --> 00:17:41.260
call as the charp[its]- charpits method so
here in this method we consider nonlinear
00:17:41.260 --> 00:17:46.550
partial differential equation of first order
which cannot be reduced to the four standard
00:17:46.550 --> 00:17:52.300
forms which we have already discussed we will
apply this method only in the cases where
00:17:52.300 --> 00:17:57.330
the first order non linear partial differential
equations cannot be reduced to the cannot
00:17:57.330 --> 00:18:01.800
be [dedu/reduce] reduce to the four standard
form because to find the complete integral
00:18:01.800 --> 00:18:07.670
by applying the charpits method its rather
more cumbersome
00:18:07.670 --> 00:18:15.680
so we prefer to solve the partial differential
equations which are of first order and nonlinear
00:18:15.680 --> 00:18:23.280
by trying to reduce them to the four standard
forms if if it is not possible to do that
00:18:23.280 --> 00:18:28.010
then we apply the charpits method so here
what we do is let us consider first order
00:18:28.010 --> 00:18:33.700
nonlinear partial differential equation as
a given by f x y z p q equal to zero we know
00:18:33.700 --> 00:18:38.760
that since there are two independent variables
x and y and z is dependent on x and y so we
00:18:38.760 --> 00:18:41.850
can write the d z equal to p d x plus q d
y
00:18:41.850 --> 00:18:49.570
now in the charpits method we try to find
relation between x y z and p q say we can
00:18:49.570 --> 00:18:55.430
which when from which and the given
partial differential equation which when we
00:18:55.430 --> 00:19:00.880
when we obtain the values of p and q and put
them in the equation d z equal to p d x plus
00:19:00.880 --> 00:19:07.570
q d y if it is integrable we get the complete
integral of the given partial differential
00:19:07.570 --> 00:19:13.950
equation so in this method we will try to
find a relation involving x y z p q equal
00:19:13.950 --> 00:19:20.570
to zero and then saw which this uh equation
three and the equation one the given partial
00:19:20.570 --> 00:19:26.140
differential equation be then solve to find
the values of p and q which when we put in
00:19:26.140 --> 00:19:32.060
d z equal to p d x plus q d y if it is integrable
we get the complete integral of the given
00:19:32.060 --> 00:19:39.170
partial differential equation
so let us assume that the three this
00:19:39.170 --> 00:19:46.360
equation z phi z phi x y z p q equal to zero
is the required relation then z p and q
00:19:46.360 --> 00:19:51.490
so p is delta z by delta x q is delta z by
delta y they are expressive they are functions
00:19:51.490 --> 00:19:58.721
of x and y so that when they are substituted
in the equations f equal to zero this equation
00:19:58.721 --> 00:20:07.890
f equal to zero and phi equal to zero ok
they are satisfied identically so now
00:20:07.890 --> 00:20:13.590
when we differentiate f equal to zero with
respect to x what we get is delta f by delta
00:20:13.590 --> 00:20:18.500
x plus delta fby delta z into delta z by delta
x which which is which is equal to p plus
00:20:18.500 --> 00:20:23.860
delta f by delta p delta p by delta x delta
f by delta q delta q by delta x equal to zero
00:20:23.860 --> 00:20:28.640
this is obtained by solving f equal to zero
with respect to x similarly when we solve
00:20:28.640 --> 00:20:33.530
differentiate phi with respect to x we will
get delta phi by delta x plus delta phi
00:20:33.530 --> 00:20:39.420
by delta z into p delta phi by delta p delta
p by delta x plus delta phi by delta q delta
00:20:39.420 --> 00:20:44.580
q by delta x is equal to zero and two similar
equations are obtained from f equal to zero
00:20:44.580 --> 00:20:50.540
phi equal to zero by differentiating them
with respect to y and they are delta f over
00:20:50.540 --> 00:20:55.810
delta y plus delta f over delta z delta z
by delta y which is q delta f over delta p
00:20:55.810 --> 00:21:00.930
delta p by delta y delta f over delta q delta
q by delta y equal to zero and delta phi over
00:21:00.930 --> 00:21:06.610
delta y plus delta phi over delta z delta
z by delta y is q delta phi over delta phi
00:21:06.610 --> 00:21:12.190
delta delta phi over delta p delta p over
delta y delta phi over delta q and delta q
00:21:12.190 --> 00:21:17.300
over delta y equal to zero
now what we do is let us eliminate delta
00:21:17.300 --> 00:21:22.070
p over delta x from the first pair there are
four equations so from the first two equations
00:21:22.070 --> 00:21:30.350
first pair of equations let us eliminate
this derivative derivate of p with respect
00:21:30.350 --> 00:21:35.880
to x so delta p over delta x let us eliminate
from these two equations and which is not
00:21:35.880 --> 00:21:41.520
difficult because it is a linear equation
so this is a so this will give you delta
00:21:41.520 --> 00:21:46.610
f over delta x delta f over delta z into p
delta f over delta q del[ta]- so the this
00:21:46.610 --> 00:21:51.570
we can obtain this equation is very simple
and then what we do is we can write it as
00:21:51.570 --> 00:21:58.010
we can write it in the form the we
collect the terms which contain p and
00:21:58.010 --> 00:22:02.900
the terms which contain delta q by delta x
and this constant this term which is free
00:22:02.900 --> 00:22:08.270
from p and delta q by delta x
so we can write those that this equation
00:22:08.270 --> 00:22:14.960
we can write this equation in the form this
one ok so we similarly let us eliminate delta
00:22:14.960 --> 00:22:20.230
q over delta y to this delta q over delta
y in let us eliminate from this pair of equations
00:22:20.230 --> 00:22:25.090
so for that you need to multiply this equation
by delta phi over delta q and this fourth
00:22:25.090 --> 00:22:31.200
equation by delta f over delta q and then
subtract so delta q over delta y term
00:22:31.200 --> 00:22:37.130
will vanish so eliminating delta q over delta
y between the last pair of equations we shall
00:22:37.130 --> 00:22:42.470
have this one delta f over delta y delta phi
by delta q minus delta phi by delta y delta
00:22:42.470 --> 00:22:48.700
f by delta q plus q times this plus delta
v over delta y times this equal to zero
00:22:48.700 --> 00:22:55.720
now let us notice one thing delta q is
delta z by delta y ok so delta q by delta
00:22:55.720 --> 00:23:02.120
x is delta square z by delta x delta y and
p is delta z by delta x so delta p by delta
00:23:02.120 --> 00:23:07.380
y is delta square z by delta x delta y and
assuming that the second order partial derivatives
00:23:07.380 --> 00:23:13.230
are continuous delta square z by delta x delta
y is same as delta square z by delta y delta
00:23:13.230 --> 00:23:19.070
x we know is we know this result so delta
q over delta x and delta p over delta y are
00:23:19.070 --> 00:23:26.010
same and this they are coefficient you can
see is also same except for the negative
00:23:26.010 --> 00:23:33.660
sign this is negative of this so when we
add these two equations ok the last term this
00:23:33.660 --> 00:23:36.170
term here and this term here they will cancel
out
00:23:36.170 --> 00:23:44.070
so now let us notice that delta q over
delta x ahis equal to delta p over delta
00:23:44.070 --> 00:23:50.450
y so we have this one delta q over delta
x equal to delta square z by delta x delta
00:23:50.450 --> 00:23:54.670
y and which is equal to delta p over delta
y because we are assuming that second order
00:23:54.670 --> 00:24:01.850
partial derivatives are continuous so when
we add the equations four and five as we have
00:24:01.850 --> 00:24:07.990
discussed just now this is negative of
this quantity and this delta q over delta
00:24:07.990 --> 00:24:13.230
x and delta p over delta y are same are
same so when we add the equations four and
00:24:13.230 --> 00:24:19.260
five and rearrange the terms what happens
is this term and this term they cancel out
00:24:19.260 --> 00:24:24.040
and what we get is this so delta f over
delta x plus p times delta f over delta z
00:24:24.040 --> 00:24:29.332
into delta phi over p plus delta f over delta
y plus q times delta f over delta z delta
00:24:29.332 --> 00:24:36.960
phi by delta q plus minus p delta f f by delta
p minus q delta f by delta q delta phi by
00:24:36.960 --> 00:24:41.750
delta z plus minus delta f by delta p into
delta phi by delta x minus delta f by delta
00:24:41.750 --> 00:24:48.400
q into delta phi over delta y equal to zero
as since the terms involving are the partial
00:24:48.400 --> 00:24:52.400
derivative of p with respect to y and partial
derivative of q with respect to x cancel
00:24:52.400 --> 00:24:58.750
now the equation six is a partial differential
equation of the first order in q no phi let
00:24:58.750 --> 00:25:06.250
us see this is a partial differential equation
of the first order in phi ok because phi
00:25:06.250 --> 00:25:12.690
is phi is a function of x y z p and q and
you can see here we have in the first term
00:25:12.690 --> 00:25:16.480
partial derivative of phi with respect to
p partial derivative of phi with respect to
00:25:16.480 --> 00:25:20.410
q partial derivative of phi with respect to
z partial derivative of phi with respect to
00:25:20.410 --> 00:25:25.350
x partial derivative of phi with respect to
y which are of first order so it is a first
00:25:25.350 --> 00:25:32.070
order linear differential equation in
phi and therefore its integrals are the integrals
00:25:32.070 --> 00:25:36.940
of the auxiliary equations as we have discussed
in the case of lagranges linear equation in
00:25:36.940 --> 00:25:43.280
p and q it is same it is of the same
so we have d p over this quantity and d
00:25:43.280 --> 00:25:49.080
q over this d z over this quantity and d x
over minus delta f by delta p d y over minus
00:25:49.080 --> 00:25:55.270
delta f by delta q equal to d phi over zero
we are we are actually solving this equations
00:25:55.270 --> 00:26:00.710
this this equation like we have sol[ved]-
solved the lagranges linear equation in p
00:26:00.710 --> 00:26:08.210
and q so it is a linear equation in the
derivatives of phi with respect to p q z x
00:26:08.210 --> 00:26:15.610
and y so we solve this auxiliary system of
equations now any of the integrals of this
00:26:15.610 --> 00:26:22.740
equation seven will satisfy six and therefore
we take the simplest relation involving
00:26:22.740 --> 00:26:29.750
atleast one of p and q for phi equal to zero
so that the values of p and q may be obtained
00:26:29.750 --> 00:26:35.010
easily once the values of p and q are obtained
easily we can take the help of the given
00:26:35.010 --> 00:26:41.540
equation f x y z p q equal to zero and determine
the values of the determine the value of
00:26:41.540 --> 00:26:48.390
the other other q r p and once p and
q are known we can use d z is equal to p d
00:26:48.390 --> 00:26:52.680
x plus q d y and determine the complete
integral
00:26:52.680 --> 00:26:57.750
so we shall be able to determine the complete
integral by writing the auxiliary system of
00:26:57.750 --> 00:27:03.940
equations and this auxiliary system of
equations will then will satis[fy]- we will
00:27:03.940 --> 00:27:09.680
find the shortest simplest relation involving
at least one of p and q for phi equal to zero
00:27:09.680 --> 00:27:17.290
and then we may use the these value this
relation for p with in p and q and
00:27:17.290 --> 00:27:22.820
determine the values of p and q so when
we put them in the equation dz equal to p
00:27:22.820 --> 00:27:29.950
d x plus q d y and integrate we shall obtain
the complete integral which will be the
00:27:29.950 --> 00:27:34.240
complete integral of the given partial
differential equation and the similar and
00:27:34.240 --> 00:27:39.380
a general integrals are then obtained in the
usual manner so with that i would like to
00:27:39.380 --> 00:27:40.570
conclude my lecture
thank you for your attention