hello friends welcome to my second lecture on solution of lagrange's equation it is in
continuation of my previous lecture on this topic here first we discuss the how
geometrical interpretation of the lagrange's linear equation which is p into p plus q into
q equals r where small p and small q represent the partial derivative subject with respect
to x and y now we may write this equation pp plus q q equal to rs pp plus qq plus
r times minus one equal to zero now we know that the direction cosines of the normal
to a surface z is equal to f x y are given by delta z over delta x delta z over
delta y and minus one so direction cosines of the normal to the surface z equal to f
x y are proportional to p q and minus one +
now if the surfaces f x y z equal to zero then the direction cosines of the normal to
the surface are proportional to its partial derivatives of f with respect to x
by z because we know that del f is a vector normal to the surface f x y z equal
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
del y plus k del f by del z so it is known that if the surface is given by f x by
z equal to zero then del f is a vector normal to the surface so the direction cosines of
the normal to the surface are proportional to partial derivatives of f with respect
to x y z that is [dual f/delta f] over delta f over delta x delta f over delta y delta
f over delta z
now or we can also say that this is there are proportional to minus delta f over
delta x over delta f over delta z minus delta f over delta y over delta f over delta z and
minus one or delta z over delta x delta z over delta y and minus one because if f x
y z equal to zero then when we differentiate it with respect to x what we get is so which
implies that and similarly
when we differentiate with respect to y the equation f x y z equal to zero we get
ok so the ratio is minus delta f by delta x over delta f by delta z becomes delta z
by delta x minus delta f over delta y over delta f by delta z becomes delta z over delta
y and so we can say that these ratios are same as the partial derivatives of z with
respect to x and y
so delta z over delta x delta z over delta y and minus one that is p q and minus one
now hence the geometrical interpretation of the equation one is that let us look at this
p into p plus q into q plus r into minus one equal to zero now we are seeing that the direction
cosines of the normal to the surface are proportional to p q and minus one and therefore we can
geometrically interpret this equation one as the normal at a point to a certain surface
is perpendicular to a line whose direction direction cosines are in the ratio p q and
r because we know that if the two lines are perpendicular to each other then the dot
product of their direction ratio is equal to zero so we have normal to a point
at a point to a certain surface is perpendicular to join whose directional ratios are in the
ratio p q and r
now let us look at the simultaneous equations d x by p equal to d y by q equal to d z by
r it represents the family of curves such that the tangent at any point has the direction
ratios proportional to p q and r ok now if u is equal to a and v is equal to b are two
particular integrals of these equations then phi u v equal to zero represents a surface
through these curves let us take a point p dash on the surface represented by phi
u v equal to zero then through this point passes a curve of the family which lies entirely
on the surface the normal to the surface at the point p dash must therefore be at right
angles to the tangent at this point to the curve that is it is perpendicular to a line
whose direction cosines are proportional to p q and r and this is exactly what we require
in the partial differential equation one
so one and two one and two this is two one and two define the same set of surfaces and
are therefore equivalent so if the solutions the solutions u and v of the equations
subsidiary equations d x over p equal to d y over q equal to d z over r can be obtained
easily when the variables are separable if the variables are not separable then we shall
express it in the form d x over p equal to d y over q equal to d z over r equal to lambda
d x plus mu d y plus delta d z over lambda p plus mu q plus delta r where lamda mu delta
are certain functions of x y z such that lambda p plus mu q plus delta r equal to zero
so we search the functions lambda mu and delta in such a way that lambda into p plus
mu q plus delta r equal to zero now if this equation one lambda p plus mu q plus delta
r equal to zero we will have lambda d x plus mu d y plus delta z equal to ze[ro] d z
equal to zero now you on choosing lambda mu new and lambda mu and delta in such a way
that lambda p plus mu q plus delta r equal to zero from here it turns out that lambda
d x plus mu d y plus delta d z equal to zero
so now if the equation lambda d x plus mu d y plus delta d z equal to zero is integrable
we will obtain a solution of the subsidiary equations then we can make another choice
of lambda mu delta and obtain another solution are from this solution using this solution
we can obtain another independent solution of the subsidiary equations so how we will
arrive at the general integral of a partial differential equation of first odd and first
degree using this lagrange's method let it is illustrated in the examples below so let
us first take the example of y square z into p minus x square z into q equal to x square
y
so y square z into p minus x square z into q equal to x square y so the subsidiary equations
are d x over p p is y square z equal to d y over minus x square z that which is q and
then r is x square y so d z upon x square y we will find two independent solutions u
and v from the subsidy equations so let us first consider d x over y square z equal
to d y over minus x square z now here so this z and z can be cancelled
out and then we see that x square d x plus y square d y equal to zero so the variables
x and y can are separated here and we can integrate this integrating this we get x cube
by three plus y cube by three equal to a constant or we can say x cube and y cube equal to a
where a is an arbitrary constant now we can find another solution of this is u
x y z equal to a we can find v x y z equal to v from the other equations d y over minus
x square z equal to d x upon x square y now here we can cancel x square and we get y d
y plus z d z equal to zero so again z d z equal to zero so again the variables are
separable and therefore here we we can integrate and we will get y square by two plus z square
by two equal to a constant
so we can write y square plus z square as some constant let us say b ok so this is b
x y z equal to b and thus the general integral is phi [pi/pi] is an arbitrary function phi
u v equal to zero so x cube plus y cube y square plus z square equal to zero so this
is how we will find the general solution of the lagrange's equation given in example
one here we have seen that the it is easy to solve this subsidiary equation because
it is the case of case where the variables are separable we can easily separate the variables
x and y so it was easy to solve now let us take the example two
so in example two z minus y into p plus x minus z into q equal to x by minus x ok so
we will have subsidiary equations as d x over z minus y equal to d y over x minus z equal
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
to d y over x minus z the variables x y z are not separable so it is a case where the
variables are not separable so we shall consider this equal to lambda d x plus mu d y plus
delta d z divided by lambda times x minus y plus mu times x minus z plus delta times
y minus x
now let us make a choice of lambda mu and delta in such a way that the denominator here
is zero ok so let us choose lamda mu and delta all to be equal to one ok let us choose lambda
mu delta equal to one then we see that x minus y plus x minus z z minus y z minus y this
is z minus y z minus y x minus z ok this is z minus y plus x minus z plus y minus x
this cancels with this y cancels with y and z cancels with z so this is equal to zero
ok and hence d x plus d y plus d z equal to zero from here and so x integrating we
get x plus y plus z equal to a constant a so we have got one solution u x y z equal
to a where u x y z is x plus y plus z
now let us make another choice of lambda mu delta now let us choose
lambda equal to x mu equal to y and delta equal to z then we notice that lambda times
z minus y plus mu times x minus z plus delta times y minus x is again zero because this
is x times z minus y plus y times x minus z plus z times y minus x so we have x y which
cancels with x y here we have x z which cancels with x z here then we have y z y z
cancels with this is equal to zero so hence we have another equation hence
x d x plus y d y plus z d z equal to zero and on integration we get x square by two
plus y square by two plus z square by two equal to a constant which can be written as
x square plus y square plus z square equal to a constant b ok so this is v x y z equal
to b and the general solution is phi x plus y plus z x square plus y square plus z square
equal to zero
so now we have seen an example of a lagrange's linear equation were the variables are
not separable and we have to make a choice of lambda mu and delta such that lambda p
plus mu q plus delta r equal to zero now by choosing two by making two different choices
of lambda mu and delta we arrive at lambda p plus mu q plus delta r equal to zero and
then integrating the two for cases on integration we get the two independent integrals u and
b u equal to a and b equal to the b of the subsidiary equations and so we can we are
obtain we obtain the general solution
now let's take one more case another case where we will see that we obtain one solution
of the auxiliary equation and using that solution we shall be getting the second
solution so in example three it is a it is a different type of example here we have
d x by one d y by three equal to d z upon ten by five z plus ten by minus three x so
this the subsidiary these are subsidiary equations now solving the equation d x over one equal
to d y over three we obtain d y minus three d x equal to zero are y minus
three x equal to a where a is some constant
now in order to find the second solution what we do is let us consider next
let us take d y over three are ok we can take d x over one equal to d z over five z plus
ten by minus three x ok now we shall in order to solve this question we shall be using
the solution y minus three x equal to a so y minus three x equal to a we can put
here so the which implies that d x by one equal to d z upon five z plus tan a now
we can this is the case of separation separation of variables on one side we have x on the
other side we have z so we can integrate so we get x equal to one by five l n five z plus
tan a plus some constant a constant ok or we can say five x minus l n five z plus
tan a equal to some constant equal to a constant
now let us replace back the value of a so five x minus l n five z plus ten by minus
three x equal to a constant let us say b ok so thus we have then thus the general solution
is phi y minus three x and then five x minus l n five z plus tan y minus three x
equal to zero so the the general integral in this case here we have seen that by
using one solution we can obtain the second independent integral of the subsidiary
equations now lets take one more example of a similar type so z minus x p minus y q equal
to a under root x square plus y square plus z square where a is some constant it is given
now we can put it in this standard form or x p plus y q equal to z minus a under root
x square plus y square plus z square
now this is in the form of the lagrange's equation p p plus q q equal to r so the subsidiary
equations are dx over x equal to d y over y equal to d z over z minus a under root x
square plus y square plus z square now d x over x equal to d y over y gives the solution
very easily so this is l n x equal to l n y plus a constant
that this can be written as y equal to c into x i can cut the constant as minus l
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
x let us say where c one is some constant
now let us try to find the second solution so we have ok so d x over x equal to d
y over y equal to d z upon z minus a under root x square plus y square plus z square
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
plus y square plus z square minus a z under root x square plus y square plus z square
so let us now take let u square b equal to x square plus y square
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
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
divided u square minus a z into u on putting where u square is equal to x square plus y
square plus z square and this is same as d u over u minus a z d u over u minus a z we
get
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
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
d z minus d u divided by z minus u minus z z minus a u plus a z so this is equal to
d z by d u divided by one plus a times z minus u so d z upon z minus a u is equal
to d u over u minus a z which is equal to d z minus d u divided by z minus u minus a
u plus a z so we get z minus u times one plus a ok now we can inte[grate]- easily
integrate this so what we will do is ok so this is equal to this so thus d x by x
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
equal to d z by d u minus upon one plus a into z minus u so we get this
and this will give you l n x equal to one plus a one over one plus a l n z minus
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
t and when integrate when we integrate we get one over one plus a ln z minus u so this
is we can put it as one plus some constant we can simplify this now you can multiply
by one plus a so one plus a l n x equal to l n z minus u plus some constant
and this can then be written as l n x to the power one plus a
so we can write it as z minus u equal to some constant c two times x to the power one plus
a this can be written as minus l n c two so we will get z minus u equal to c two
x to the power of one plus a now r z minus under root x square plus y square plus z square
equal to c two times x to the power one plus a and so thus we get the following solutions
general so integral we get as follows
so one solution was y equal to c one x the the other solution is i can say x to
the power minus one plus a into z minus under root x square plus y square plus z square
equal to c two so the general integral is phi y over x and x to the power minus one
plus a z minus under root x square plus y square plus z square equal to zero
so this the general integral in this case so we have discussed four different
types of examples in the first case we have the case of a suppression of variables and
the second case we had the example where we had to make two different choices of lambda
mu and delta so that lambda p plus mu q plus delta r equal to zero in the third case we
had an example where we use the one solution y minus three x equal to a to find
the second solution independent solution in the fourth case in the fourth case it was
a typical example where we one the solution bar very bar very easy to find which was y
equal to c minus x but for the second solution we had to do i mean several
this thing substitutions and all so the second solution was not so easy to obtain
but that is how we get the second solution and the general solution is therefore given
by phi u v equal to zero so with that i would will like to conclude my lecture and
i
thank you all for attention