hello friends welcome to my lecture on solution of first order nonlinear equation
so this is our first the lecture on the how to find the solution of nonlinear equations
that are of first degree but not of that are of order but not of first degree so let
us see we have so far discussed how to determine the solution of a lagrange's linear
partial differential equation which is of the form p into p plus q into q equal to r
where p q and r as you know they are functions of x by z and so this is a first order and
first degree linear partial differential equation i mean partial differential equation which
is linear in p and q
now we shall be considering first order partial differential equations which are not
of first degree so if there are not of first degree then they will be nonlinear partial
differential equations so let us see how we solve first order nonlinear partial differential
equations and such solution of nonlinear partial differential equations present considerable
difficulties so first we will focus on before giving the general method that
is the char[pit] charpit method of solving such equations we shall present some
special types of equations whose solutions can be easily determined so many equations
we shall see can be reduced to few standard forms and therefore can be [jolved/solved]
solved generally which by methods which are shorter than charpit method
so let us first discuss the equations of the form f p q equal to zero such partial
differential equations where the partial derivatives p and q that is delta z by delta x and
delta z delta y occur and they do not involve the variables x y and z they consider they
they come under this category of the equations of the form f p q equal to zero
so in the case of these equations f p equal to zero we see that the complete integral
is given by z equal to a x plus b y plus c now let i want to remind you that the complete
integral is defined as the solution of the differential equation which contains as many
arbitrary constants as there are independent variables in the differential equations so
there are two independent variables here x and y so this is a solution of the partial
differential equation f p q equal to zero because here a and b are connected by
the relation f a b equal to zero if you find the partial derivatives of this equation
z equal to a x plus b y plus c with respect to x and y then what we get is the derivative
of z with respect to x which is p we have denoted by p comes out to be a and the derivative
of z with respect to y which we have denoted by q comes out to be b which when substituted
in equation two that is f a b equal to zero give us f p q equal to zero
so z equal to a x plus b y plus c is a solution of f p q equal to zero and moreover it
involves two arbitrary constants they are a and c b is not an arbitrary constant b is
connected to a y by the relation f a b equal to zero so b can be said as the function
of a so there are two arbitrary constants a and c and because of these two arbitrary
constants and this is a solution of the equation f p q equal to zero which contains two two
independent variables it can be called as these complete integral of f p q equal
to zero
now so now as be if you remember of to from the complete integral one can determine
the general solution if the complete integral is given by
f x y z a b equal to zero where a and b are arbitrary constants
then to determine the general general integral what we do is let us say let b be an arbitrary
function of a where phi is arbitrary so then we have the equation one then we have
from one f x y z a phi a equal to zero now by eliminating
a the arbitrary constant a between the equations f x y z a phi a equal to zero and the partial
derivative of f with respect to a equal to zero between these two equations b obtained
the general integral of the partial differential equation of first order differential equation
of f of f x y z p q equal to zero
so this we have known already ok so in order to find the general solution from the
complete integral ok the complete integral is z equal to a x plus b y plus c ok f
a b equal to zero gives you since f a b is equal to zero we can say we may take b
as a function of a b as the function a and the arbitrary constant c and c is equal to
phi a where phi is an arbitrary function
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
equation a x plus h a into y minus z plus phi a equal to zero thus we have this equation
so this equation is same as the equation f x y z a phi a equal to zero we differentiate
it with respect to a so differentiating it we get
x plus h dash a into y plus phi dash a equal to zero so we then on eliminating a
between this equation let me call it as one and this as two so when we eliminate a
on eliminating a a between one and two we get the general solution
the general integral
so this how we will obtain the general integral from the complete integral we have already
said that a partial differential equation is said to be completely solved if we get
the complete integral general integral and the similar solution of the partial differential
equation so we have obtained the complete integral general integral we can obtain from
here and let us now discuss how we can obtain the complete integral now if you so now
how to obtain the singular integral to obtain the singular integral from the complete
[ingra/integral] integral let us recall that if the complete integral is f x y
z a b equal to zero of the if the complete integral is this of the partial differential
equation f x y z p q equal to zero then the complete
integral is obtained then the similar int[egral]- integral is obtained by eliminating a and
b between the equations f x by z a b equal to zero delta f over delta a equal to zero
and delta f over delta b equal to zero ok
so by eliminating the two arbitrary constants a and b from these three equations we arrive
at the singular integral now in this case what happens is let us look at the complete
integral the complete integral is z equal to so here the complete integral is z equal
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
constants a and c we can write it in the form f z f x by z is equal to zero so are z
minus a x minus h a into y minus c equal to zero
so this equation is of the form f x y z a b equal to zero here the arbitrary constants
are a and c now let us differentiate these with respect to a so differentiating
with respect to a we have ok so minus x minus h dash a into y equal to zero and when we
differentiate this with respect to c and next differentiating with respect to the other
or other arbitrary constant c we find minus one equal to zero
so we have to know eliminate the two arbitrary constants a and c between this equation this
equation and this equation now minus one equal to zero is not possible so therefore there
is no singular integral in this case so in this case of the partial differential
equation we have only complete integral and general integral
now let us go the example one p q plus p plus q equal to zero so let us consider
p q plus p plus q equal to zero now we can see that this differential equation is
a nonlinear differential equation because the partial derivatives p and q they occur
as a product in the term first term so it is it is a first order differential
equation but it is not linear in p and q now so it end more over there there only here
p and q occur x y z do not occur so it is of the type one it is a partial differential
equation of type one
so its complete integral could be written as z equal to a x plus b y plus c where
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
to zero so f a b equal to zero or you can say a b plus a plus b equal to zero so from
here i can get the value of the b equal to minus a upon a plus one and hence the complete
integral is z equal to a x minus a over a plus one into y plus c the from the complete
integral we can obtain the general integral so to obtain the general integral
let us take c two b some arbitrary function of a where phi is arbitrary where phi is arbitrary
now then then we can say then one can be one is one can be written as
a x minus z minus a over a plus one into y plus phi a equal to zero and differentiating
this with respect to a we have x minus now a over a plus one is nothing but one minus
one upon a plus one so when you differentiate a over a plus one with respect to a what you
get is d over d a of a over a plus one this giv[es]- this gives you one over a plus one
whole square so here we shall have minus one upon a plus one whole square into y plus phi
dash a equal to zero so let me call it as one equation number one and this is equation
number two eliminating a between one and two we
obtain the general integral
ok now to obtain the singular integral as we did said earlier a and c are two
arbitrary constants in this complete integral so we shall differentiate it partially
with respect to a and with respect to c and when we differentiate this equation with respect
to c what we will have is one equal to zero if we write it as a x minus a over a plus
one by plus c minus z equal to zero and then differentiate it with respect to c we obtain
one equal to zero and one equal to zero is not possible so this is the singular sol[ution]-
solution are singular integral does not exist
again i repeat we have the complete integral z equal to a x minus a over a plus one
y plus c so you can write it as a x minus a over a plus one y plus c minus z equal to
zero then we have to differentiate partially with respect to a and with respect to c when
we then differentiated with respect to c we get one equal to zero which is not possible
and so this singular solution does not exist in this case
now let us go over two next question where we will see that by suitable substitutions
we shall be able to bring this differential equation partial differential equation of
first order which is non linear to the form f p q equal to zero so let us analyze this
equation closely we have x square p square plus y square q square equal to z square
let us divide this equation by z square assuming that z is not equal to zero so we have x p
over z whole square plus y q divided by z whole square equal to one if z is not zero
ok now this is what x over z p is delta z by delta x plus this is y over z we can write
like this now this can also expressed as or one over z delta z divided by one by x delta
x whole square plus one by z so let us define capital x equal to small capital x equal to
l n small x capital y equal to l n small y and capital z equal to l n small z this small
z
so then d x over x d x will be one over x d y over
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
and so this can be written as delta l n z i can write like this delta l n x whole
square plus delta l n z over delta l n y whole square equal to one or we can say delta z
over delta x whole square plus delta z over delta y whole square equal to one
now this is of the type this equation is of the type one
so its complete integral is z equal to a x plus b y plus c let me write small c a x plus
b y plus c where b where we have so where if delta z over delta x this is a
so then b square is equal to one minus a square so where b is equal to plus minus under root
one minus a square so thus we have this is l n z and then a times l n x plus minus under
root one minus a square into l n by and this c we can write as l n c dash let us write
c as where c is equal to l n c dash
so then we can write z equal to x to the power a y to the power plus minus under root one
minus a square into c dash let me write it as z equal to c dash times x to the power
a y to the power plus minus under root one minus a square where a and c dash are two
arbitrary constants so this is complete integral of the given partial differential equation
now what we will do is to determine the general integral from this equation we have to c dash
equal to phi into a so to determine the general integral from the complement integral from
let me it equation one from one we let c dash equal to some arbitrary function of a where
phi is arbitrary ok so we have phi a into x to the power a y to the power plus minus
under root one minus a square this is z this is equal to this is equal to z
now we can make it simpler ok by taking say a equal to suppose a equal to cos alpha
ok a is equal to cos alpha then we shall have z equal to phi of cos alpha so i can write
it as some psi alpha ok where psi alpha is phi of cos alpha x to the power cos alpha
and then y to the power plus minus psi alpha where psi alpha is phi of cos alpha
now we differentiate this equation with respect to alpha ok so we shall have or
we can say psi alpha minus x to the power cos alpha sorry psi alpha into x to the power
cos alpha y to the power plus minus sign alpha minus z equal to zero so so what we will do
is let me call it as equation number let me call it as equation number two ok and then
we differentiate this with respect to alpha so the [five/psi] psi dash alpha into x to
the power cos alpha y to the power plus minus sign alpha plus psi alpha x to the power
cos alpha when we differentiate x to the power cos alpha with respect to alpha what
we get is x to the power cos alpha and then x to the power a if we differentiate with
respect to a to the power x if we differentiate with respect to x we get a to power x log
a
so x to the power this we are differentiate a minus sign alpha and then we get x to
the power alpha we are differentiating with respect to alpha so x to the power cos alpha
into minus sign alpha into l n x we shall have y to the power plus minus sign alpha
and then psi alpha x to the power cos alpha y to the power plus minus sign alpha into
plus minus cos alpha l n y minus equal to zero so this our equation number let me call
it equation number three
so by eliminating alpha between equation number two or this equation and three we get the
general integral so general integral is obtained this singular integral is obtained by eliminating
a and c dash or you can say by eliminating yeah by eliminating a and c dash between
equation one and then its partial differentiation with respect to a and c dash will give you
two more equations by eliminating a and c dash
so by differentiating by differentiating equation one with
respect to a and c dash we will we where have two more equations we have we will two more
equations will have two equations if we take it as the if we write f x y z a
c dash equal to z minus c dash times x to the power a y to the power plus minus under
root one minus a square then equal to zero then delta f by delta a equal to zero and
delta f by delta c equal to zero
ok so if we write f x y z c dash to be equal to this then by differentiating this with
respect to a and with respect to c dash we shall have these two equations so eliminating
a and c dash between these three equations we shall see that
three equations i can call them as [say/if] four five six four to six we shall have
we shall obtain z equal to zero z equal to zero therefore the singular integral so
in this case z is equal to zero is the singular integral we can see that z equal to zero satisfies
this equation x square p square plus y square q square equal to z square because z equal
to zero means p and q both are zeros so both sides are zero zero
so let's take third question where we have x plus y times p plus q whole square and
then x minus y p minus q whole square so we shall make a certain subst[itution] some
substitution and will reduce it to the form which we have considered in type one
so let us say let us let us define x equal to root x plus y and y equal to root x
minus y let us define capital x equal to this and capital y equal to this then delta x over
delta x is one over two root x plus y delta x over delta y is one over two root x plus
y and here when we differentiate y with respect to x
partially we get one over two root x minus y and here we get delta y over delta y as
minus one over two root x minus y
now p is delta z over delta x we can write it as delta z over delta x into delta x by
delta x plus delta z over delta y into delta y over delta x because capital x capital y
both are functions of x so we have here delta x over delta x is one over two root
x plus y is one over two x so one over two x and then delta z y over delta x delta y
delta x one over two root x minus y and one over two that x minus y is one over two y
this is one over two x so we have one over two y and similarly q which is delta z
by delta y it can be written as yes ok and this is equal to delta x over delta y this
is one over two x and delta y over do delta y is minus one over two y so we will have
one over
ok now let us add the equations one and two so adding one and two we
get p plus q p plus q equal to one over two x delta z by delta x we get one
over x and p minus q similarly and subtracting two
from one we get p minus q equal to one over y delta z over delta y ok now so we will
put the values here so x plus y x plus y is x square so thus we thus the given equation
becomes partial differential equation reduces to we have x square into p plus q whole
square so one over x square delta z by delta x square plus x minus y x minus y is y square
into p minus q whole square so one over y square
this equal to one so here we have x square so this cancel with
this and this cancels with this and we have which is a partial differential equation of
the form one where z is the dependent variable capital x capital y are the independent variables
so it is of type one so the complete integral is z equal to a x plus minus under root
one minus a square this is capital x here capital y plus some constant let us say c
let us put the value of capital x so a times square root x plus y plus minus under
root one minus a square then y is a square root x minus y plus c so z equal to a times
square root x plus y plus minus square root one minus a square square root x minus y plus
c this is a square plus b square equal to one so b is plus minus under root one minus
a square which we have put here so this is the complete integral
now from the complete integral as we have seen on the earlier examples we will for
the gen[eral] general integral we will write c as five a and then differentiate the
equation with respect to a and then eliminate a between this equation where c is replaced
by five a and its derived equation where we differentiate this equation with respect to
a partially and then in order to find the and then we eliminate a so to get the general
integral and when we want to find the singular integral we eli[minate] eliminate
the two arbitrary constants a and c between this equation and its derived equations which
are the partial derivatives of this equation with respect to a and c
so by eliminating a a and c between this equation and those equations we can obtain the singular
integrals that the usual that is means by the usual method of obtaining the
singular integral we can do this so once we have the complete integral we can obtain
the general integral and the singular singular integral for the given partial differential
equation so this how we solve the problems which occur in the case f p q equal
to zero that is of type one where the differential equation only contains the partial derivatives
p and q the variables x y z do not occur but that i would like to conclude my lecture
thank you very much for your attention