hello friends welcome to my third lecture on solution of first order nonlinear equations
in previous two lectures we have discussed first order nonlinear equations of type
one two and three in this lecture we shall discuss the forth type of standard type
of first order nonlinear equations and then we shall discuss the general method of
solving such nonlinear equations first of first order which cannot be reduced
to any those four standard forms so let us begin with fourth order of first order
nonlinear equations of type four so in the in this category we consider
those kind of equations which are of the form z equal to p x plus q y plus f p q and as
you remember p and q denotes the partial derivatives of z with respect to x and y so if z
is p x plus q y plus f p q then this equation you can see analogous to clairauts ordinary
differential equation the clalelout clairauts clairauts ordinary differential equation
is y plus p x plus f p where p is d z by dx in the ordinary differential equation clairauts
form is p x plus f p where p is dy by d x so so the partial differential equation
z equal to p x plus q y plus f p q is analogous to the clairauts ordinary differential
equation y equal to p x plus f p the complete integral of equation one is then given
by z equal to a x plus b y plus f a b because we can you can see that when z equal to
a x plus b y plus f a b then your partial derivative with of z with respect to x which
is which we we have denoted by p it will be equal to a and q which is the partial derivative
of z with respect to y it will be equal to b and therefore y equal therefore z
equal to a x plus b y plus f a b this is a solution of equation one because when
you put z equal to a x plus b y plus f a b into one it is satisfied due to the fact that
the value of p is a and the value of q is b
now so this complete solution you can see z equal to p a x plus b y plus f a b this
solution it represents plane and it is at two para[meter]- plane family
of two two parameter family of planes because there are two parameters a and b so
it is at two parameters family of planes now inorder to obtain the general integral as
we have earlier discussed let us put b equal to phi a in this solution and where
phi denotes an arbitrary function then z will be equal to a z equal to a x plus y into phi
a plus y into phi a plus f a phi a we differentiate this equation with respect to a to get z d
zero equal to x plus y phi dash a plus d over d a of f a phi a and then eliminate a between
equation three this equation and the equation four to get the general integral
now inorder to obtain the singular integral we differentiate equation two this one
z equal to a x plus b y plus f a b which is the complete integral with respect v we differentiate
this with respect to a so we will get zero equal to x plus d over delta over delta
a of f a b zero equal to x x plus delta f over delta a and the second one will be when
we differentiate with respect to b we will get zero equal to y plus delta f over delta
v so we have three equations now equation two which is the complete integral this one
and then the equations three sorry f[ive]- five z zero equal to x plus delta f over delta
a and six that is zero equal to y plus delta f over delta v so we eliminate a and b between
these three equations two reach the singular integral
let us see how we apply this method to some differential equations let us begin with
the first example z equal to p x plus q y plus f p q so let us consider z equal to p
x plus q y plus f p q and we can see of the clairauts form z equal to p x plus sorry
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
pq is equal to p q so it is a equation of type four now let us so the complete integral
is z equal to a x plus b y plus a b so this is the complete integral to find the singular
integral let us differentiate one with respect to a differentiating
with respect to a and b we obtain zero equal to x plus b and
when we when we differentiate with respect to v we will get zero equal to x plus b and
when we differentiate with respect to b we will get zero plus y plus a so which imply
x is equal to minus b b equal to minus x
lets write like this b equal to minus x anda equal to minus y so now eliminating
eliminating a and b between equation one and these two equations we have eliminating
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
minus x into minus y so we get minus x y see this is the which is the singular integral
in this case to determine the general integral let us put
b equal to phi a where phi is an arbitrary function
ok so then z will be equal to a x plus y into five a plus a into phi a now we differentiate
with respect to a so differentiating
we get zero equal to x plus y times phi dash a plus phi a into a phi dash a let
let me call it as equation number two and this as equation number three so then we eliminate
a between equations two and three to determine the general integral so eliminating
a between two and three we ob[tain]- obtain the desired result so we will get the general
integral so that is how we solve the problem given in example one now lets take up the
problem number two so we have z equal to p x plus q y plus under root one plus p square
plus q square now we can see that here one plus f p q is
equal to under root one plus p square plus q square so it is of the clairauts form
so this the complete integral is therefore a z equal to a x plus b y plus under root
one plus a square plus b square to determine the complete integral to determine the this
is the complete detail to obtain the singular integral we differentiate with respect to
a and b differentiating we get zero equal to x plus a upon under root one plus a square
plus b square and when we differentiate with respect to b we get zero equal to y plus
b upon under root one plus a square plus b square
now from these two equations we get x equal minus a upon under root one plus a square
plus b square and y equal to minus b upon under root a square plus plus one plus a square
plus b square so x square plus y square is equal to a square plus b square divided by
one plus a square plus b square which can be written as one minus one upon one plus
a square plus b square and so one upon one plus a square plus b square is equal to
one minus x square minus y square or we can say one plus a square plus b square
is equal to one over one minus x square minus y square
now x is equal to minus a upon under root one plus a square plus b sq[uare]- we see
so x is equal to minus a upon under root one plus a square plus b square we will
get a as a equal to minus x into under root one plus a square plus b square so minus
x times
ok so a is equal to minus x into under root one plus a sq[uare]- which is this equal to
this and similarly b is equal to b is equal to minus y minus y upon under root one
minus x square minus b square yeah so now eliminating a and b between equation one
and ok so eliminating a and b between equations one two three we
get the singular integral we get the singular integral and for general integral as we
have discussed earlier so for the general integral
let us put b equal to phi a then one becomes z equal to a x plus
b phi plus by phi a plus under root one plus a square plus phi a whole square we differentiate
this equation with respect to a and get zero equal to this implies zero equal to x plus
y phi dash a plus one over two times under root one plus a square plus phi a whole square
and then we have two a plus two times phi a into phi dash a ok this two can be cancelled
so we have this equation so eliminating a between this equation and this equation
this we call it as this as four and this as five so eliminating a between four and five
we get the
desired inte[gral]- desired integral ok this is how we solve these two equations now
we go to the general method of solving first order nonlinear equations which we
call as the charp[its]- charpits method so here in this method we consider nonlinear
partial differential equation of first order which cannot be reduced to the four standard
forms which we have already discussed we will apply this method only in the cases where
the first order non linear partial differential equations cannot be reduced to the cannot
be [dedu/reduce] reduce to the four standard form because to find the complete integral
by applying the charpits method its rather more cumbersome
so we prefer to solve the partial differential equations which are of first order and nonlinear
by trying to reduce them to the four standard forms if if it is not possible to do that
then we apply the charpits method so here what we do is let us consider first order
nonlinear partial differential equation as a given by f x y z p q equal to zero we know
that since there are two independent variables x and y and z is dependent on x and y so we
can write the d z equal to p d x plus q d y
now in the charpits method we try to find relation between x y z and p q say we can
which when from which and the given partial differential equation which when we
when we obtain the values of p and q and put them in the equation d z equal to p d x plus
q d y if it is integrable we get the complete integral of the given partial differential
equation so in this method we will try to find a relation involving x y z p q equal
to zero and then saw which this uh equation three and the equation one the given partial
differential equation be then solve to find the values of p and q which when we put in
d z equal to p d x plus q d y if it is integrable we get the complete integral of the given
partial differential equation so let us assume that the three this
equation z phi z phi x y z p q equal to zero is the required relation then z p and q
so p is delta z by delta x q is delta z by delta y they are expressive they are functions
of x and y so that when they are substituted in the equations f equal to zero this equation
f equal to zero and phi equal to zero ok they are satisfied identically so now
when we differentiate f equal to zero with respect to x what we get is delta f by delta
x plus delta fby delta z into delta z by delta x which which is which is equal to p plus
delta f by delta p delta p by delta x delta f by delta q delta q by delta x equal to zero
this is obtained by solving f equal to zero with respect to x similarly when we solve
differentiate phi with respect to x we will get delta phi by delta x plus delta phi
by delta z into p delta phi by delta p delta p by delta x plus delta phi by delta q delta
q by delta x is equal to zero and two similar equations are obtained from f equal to zero
phi equal to zero by differentiating them with respect to y and they are delta f over
delta y plus delta f over delta z delta z by delta y which is q delta f over delta p
delta p by delta y delta f over delta q delta q by delta y equal to zero and delta phi over
delta y plus delta phi over delta z delta z by delta y is q delta phi over delta phi
delta delta phi over delta p delta p over delta y delta phi over delta q and delta q
over delta y equal to zero now what we do is let us eliminate delta
p over delta x from the first pair there are four equations so from the first two equations
first pair of equations let us eliminate this derivative derivate of p with respect
to x so delta p over delta x let us eliminate from these two equations and which is not
difficult because it is a linear equation so this is a so this will give you delta
f over delta x delta f over delta z into p delta f over delta q del[ta]- so the this
we can obtain this equation is very simple and then what we do is we can write it as
we can write it in the form the we collect the terms which contain p and
the terms which contain delta q by delta x and this constant this term which is free
from p and delta q by delta x so we can write those that this equation
we can write this equation in the form this one ok so we similarly let us eliminate delta
q over delta y to this delta q over delta y in let us eliminate from this pair of equations
so for that you need to multiply this equation by delta phi over delta q and this fourth
equation by delta f over delta q and then subtract so delta q over delta y term
will vanish so eliminating delta q over delta y between the last pair of equations we shall
have this one delta f over delta y delta phi by delta q minus delta phi by delta y delta
f by delta q plus q times this plus delta v over delta y times this equal to zero
now let us notice one thing delta q is delta z by delta y ok so delta q by delta
x is delta square z by delta x delta y and p is delta z by delta x so delta p by delta
y is delta square z by delta x delta y and assuming that the second order partial derivatives
are continuous delta square z by delta x delta y is same as delta square z by delta y delta
x we know is we know this result so delta q over delta x and delta p over delta y are
same and this they are coefficient you can see is also same except for the negative
sign this is negative of this so when we add these two equations ok the last term this
term here and this term here they will cancel out
so now let us notice that delta q over delta x ahis equal to delta p over delta
y so we have this one delta q over delta x equal to delta square z by delta x delta
y and which is equal to delta p over delta y because we are assuming that second order
partial derivatives are continuous so when we add the equations four and five as we have
discussed just now this is negative of this quantity and this delta q over delta
x and delta p over delta y are same are same so when we add the equations four and
five and rearrange the terms what happens is this term and this term they cancel out
and what we get is this so delta f over delta x plus p times delta f over delta z
into delta phi over p plus delta f over delta y plus q times delta f over delta z delta
phi by delta q plus minus p delta f f by delta p minus q delta f by delta q delta phi by
delta z plus minus delta f by delta p into delta phi by delta x minus delta f by delta
q into delta phi over delta y equal to zero as since the terms involving are the partial
derivative of p with respect to y and partial derivative of q with respect to x cancel
now the equation six is a partial differential equation of the first order in q no phi let
us see this is a partial differential equation of the first order in phi ok because phi
is phi is a function of x y z p and q and you can see here we have in the first term
partial derivative of phi with respect to p partial derivative of phi with respect to
q partial derivative of phi with respect to z partial derivative of phi with respect to
x partial derivative of phi with respect to y which are of first order so it is a first
order linear differential equation in phi and therefore its integrals are the integrals
of the auxiliary equations as we have discussed in the case of lagranges linear equation in
p and q it is same it is of the same so we have d p over this quantity and d
q over this d z over this quantity and d x over minus delta f by delta p d y over minus
delta f by delta q equal to d phi over zero we are we are actually solving this equations
this this equation like we have sol[ved]- solved the lagranges linear equation in p
and q so it is a linear equation in the derivatives of phi with respect to p q z x
and y so we solve this auxiliary system of equations now any of the integrals of this
equation seven will satisfy six and therefore we take the simplest relation involving
atleast one of p and q for phi equal to zero so that the values of p and q may be obtained
easily once the values of p and q are obtained easily we can take the help of the given
equation f x y z p q equal to zero and determine the values of the determine the value of
the other other q r p and once p and q are known we can use d z is equal to p d
x plus q d y and determine the complete integral
so we shall be able to determine the complete integral by writing the auxiliary system of
equations and this auxiliary system of equations will then will satis[fy]- we will
find the shortest simplest relation involving at least one of p and q for phi equal to zero
and then we may use the these value this relation for p with in p and q and
determine the values of p and q so when we put them in the equation dz equal to p
d x plus q d y and integrate we shall obtain the complete integral which will be the
complete integral of the given partial differential equation and the similar and
a general integrals are then obtained in the usual manner so with that i would like to
conclude my lecture thank you for your attention