hello friends welcome to my lecture on solution of second order linear differential
equation by changing the dependent variable so we begin with a differential equation
linear differential equation of second order in the standard form that is y double
dash plus p y dash plus q y equal to r where p q r are continuous functions of x on an
interval say j ok in this method what we do is when in my previous lecture we studied
how to find the solution of this second order linear differential equation with variable
coefficients when one integral included in the complimentary function is known
now there are cases of second order differential equations where one integral included in
the complimentary function is not easy to find so if an integral included in the complimentary
function is not obvious by instruction or we cannot determine it easily then what we
will do is we reduce the given differential equation of second order to the normal form
by normal we means we will reduce it to a form where the first order derivative term
is missing so how we do it let us see in the second slide let us put y equal to
u v in the equation one
so we will put y equal to u in equation one then what do we get so we have y double dash
plus p y dash plus q y equal to r let us put y equal to u v then y dash is equal to u dash
v plus u v dash y double dash will be equal to y double dash v plus u dash v dash n plus
u dash v dash plus u v double dash let us replace these values in the differential equations
one so y double dash plus p y dash plus q y equal
to r will then become u double dash v plus two u dash v dash plus u v double dash plus
p into y dash so u dash v plus u v dash plus q into u v equal to r ok now let us write
it as a second order differential equation in v so we have u v double dash first we write
this term u v double dash then we write v dash times so v dash times what we get two
u dash plus p u and then we write the coefficient of v the coefficient of v is u double dash
plus p u dash plus q u equal to r ok let us divide it by u ok the coefficient of
v double dash so r we can write v double dash plus v dash times two u dash by u plus
p plus v times u double dash by u plus p times u dash by u plus q equal to r by u
ok now we are going to reduce this second order
differential equation in v to the normal form normal form means where the first order derivative
term is missing so let us put two u dash by u plus p equal to zero ok so putting
we get u dash by u equal to minus half p when we integrate this with respect to x what we
get is l n u equal to minus half integral p d x ok so this will give you u equal to
e to the power minus half integral p d x ok now let us say suppose so this term is
now zero ok with this choice of u now let us put this as say i ok so if i call i let
i be equal to u double dash by u plus p times u dash by u plus q then let us simplify this
expression for i we can see here that u dash by u equal to we know that u dash by
u equal to minus half p so if we differentiate with respect to x what
do we get u double dash into u minus u dash into u dash divided by u square equal to minus
half d p by d x [or/r] if you divide by u square you get u double dash by u minus
u dash by u whole square equal to minus half d p by d x now u dash by u equal minus half
p so let us put here so u double dash by u is equal to minus one by four p square equal
to minus half d p by d x so let us replace the value of u double dash by u here and the
value of u dash by u and let us find find a simplified expression for i ok
so i will be equal to one by four p square minus half d p by d x ok p times u dash by
u is minus half p plus q so this is minus half p square that is one by four p square
so we get minus one by four p square so this q minus half d p by d x minus one by four
p square so we get a simplified expression for i and let us denote this r by u by
x ok so let us say further let r by u equal to s so then we shall have
d square p by d x square plus i into v equal to s where i is again i write i equal to
q minus half d p by d x minus one by four p square and s is equal to r by u and u is
e to the power minus half integral p d x so r times e to the power half integral p
d x now let us note here that this equation this equation is called as the normal form
let us say let me call it as equation two the equation two
is called the normal form now we can integrate this equation or
we can find a solution of this general solution of this equation provided r is a constant
if r is a constant then this will be linear differential equation of second order with
constant coefficients r i is a some constant divided by x square which if you assume
that i is some constant divided by x square then multiplying the equation by x square
it will reduce to a quasi euler equation and you know how and we know how to solve
a quasi euler equation so we have a catch here so let us note that
the equation two can can be solved if either i is a constant or i is a constant divided
by x square so yes just as we said in the beginning that
only a special class of second order linear differential equation can be solved
with variable coefficients so there is a catch here such equations where
i is either a constant or a constant divided by x square can be solved by reducing the
given equation to the normal form so let us see how this is these to the
this this the normal form we are given the value of or we have the value of i and s and
this is what we have noted just now if the value of i is a constant or a constant divided
by x square then the constant becomes readily integrable this form is said to be the normal
form now let us take an example and see how we
will use this method so ok so let us say d square y over d x over that is y double
dash we have y double dash minus four x y dash plus four x square minus one into
y equal to minus three e to the power x square sin two x so when we compare it with the standard
form we have the value of p equal to minus four x q equal to four x square minus one
and r equal to minus three e to the power x square sin two x
so let us find first when we reduce this equation to the normal form the value of u
ok u is equal to e to the power minus half integral p d x so e to the power minus half
integral minus four x d x so this is e to the power integral two x d x and which gives
you e to the power x square so we know the value of u now let us find
i i is equal to q minus half d p by d x minus one by four p square so q is four x square
minus one minus half d p by d x when you find d p by d x here you get minus four and then
minus one by four p square so you get minus four x square whole square so what we get
is four x square minus one and here we get plus two and here we get minus four
x square so this cancels with this two minus one is
one so we get i equal to a constant and therefore we can see from this article that this
given equation is readily integrable the normal form is now
d square v over d x square plus i into v equal to x so i is equal to one and s is r over
u ok so s is r over u and r over u is minus three e to the power x square sin two x divided
by u u is e to the power x square so e to the power x square gets cancelled
and we get minus three sin two x so we get d square v by d x square i is one so plus
v equal to minus three sin two x now this is a linear differential equation of second
order with constant coefficients so we can write auxiliary equation auxiliary equation
is m square plus one equal to zero so we get m equal to plus minus one plus minus i
sorry so complimentary function y c x is equal to a cos x plus v sin x where a
and v are arbitrary constants and particular integral y p x equal to one over d square
plus one where d represents d over d x operating on minus three sin two x ok so this is minus
three times one over d square plus one operating on sin two x when we replace d square by
minus two square the denominator does not become zero so this is minus three one over
minus two square that is minus four plus one sin two x so this is minus three minus three
cancel and you get sin two x so this way we get the value of v s so hence
v is equal to y c x plus y p x so a cos x plus v sin x plus sin two x now general
solution is y equal to u into v that is u is e to the power x square so e to the
power x square into a cos x plus v sin x plus sin two x so this is the general solution
in the case of example one let us take one more example to make the
things clear so we have y double dash in example two minus two tan x plus y into y dash
plus five by equal to e to the power x into sec x so here p is equal to minus two tan
x q equal to five and r equal to e to the power x into sec x now u is equal to y over
formula e equal to u equal to e to the power minus half integral p d x so this is e to
the power minus half integral minus two tan x d x which is e to the power integral
now integral of tan x is log sec x so we have e to the power l n sec x ok so this is equal
to sec x so we have found the value of u let us now determine i i equal to q minus half
d p by d x minus one by four p square so q is five minus half d p by d x will be minus
two sec square x the derivative what tan x is sec square x minus one by four p square
p square is four tan square x ok so this will cancel with this and we have five this
will cancel with this we have five plus sec square x minus ten square x sec square x equal
to one plus ten square x so this is one sec square x minus ten square x is one so this
is six so i is again a constant and therefore we can easily integrate the normal form
that is d so thus we have the normal form as
i into v that is six into v equal to s s is r by u so s is equal to r by u which
is equal to or e to the power x sec x divided by sec x so we cancel this and now this is
easy to integrate auxiliary equation is m square plus six equal to zero it is our auxiliary
equation so m equal to plus minus i root six and therefore complimentary function is
c one cos root six x plus c two sin root six x and particular integral y p x is equal to
one over d square plus six operating on e to the power x
now we can apply one over f d e to the power a x formula because here when d is replaced
by a a is one here f d does not become zero so one over one square plus six e to the
power x so we have e to the power x by seven so we have thus v equal to c f so c one cos
root six into x plus c two sin root six into x plus one by seven e to the power x this
is the and we multiply it by u to get the general solution u is sec x so hence the general
solution is y equal to u into v equal to sec x times c one cos root six x plus c two
sin root six x plus one by seven e to the power x
so this is the general solution of equation given in the case of example two in my
next lecture i will discuss how to solve second order linear differential equation
with variable coefficients by changing changing the independent variable in this article we
have changed the dependent variable from y to v in that next lecture we shall be changing
the independent variable from x to z now how how we get the relationship between x and
z that we shall see in the net lecture ok thank you for your attention