WEBVTT
Kind: captions
Language: en
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hello friends welcome to my lecture on
solution of first order nonlinear equation
00:00:23.770 --> 00:00:31.239
so this is our first the lecture on the
how to find the solution of nonlinear equations
00:00:31.239 --> 00:00:37.070
that are of first degree but not of that
are of order but not of first degree so let
00:00:37.070 --> 00:00:43.530
us see we have so far discussed how to
determine the solution of a lagrange's linear
00:00:43.530 --> 00:00:49.610
partial differential equation which is of
the form p into p plus q into q equal to r
00:00:49.610 --> 00:00:56.640
where p q and r as you know they are functions
of x by z and so this is a first order and
00:00:56.640 --> 00:01:00.920
first degree linear partial differential equation
i mean partial differential equation which
00:01:00.920 --> 00:01:02.530
is linear in p and q
00:01:02.530 --> 00:01:07.509
now we shall be considering first order
partial differential equations which are not
00:01:07.509 --> 00:01:13.700
of first degree so if there are not of
first degree then they will be nonlinear partial
00:01:13.700 --> 00:01:19.939
differential equations so let us see how we
solve first order nonlinear partial differential
00:01:19.939 --> 00:01:25.490
equations and such solution of nonlinear partial
differential equations present considerable
00:01:25.490 --> 00:01:31.920
difficulties so first we will focus
on before giving the general method that
00:01:31.920 --> 00:01:37.440
is the char[pit] charpit method of solving
such equations we shall present some
00:01:37.440 --> 00:01:44.229
special types of equations whose solutions
can be easily determined so many equations
00:01:44.229 --> 00:01:49.619
we shall see can be reduced to few standard
forms and therefore can be [jolved/solved]
00:01:49.619 --> 00:01:54.049
solved generally which by methods which
are shorter than charpit method
00:01:54.049 --> 00:02:00.880
so let us first discuss the equations of
the form f p q equal to zero such partial
00:02:00.880 --> 00:02:06.490
differential equations where the partial derivatives
p and q that is delta z by delta x and
00:02:06.490 --> 00:02:13.900
delta z delta y occur and they do not involve
the variables x y and z they consider they
00:02:13.900 --> 00:02:20.090
they come under this category of the equations
of the form f p q equal to zero
00:02:20.090 --> 00:02:26.480
so in the case of these equations f p equal
to zero we see that the complete integral
00:02:26.480 --> 00:02:33.060
is given by z equal to a x plus b y plus c
now let i want to remind you that the complete
00:02:33.060 --> 00:02:37.620
integral is defined as the solution of the
differential equation which contains as many
00:02:37.620 --> 00:02:42.090
arbitrary constants as there are independent
variables in the differential equations so
00:02:42.090 --> 00:02:48.260
there are two independent variables here x
and y so this is a solution of the partial
00:02:48.260 --> 00:02:53.460
differential equation f p q equal to zero
because here a and b are connected by
00:02:53.460 --> 00:02:58.690
the relation f a b equal to zero if you
find the partial derivatives of this equation
00:02:58.690 --> 00:03:05.470
z equal to a x plus b y plus c with respect
to x and y then what we get is the derivative
00:03:05.470 --> 00:03:12.080
of z with respect to x which is p we have
denoted by p comes out to be a and the derivative
00:03:12.080 --> 00:03:18.349
of z with respect to y which we have denoted
by q comes out to be b which when substituted
00:03:18.349 --> 00:03:24.120
in equation two that is f a b equal to zero
give us f p q equal to zero
00:03:24.120 --> 00:03:31.220
so z equal to a x plus b y plus c is a solution
of f p q equal to zero and moreover it
00:03:31.220 --> 00:03:36.569
involves two arbitrary constants they are
a and c b is not an arbitrary constant b is
00:03:36.569 --> 00:03:42.110
connected to a y by the relation f a b
equal to zero so b can be said as the function
00:03:42.110 --> 00:03:47.150
of a so there are two arbitrary constants
a and c and because of these two arbitrary
00:03:47.150 --> 00:03:52.890
constants and this is a solution of the equation
f p q equal to zero which contains two two
00:03:52.890 --> 00:03:58.690
independent variables it can be called
as these complete integral of f p q equal
00:03:58.690 --> 00:03:59.750
to zero
00:03:59.750 --> 00:04:07.740
now so now as be if you remember of
to from the complete integral one can determine
00:04:07.740 --> 00:04:24.520
the general solution if the complete integral
is given by
00:04:24.520 --> 00:04:42.240
f x y z a b equal to zero where a and b are
arbitrary constants
00:04:42.240 --> 00:04:49.580
then to determine the general general integral
what we do is let us say let b be an arbitrary
00:04:49.580 --> 00:05:07.759
function of a where phi is arbitrary
so then we have the equation one then we have
00:05:07.759 --> 00:05:25.849
from one f x y z a
phi a equal to zero now by eliminating
00:05:25.849 --> 00:05:58.110
a the arbitrary constant a between the equations
f x y z a phi a equal to zero and the partial
00:05:58.110 --> 00:06:03.689
derivative of f with respect to a equal to
zero between these two equations b obtained
00:06:03.689 --> 00:06:21.491
the general integral of the partial differential
equation of first order differential equation
00:06:21.491 --> 00:06:45.099
of f of f x y z p q equal to zero
00:06:45.099 --> 00:06:50.929
so this we have known already ok so in
order to find the general solution from the
00:06:50.929 --> 00:06:58.020
complete integral ok the complete integral
is z equal to a x plus b y plus c ok f
00:06:58.020 --> 00:07:16.639
a b equal to zero gives you since f a b
is equal to zero we can say we may take b
00:07:16.639 --> 00:07:26.860
as a function of a b as the function a and
the arbitrary constant c and c is equal to
00:07:26.860 --> 00:07:38.550
phi a where phi is an arbitrary function
00:07:38.550 --> 00:07:46.169
so then z will be equal to a x plus h by a
h a into y plus phi a ok so that we have the
00:07:46.169 --> 00:08:05.899
equation a x plus h a into y minus z plus
phi a equal to zero thus we have this equation
00:08:05.899 --> 00:08:12.719
so this equation is same as the equation f
x y z a phi a equal to zero we differentiate
00:08:12.719 --> 00:08:36.459
it with respect to a so differentiating it
we get
00:08:36.459 --> 00:08:51.940
x plus h dash a into y plus phi dash a
equal to zero so we then on eliminating a
00:08:51.940 --> 00:08:59.529
between this equation let me call it as one
and this as two so when we eliminate a
00:08:59.529 --> 00:09:22.970
on eliminating a a between one and two we
get the general solution
00:09:22.970 --> 00:09:36.639
the general integral
00:09:36.639 --> 00:09:41.800
so this how we will obtain the general integral
from the complete integral we have already
00:09:41.800 --> 00:09:46.910
said that a partial differential equation
is said to be completely solved if we get
00:09:46.910 --> 00:09:52.130
the complete integral general integral and
the similar solution of the partial differential
00:09:52.130 --> 00:09:56.629
equation so we have obtained the complete
integral general integral we can obtain from
00:09:56.629 --> 00:10:04.000
here and let us now discuss how we can obtain
the complete integral now if you so now
00:10:04.000 --> 00:10:14.500
how to obtain the singular integral
to obtain the singular integral from the complete
00:10:14.500 --> 00:10:29.610
[ingra/integral] integral let us recall
that if the complete integral is f x y
00:10:29.610 --> 00:10:39.750
z a b equal to zero of the if the complete
integral is this of the partial differential
00:10:39.750 --> 00:10:56.790
equation
f x y z p q equal to zero then the complete
00:10:56.790 --> 00:11:03.689
integral is obtained then the similar int[egral]-
integral is obtained by eliminating a and
00:11:03.689 --> 00:11:49.089
b between the equations f x by z a b equal
to zero delta f over delta a equal to zero
00:11:49.089 --> 00:11:56.060
and delta f over delta b equal to zero ok
00:11:56.060 --> 00:12:01.910
so by eliminating the two arbitrary constants
a and b from these three equations we arrive
00:12:01.910 --> 00:12:08.029
at the singular integral now in this case
what happens is let us look at the complete
00:12:08.029 --> 00:12:21.430
integral the complete integral is z equal
to so here the complete integral is z equal
00:12:21.430 --> 00:12:39.120
to a a x plus b y b is equal to h a so h into
a into y plus c so there are two arbitrary
00:12:39.120 --> 00:12:48.290
constants a and c we can write it in the form
f z f x by z is equal to zero so are z
00:12:48.290 --> 00:12:55.540
minus a x minus h a into y minus c equal to
zero
00:12:55.540 --> 00:13:01.720
so this equation is of the form f x y z a
b equal to zero here the arbitrary constants
00:13:01.720 --> 00:13:21.389
are a and c now let us differentiate these
with respect to a so differentiating
00:13:21.389 --> 00:13:39.160
with respect to a we have ok so minus x minus
h dash a into y equal to zero and when we
00:13:39.160 --> 00:13:54.709
differentiate this with respect to c and next
differentiating with respect to the other
00:13:54.709 --> 00:14:04.430
or other arbitrary constant c we find minus
one equal to zero
00:14:04.430 --> 00:14:11.860
so we have to know eliminate the two arbitrary
constants a and c between this equation this
00:14:11.860 --> 00:14:16.800
equation and this equation now minus one equal
to zero is not possible so therefore there
00:14:16.800 --> 00:14:23.579
is no singular integral in this case so in
this case of the partial differential
00:14:23.579 --> 00:14:29.310
equation we have only complete integral
and general integral
00:14:29.310 --> 00:14:39.820
now let us go the example one p q plus
p plus q equal to zero so let us consider
00:14:39.820 --> 00:14:49.800
p q plus p plus q equal to zero now we can
see that this differential equation is
00:14:49.800 --> 00:14:55.610
a nonlinear differential equation because
the partial derivatives p and q they occur
00:14:55.610 --> 00:15:02.220
as a product in the term first term so
it is it is a first order differential
00:15:02.220 --> 00:15:08.370
equation but it is not linear in p and q now
so it end more over there there only here
00:15:08.370 --> 00:15:16.310
p and q occur x y z do not occur so it is
of the type one it is a partial differential
00:15:16.310 --> 00:15:18.519
equation of type one
00:15:18.519 --> 00:15:44.600
so its complete integral could be written
as z equal to a x plus b y plus c where
00:15:44.600 --> 00:15:56.769
if i write it is as f f p q f p q equal to
p q plus p plus q then here f a b equal
00:15:56.769 --> 00:16:06.389
to zero so f a b equal to zero or you can
say a b plus a plus b equal to zero so from
00:16:06.389 --> 00:16:14.330
here i can get the value of the b equal to
minus a upon a plus one and hence the complete
00:16:14.330 --> 00:16:39.019
integral is z equal to a x minus a over a
plus one into y plus c the from the complete
00:16:39.019 --> 00:16:55.119
integral we can obtain the general integral
so to obtain the general integral
00:16:55.119 --> 00:17:13.069
let us take c two b some arbitrary function
of a where phi is arbitrary where phi is arbitrary
00:17:13.069 --> 00:17:29.409
now then then we can say then one can be
one is one can be written as
00:17:29.409 --> 00:17:42.299
a x minus z minus a over a plus one into y
plus phi a equal to zero and differentiating
00:17:42.299 --> 00:18:12.970
this with respect to a we have x minus now
a over a plus one is nothing but one minus
00:18:12.970 --> 00:18:18.899
one upon a plus one so when you differentiate
a over a plus one with respect to a what you
00:18:18.899 --> 00:18:30.150
get is d over d a of a over a plus one this
giv[es]- this gives you one over a plus one
00:18:30.150 --> 00:18:39.050
whole square so here we shall have minus one
upon a plus one whole square into y plus phi
00:18:39.050 --> 00:18:46.710
dash a equal to zero so let me call it as
one equation number one and this is equation
00:18:46.710 --> 00:19:02.940
number two eliminating a between one and two
we
00:19:02.940 --> 00:19:16.830
obtain the general integral
00:19:16.830 --> 00:19:24.210
ok now to obtain the singular integral
as we did said earlier a and c are two
00:19:24.210 --> 00:19:29.050
arbitrary constants in this complete integral
so we shall differentiate it partially
00:19:29.050 --> 00:19:33.880
with respect to a and with respect to c and
when we differentiate this equation with respect
00:19:33.880 --> 00:19:39.690
to c what we will have is one equal to zero
if we write it as a x minus a over a plus
00:19:39.690 --> 00:19:46.120
one by plus c minus z equal to zero and then
differentiate it with respect to c we obtain
00:19:46.120 --> 00:19:50.880
one equal to zero and one equal to zero is
not possible so this is the singular sol[ution]-
00:19:50.880 --> 00:19:53.610
solution are singular integral does not exist
00:19:53.610 --> 00:20:08.409
again i repeat we have the complete integral
z equal to a x minus a over a plus one
00:20:08.409 --> 00:20:14.620
y plus c so you can write it as a x minus
a over a plus one y plus c minus z equal to
00:20:14.620 --> 00:20:19.070
zero then we have to differentiate partially
with respect to a and with respect to c when
00:20:19.070 --> 00:20:24.090
we then differentiated with respect to c we
get one equal to zero which is not possible
00:20:24.090 --> 00:20:30.970
and so this singular solution does not exist
in this case
00:20:30.970 --> 00:20:38.690
now let us go over two next question
where we will see that by suitable substitutions
00:20:38.690 --> 00:20:44.010
we shall be able to bring this differential
equation partial differential equation of
00:20:44.010 --> 00:20:51.850
first order which is non linear to the form
f p q equal to zero so let us analyze this
00:20:51.850 --> 00:21:05.130
equation closely we have x square p
square plus y square q square equal to z square
00:21:05.130 --> 00:21:11.179
let us divide this equation by z square assuming
that z is not equal to zero so we have x p
00:21:11.179 --> 00:21:23.149
over z whole square plus y q divided by z
whole square equal to one if z is not zero
00:21:23.149 --> 00:21:38.760
ok now this is what x over z p is delta z
by delta x plus this is y over z we can write
00:21:38.760 --> 00:21:51.210
like this now this can also expressed as or
one over z delta z divided by one by x delta
00:21:51.210 --> 00:22:16.620
x whole square plus one by z so let us define
capital x equal to small capital x equal to
00:22:16.620 --> 00:22:30.990
l n small x capital y equal to l n small y
and capital z equal to l n small z this small
00:22:30.990 --> 00:22:35.650
z
00:22:35.650 --> 00:22:51.130
so then
d x over x d x will be one over x d y over
00:22:51.130 --> 00:23:07.659
d y will be one by y and d z over d z this
capital z this is big z equal to one by z
00:23:07.659 --> 00:23:22.590
and so this can be written as delta
l n z i can write like this delta l n x whole
00:23:22.590 --> 00:23:38.019
square plus delta l n z over delta l n y whole
square equal to one or we can say delta z
00:23:38.019 --> 00:23:48.970
over delta x whole square plus delta z over
delta y whole square equal to one
00:23:48.970 --> 00:24:03.340
now this is of the type this equation is of
the type one
00:24:03.340 --> 00:24:28.690
so its complete integral is z equal to a x
plus b y plus c let me write small c a x plus
00:24:28.690 --> 00:24:39.789
b y plus c where b where we have
so where if delta z over delta x this is a
00:24:39.789 --> 00:24:45.230
so then b square is equal to one minus a square
so where b is equal to plus minus under root
00:24:45.230 --> 00:25:02.139
one minus a square so thus we have this is
l n z and then a times l n x plus minus under
00:25:02.139 --> 00:25:14.809
root one minus a square into l n by and this
c we can write as l n c dash let us write
00:25:14.809 --> 00:25:21.419
c as where c is equal to l n c dash
00:25:21.419 --> 00:25:30.080
so then we can write z equal to x to the power
a y to the power plus minus under root one
00:25:30.080 --> 00:25:41.100
minus a square into c dash let me write
it as z equal to c dash times x to the power
00:25:41.100 --> 00:25:49.430
a y to the power plus minus under root one
minus a square where a and c dash are two
00:25:49.430 --> 00:25:55.350
arbitrary constants so this is complete integral
of the given partial differential equation
00:25:55.350 --> 00:26:05.590
now what we will do is to determine the general
integral from this equation we have to c dash
00:26:05.590 --> 00:26:22.500
equal to phi into a so to determine the general
integral from the complement integral from
00:26:22.500 --> 00:26:34.529
let me it equation one from one we let c dash
equal to some arbitrary function of a where
00:26:34.529 --> 00:26:50.770
phi is arbitrary ok so we have phi a into
x to the power a y to the power plus minus
00:26:50.770 --> 00:26:57.220
under root one minus a square this is z this
is equal to this is equal to z
00:26:57.220 --> 00:27:09.510
now we can make it simpler ok by taking
say a equal to suppose a equal to cos alpha
00:27:09.510 --> 00:27:23.100
ok a is equal to cos alpha then we shall have
z equal to phi of cos alpha so i can write
00:27:23.100 --> 00:27:30.470
it as some psi alpha ok where psi alpha is
phi of cos alpha x to the power cos alpha
00:27:30.470 --> 00:27:47.840
and then y to the power plus minus psi alpha
where psi alpha is phi of cos alpha
00:27:47.840 --> 00:27:53.169
now we differentiate this equation with
respect to alpha ok so we shall have or
00:27:53.169 --> 00:28:03.320
we can say psi alpha minus x to the power
cos alpha sorry psi alpha into x to the power
00:28:03.320 --> 00:28:13.409
cos alpha y to the power plus minus sign alpha
minus z equal to zero so so what we will do
00:28:13.409 --> 00:28:24.309
is let me call it as equation number let me
call it as equation number two ok and then
00:28:24.309 --> 00:28:31.130
we differentiate this with respect to alpha
so the [five/psi] psi dash alpha into x to
00:28:31.130 --> 00:28:41.190
the power cos alpha y to the power plus minus
sign alpha plus psi alpha x to the power
00:28:41.190 --> 00:28:55.900
cos alpha when we differentiate x to the
power cos alpha with respect to alpha what
00:28:55.900 --> 00:29:07.020
we get is x to the power cos alpha and then
x to the power a if we differentiate with
00:29:07.020 --> 00:29:12.899
respect to a to the power x if we differentiate
with respect to x we get a to power x log
00:29:12.899 --> 00:29:14.159
a
00:29:14.159 --> 00:29:30.429
so x to the power this we are differentiate
a minus sign alpha and then we get x to
00:29:30.429 --> 00:29:35.279
the power alpha we are differentiating with
respect to alpha so x to the power cos alpha
00:29:35.279 --> 00:29:47.809
into minus sign alpha into l n x we shall
have y to the power plus minus sign alpha
00:29:47.809 --> 00:30:00.460
and then psi alpha x to the power cos alpha
y to the power plus minus sign alpha into
00:30:00.460 --> 00:30:11.120
plus minus cos alpha l n y minus equal to
zero so this our equation number let me call
00:30:11.120 --> 00:30:14.160
it equation number three
00:30:14.160 --> 00:30:21.520
so by eliminating alpha between equation number
two or this equation and three we get the
00:30:21.520 --> 00:30:54.820
general integral so general integral is obtained
this singular integral is obtained by eliminating
00:30:54.820 --> 00:31:01.900
a and c dash or you can say by eliminating
yeah by eliminating a and c dash between
00:31:01.900 --> 00:31:08.720
equation one and then its partial differentiation
with respect to a and c dash will give you
00:31:08.720 --> 00:31:14.120
two more equations by eliminating
a and c dash
00:31:14.120 --> 00:31:28.149
so by differentiating
by differentiating equation one with
00:31:28.149 --> 00:31:43.090
respect to a and c dash we will we where have
two more equations we have we will two more
00:31:43.090 --> 00:32:06.009
equations will have two equations if we take
it as the if we write f x y z a
00:32:06.009 --> 00:32:16.049
c dash equal to z minus c dash times x to
the power a y to the power plus minus under
00:32:16.049 --> 00:32:27.889
root one minus a square then equal to zero
then delta f by delta a equal to zero and
00:32:27.889 --> 00:32:32.250
delta f by delta c equal to zero
00:32:32.250 --> 00:32:37.470
ok so if we write f x y z c dash to be equal
to this then by differentiating this with
00:32:37.470 --> 00:32:42.419
respect to a and with respect to c dash we
shall have these two equations so eliminating
00:32:42.419 --> 00:33:04.460
a and c dash between these three equations
we shall see that
00:33:04.460 --> 00:33:19.039
three equations i can call them as [say/if]
four five six four to six we shall have
00:33:19.039 --> 00:33:29.730
we shall obtain z equal to zero z equal
to zero therefore the singular integral so
00:33:29.730 --> 00:33:52.080
in this case z is equal to zero is the singular
integral we can see that z equal to zero satisfies
00:33:52.080 --> 00:33:57.880
this equation x square p square plus y square
q square equal to z square because z equal
00:33:57.880 --> 00:34:03.200
to zero means p and q both are zeros so
both sides are zero zero
00:34:03.200 --> 00:34:14.020
so let's take third question where we have
x plus y times p plus q whole square and
00:34:14.020 --> 00:34:27.100
then x minus y p minus q whole square so we
shall make a certain subst[itution] some
00:34:27.100 --> 00:34:33.660
substitution and will reduce it to the
form which we have considered in type one
00:34:33.660 --> 00:34:46.781
so let us say let us let us define x equal
to root x plus y and y equal to root x
00:34:46.781 --> 00:34:54.320
minus y let us define capital x equal to this
and capital y equal to this then delta x over
00:34:54.320 --> 00:35:07.170
delta x is one over two root x plus y delta
x over delta y is one over two root x plus
00:35:07.170 --> 00:35:17.330
y and here
when we differentiate y with respect to x
00:35:17.330 --> 00:35:23.380
partially we get one over two root x minus
y and here we get delta y over delta y as
00:35:23.380 --> 00:35:28.890
minus one over two root x minus y
00:35:28.890 --> 00:35:42.460
now p is delta z over delta x we can write
it as delta z over delta x into delta x by
00:35:42.460 --> 00:35:52.220
delta x plus delta z over delta y into delta
y over delta x because capital x capital y
00:35:52.220 --> 00:35:58.640
both are functions of x so we have here
delta x over delta x is one over two root
00:35:58.640 --> 00:36:08.700
x plus y is one over two x so one over two
x and then delta z y over delta x delta y
00:36:08.700 --> 00:36:14.870
delta x one over two root x minus y and one
over two that x minus y is one over two y
00:36:14.870 --> 00:36:32.760
this is one over two x so we have one over
two y and similarly q which is delta z
00:36:32.760 --> 00:36:52.590
by delta y it can be written as yes ok and
this is equal to delta x over delta y this
00:36:52.590 --> 00:37:00.480
is one over two x and delta y over do delta
y is minus one over two y so we will have
00:37:00.480 --> 00:37:10.760
one over
00:37:10.760 --> 00:37:22.820
ok now let us add the equations one and two
so adding one and two we
00:37:22.820 --> 00:37:38.540
get p plus q p plus q equal to
one over two x delta z by delta x we get one
00:37:38.540 --> 00:37:55.090
over x
and p minus q similarly and subtracting two
00:37:55.090 --> 00:38:12.420
from one we get p minus q equal to one
over y delta z over delta y ok now so we will
00:38:12.420 --> 00:38:18.890
put the values here so x plus y x plus y is
x square so thus we thus the given equation
00:38:18.890 --> 00:38:33.550
becomes partial differential equation reduces
to we have x square into p plus q whole
00:38:33.550 --> 00:38:46.400
square so one over x square delta z by delta
x square plus x minus y x minus y is y square
00:38:46.400 --> 00:38:56.130
into p minus q whole square so one over y
square
00:38:56.130 --> 00:39:03.360
this equal to one
so here we have x square so this cancel with
00:39:03.360 --> 00:39:16.730
this and this cancels with this and we have
which is a partial differential equation of
00:39:16.730 --> 00:39:23.170
the form one where z is the dependent variable
capital x capital y are the independent variables
00:39:23.170 --> 00:39:40.880
so it is of type one so the complete integral
is z equal to a x plus minus under root
00:39:40.880 --> 00:39:52.280
one minus a square this is capital x here
capital y plus some constant let us say c
00:39:52.280 --> 00:40:01.990
let us put the value of capital x so
a times square root x plus y plus minus under
00:40:01.990 --> 00:40:14.091
root one minus a square then y is a square
root x minus y plus c so z equal to a times
00:40:14.091 --> 00:40:20.580
square root x plus y plus minus square root
one minus a square square root x minus y plus
00:40:20.580 --> 00:40:26.590
c this is a square plus b square equal
to one so b is plus minus under root one minus
00:40:26.590 --> 00:40:30.640
a square which we have put here so this
is the complete integral
00:40:30.640 --> 00:40:35.620
now from the complete integral as we have
seen on the earlier examples we will for
00:40:35.620 --> 00:40:40.730
the gen[eral] general integral we will write
c as five a and then differentiate the
00:40:40.730 --> 00:40:46.720
equation with respect to a and then eliminate
a between this equation where c is replaced
00:40:46.720 --> 00:40:51.860
by five a and its derived equation where we
differentiate this equation with respect to
00:40:51.860 --> 00:40:57.590
a partially and then in order to find the
and then we eliminate a so to get the general
00:40:57.590 --> 00:41:04.110
integral and when we want to find the
singular integral we eli[minate] eliminate
00:41:04.110 --> 00:41:09.460
the two arbitrary constants a and c between
this equation and its derived equations which
00:41:09.460 --> 00:41:13.800
are the partial derivatives of this equation
with respect to a and c
00:41:13.800 --> 00:41:20.880
so by eliminating a a and c between this equation
and those equations we can obtain the singular
00:41:20.880 --> 00:41:28.060
integrals that the usual that is means
by the usual method of obtaining the
00:41:28.060 --> 00:41:35.620
singular integral we can do this so once we
have the complete integral we can obtain
00:41:35.620 --> 00:41:42.130
the general integral and the singular singular
integral for the given partial differential
00:41:42.130 --> 00:41:51.290
equation so this how we solve the problems
which occur in the case f p q equal
00:41:51.290 --> 00:41:57.810
to zero that is of type one where the differential
equation only contains the partial derivatives
00:41:57.810 --> 00:42:05.000
p and q the variables x y z do not occur
but that i would like to conclude my lecture
00:42:05.000 --> 00:42:06.620
thank you very much for your attention