welcome friends to my lecture on solution of second order differential equations
by changing independent variable now we are going to discuss another method sometimes
this method is found to be very useful and this in this method we change the independent
variable so far the differential equations that we are considering they are of second
order linear differential equation with variable coefficients where x is the independent
variable so in this method we will change the independent variable from x to say another
independent variable that is z so let us consider a linear differential equation of second order
that is y double dash plus p y dash plus q y is equal to r where p q r are continuous
functions of x on an on some interval i we will be changing as i said we will be changing
the independent variable from x to z the relationship between x to z will be found from the
fact that i the changed equation it becomes readily integrable
so once we have that aim in mind that the changed equation has i mean is readily
integrable from there we shall be able to derive the relationship between x and z so
let us see what we do since x is related to z we can write d y by d x equal to d y
by d z over into d z by d x and d square by d x square if you find d square by d x square
then d square by d x square will be equal to d over d x of d y by d x which is equal
to d over d x of d y by d z into d z by d x so differentiating this product of functions
of x we shall be having d over d x of d y by d z into d z by d x plus d y by d z into
d square z by d x square now this can be further expressed as d over
d z of because z is a function of x we can write it as this into d z by d x into d z
by d x plus d y by d z into d square z by d x square so this will be equal to d square
y over d z square into d z by d x whole square plus d y by d z into d square z y d x square
so the second derivative of y with respect to x is equal to d two y over d z two into
d z by d x whole square plus d y by d z into d square z by d d x square
now substituting these values of y dash and y double dash in the equation one in this
equation one we shall have this expression d square y over d z square into d z by d x
whole square plus d square x by d x square plus p d z by d x into d y by d z plus q y
equal to r ok so now let us denote let us write this equation in the form
d square y over d z in this standard form d square y by d z square plus p one d y by
d z plus q one by y equal to r one where we write p one for this expression d square
z by d x square plus p d z by d x over d z by d x whole square and q one we write for
d q over d z by d x whole square and r one denotes r over d z by d x whole square so
these p p one q one r one p one q one r one are some functions of x and can
they can be easily expressed as functions of z by the relationship between z and x which
we have yet to establish so now what we do is let us discuss various
possibilities which will give us a function of z i mean z as a function of x and
with those possibilities we shall be able to integrate the equation two so the first
possibility is that let us equate q over d z by d x whole square q over d z by d x whole
square is q one so let us equate q one to a constant so if we do that so what do we
have q one equal to q y d z by d x whole square suppose it is some constant then if its so
happens that p one also becomes a constant if p one also becomes a constant then the
equation two will be a linear differential equation of second order in the independent
variable z the dependent variable y with constant coefficients and we have seen earlier how
we solve a second order linear differential equation with constant coefficients
so if the function if z is a function of x in such way that q over d z by d x whole
square is a constant then if p one also becomes a constant the equation two is at
once integrable so this is one possibility where we can find the general solution
of equation two now since z is quite arbitrary it can be chosen to satisfy any assignable
condition s one condition that we put is that q over d z by d x whole square is equal
to a constant let us look at other possibilities in the other possibility what we do is
we choose z in such a way that the coefficient of d y by d z in the equation two vanishes
so coefficient of d y by z in the equation two is p one so let us put one equal to zero
so if p one equal to zero we get d square z by d z square plus p d z by d x equal to
zero ok so let us say suppose u is equal to d z by d x then this a question the question
can be expressed as d u by d x plus p into u equal to zero we can separate the variables
u and x are now let us integrate so integrating both sides we have l n u equal to minus integral
p d x ok are u equal to e to the power minus integral p d x ok
so thus d z by d x is equal to e to the power minus integral p d x now integrating
again with respect to x we get z as integral e to the power minus integral p d x d x
so so if we put the coefficient of d y by d z that is p y equal to zero we get
z equal to integral of e to the power minus integral p d x d x and after that if q
one with this choice of z as a function of x if it so happens that q one becomes a constant
then it is a second order linear differential equation with constant coefficients it will
be d square y over d z square plus a constant times by equal to r one so it will be a linear
differential equation of second order with constant coefficients and therefore we can
solve it we can find the general solution of this equation
now the other possibility is that suppose q one is a constant divided by z square
then z square times d square y by d x d square plus a constant times y will be equal
to r one times z square and so we will have quasi euler equation and we know how
to solve quasi euler equation so we can again get the solution of this differential equation
so if p one is taken equal to zero and after that p one equal to zero we get z as
a function of x with this choice of z as a function of x if q one either becomes a
constant or q one is a constant divided by z square then we can find the general
solution of equation two and hence we can find the general solution of equation one
so that is one more possibility and so we can see here if the equation two will
reduce to this form if the curve given comes out to be a constant or a constant divide
by z square the equation three becomes readily integrable
now let us see let us see when we put q one as a constant ok say suppose q one we
take as some constant a square ok we just want z as a function of x so this constant
can be chosen according to our requirement so we we are we will be choosing q one equal
to a square so then a q upon d z by d x whole square this is equal to q one and q one is
a square so we shall have d z by d x equal to q upon a square r or root q divided by
a ok or we can say z is a z is equal to one upon a integral root q d x we will get z as
a function of x now so this is what we will get and the question
will reduce to d square y over d z square plus p one d y by d z plus a square y equal
to r one if p one is a constant then the question can be easily integrated ok because it will
be a linear differential equation of second order with constant coefficients ok now
let us see some examples on this ok so let us consider the question given in
example six first we will write this equation in the standard form so the coefficient of
d square y by d x square you will get unity which means that we will divide the equation
by x to the power six
so we shall have y double dash plus three by x y dash plus a square by x to the power
six into y equal to one by x to the power eight now so when we compare this equation
with the standard form we get p equal to three by x q equal to a square by x to the power
six and r equal to one by x to the power eight so let us first explore the possibility when
we put q as a constant ok so let us put q as a constant so let q be equal to some
constant say ok let us take it as one ok then what we q by is q by d z by d x whole square
we put as a constant so wait i think wait a minute sorry so let us let q one be equal
to q one is equal to q by d z by d x whole square ok after we change the variable from
x to z we get this q one ok so after we change the variable from x to z we will
get q one as this so this u let us put as equal to this is equal to a square by x to
the power six into one by d z by d x whole square let us put it as a constant say one
ok we can take any value of the constant if you do not take here one you take some other
constant then the relationship between x and z will be changed accordingly so r d z by
d x whole square equal to a square by x to the power six which implies that d z by d
x equal to a by x cube and this gives you when we integrate with respect to z we get
z equal to a x to the power minus two divided by minus two ok
so this is a upon minus a upon two x square ok now let us find the value of p one p
one equal to p d z by d x plus d z by d x whole square d square z by d x square upon
d z by d x whole square so p is equal to three by x d z by d x is a by x cube plus d square
z by d x square we can find from here so d square z by d x square
we differentiated once more see this is a times x to the power minus three so we have
minus three a x to the power minus four so minus three a by x to the power four so we
get minus three a over d z by d x whole square which is a square by x to the power six
now here we have three a by x to the power four here minus three a by x to the power
four so this is equal to zero so when q one is assumed as a constant which is which
we have taken here as one p one also comes out to be a constant which is zero ok so we
have the equation in in the independent variable as d square y over d z square plus
p one d y by d z plus q one into y equal to r one ok this becomes d square y over d z
square plus q one we have taken as one so y and r one is r over d z by d x whole square
so r one is one over x to the power eight r one divided by d z by d x whole square
so divided by a square over x t the power six so we get here x to the power six over
a square so we get here d square y over d z square
plus y equal to one by a square x square now we have to change the variable independent
from x to z and the relationship is z equal to minus a by two x square so one by a
one by x square is two z over minus a ok so this is one by a square two z over minus
a one by x square equal to two z over minus a ok so this will be minus two z over a
cube ok now this is a second order linear differential equation with constant coefficients
so we can or write the complimentary function for this here the auxiliary equation
will be m square plus one equal to zero so the roots will be complex conjugate m equal
to plus minus i so complementary function will be c one cos z plus c two sin z and z
we see is equal to minus a by two x square so c one cos minus a by two x square plus
c two sin minus a by two x square cos minus theta is equal to cos theta sin minus theta
is minus sin theta so we can write it as c one cos a by two x square and here minus c
two sin a by two x square minus c two can be replaced by another constant say say
c two dash and we can write c one cos a by two x square plus c two dash sin a by two
x square so that is ok now a particular integral let us find so this
y p x y p z let us find first find ok lets find y p x y p x will be one over d
square plus one one over d square plus one acting on minus two z by a cube so this
is equal to minus two by a cube one over d square plus one acting on ok now this is a
polynomial z is a polynomial of degree one so one over z d square plus one we shall expand
as in a in the form of binomial expansion so minus two by a cube one plus d square raise
to the power minus one operating on z when we write binomial expansion for this we shall
have one minus d square plus d four minus d six and so on but the second derivative
of z is zero so we will get minus two by a cube one minus d square plus d four and so
on operating on z which will give you minus two by a cube minus two by a cube d square
z d four z all are zero so we have minus two z by a cube and z is equal to minus a by two
x square so this is
minus a by two x square so this will give you one by a square x square and thus the
general solution is y equal to y c x plus y p x this is y c x and this one is y p x
so we take the sum of the two and it write it equal to y so that is the general solution
in this case now lets discuss one more problem on this to make it more clear so let us take
up another problem here in the example two we are given the equation x d square y
by d x square minus d y by d x minus four x cube by equal to eight x cube sin x square
so again we divide the equation by x to make the coefficient of y double dash unity
so we shall have y double dash minus one over y dash minus four x square into y equal
to eight x square sin x square so here p is equal to minus one over x q equal to minus
four x square and r is equal to eight x square sin x square ok so again we try the
method where we put q one as a q q one q y q equal to a sorry q one equal to a constant
so after we change the variable from x to z we get the equation d square y by d z square
plus p one d y by d z plus q one y equal to r one so here q one is q y d z by d x whole
square so let us put q one as a constant say q one will be equal to minus four x
square upon d z by d x whole square so q one we can we can take as a constant let us take
the constant as minus one to then it will be simple so lets take q one equal to minus
one then d z by d x whole square will be equal to four x square so which will give us d z
by d x equal to two x one can take d z by d x equal to minus two x also that will give
you other relation between x and z so the equation will change accordingly so we are
taking positive here sin here so d z by d x equal to two x this gives you z as x square
ok so does z has x x square let us find the value of p one
so d z by d x is two x p here is minus one over x and d square z by d x square is two
d z by d x whole square is four x square so again this cancels with this minus two
plus two is zero so we get p one equal to zero and thus the equation one one reduces
to d square y by d z square p one is zero q is minus one so minus y equal to r
one r one is r over d z by d x whole square so a eight x square sin x square divided by
d z by d x whole square which is four x square so we shall have two sin x square ok so complementary
function here is this is auxiliary equation is i mean x minus one equal to zero so m is
equal to plus minus one so the both the roots of the auxiliary equation are real and distinct
and therefore we will have c one e to the power z plus c two e to the power minus z
and z is equal to x square so c one e to the power x square plus c two e to the power
minus x square now here this in order to find the particular
integral the right hand side of this equation which is linear differential equation with
constant coefficients in the right hand side x has to be changed to z so we have two sin
z because z is equal to x square so particular integral will be one by d square minus one
d d here represents d over d z so one over d square minus one operating on two sin z
and this will be equal to two times one over d square minus one operating on sin z so when
one over d square plus alpha s square operates on sin a at z we replace d square by minus
a square so this will will be two times one over minus one square minus one sin z this
is minus sin z ok so so we have got the particular solution as minus sin x square
so thus the general solution is
the other solution is y equal to c one e to the power x square plus c two e to the power
minus x square minus sin x square so that is how we solve this equation
given in example two so we have seen in our in the past three lectures we have
seen how we can solve a linear differential equation of second order with variable coefficients
when it is given in a when it when we are given some special kinds of such equations
as i said earlier such i mean a general solution is not a well known for all second order
linear differential equation with variable coefficients only a special class of such
equations can be solved so we have seen those equations which can be solved
by in our past three lectures now in my next lecture we shall be discussing how
to find the general solution of a higher order homogenous linear differential equation
with constant coefficients thank you for your attention