MAT Exam > MAT Videos > Calculus for MAT > Linear Differential equation of Second Order-Part-2
Linear Differential equation of Second Order-Part-2 Video Lecture | Calculus for MAT
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FAQs on Linear Differential equation of Second Order-Part-2 Video Lecture - Calculus for MAT
| 1. What is a linear differential equation of second order? |  |
Ans. A linear differential equation of second order is an equation that involves the derivatives of a function of two variables, where the highest derivative involved is of order two and the equation can be written in the form a(x)y'' + b(x)y' + c(x)y = f(x), where a(x), b(x), c(x), and f(x) are known functions of x.
| 2. How is the general solution of a linear differential equation of second order obtained? | |
Ans. The general solution of a linear differential equation of second order is obtained by finding the complementary function and the particular integral. The complementary function is the general solution of the homogeneous equation (when f(x) = 0), and the particular integral is a specific solution of the non-homogeneous equation (when f(x) ≠ 0). The general solution is the sum of the complementary function and the particular integral.
| 3. What is the characteristic equation of a linear differential equation of second order? | |
Ans. The characteristic equation of a linear differential equation of second order is obtained by substituting y = e^(mx) into the homogeneous equation, where m is a constant. This substitution gives a quadratic equation in m, known as the characteristic equation. The roots of the characteristic equation determine the form of the complementary function.
| 4. How does the nature of the roots of the characteristic equation affect the solutions of a linear differential equation of second order? | |
Ans. The nature of the roots of the characteristic equation determines the form of the complementary function and, consequently, the solutions of the linear differential equation of second order. If the roots are real and distinct, the complementary function consists of two exponential terms. If the roots are real and equal, the complementary function consists of a single exponential term. If the roots are complex conjugates, the complementary function consists of sinusoidal terms.
| 5. How is the particular integral determined in a linear differential equation of second order with non-constant coefficients? | |
Ans. In a linear differential equation of second order with non-constant coefficients, the particular integral is determined using the method of undetermined coefficients or variation of parameters. The method of undetermined coefficients involves assuming a particular form for the particular integral based on the form of f(x) and solving for the unknown coefficients. The method of variation of parameters involves assuming a particular form for the particular integral in terms of unknown functions and then solving a system of equations to determine these functions.
Text Transcript from Video
welcome to lecture series on
differential equations for undergraduate
students today's lecture is in
continuation of linear differential
equations of second order in last
lecture we had learned about these
equations moreover we have learned how
to solve linear differential equation
with constant coefficient of second
order we will learn something more about
these equations as well we will also
learn in today's lecture about non
homogeneous equations so first we will
learn about the existence and uniqueness
of the solution of initial value problem
here we are not going to find out that
is what are these conditions rather we
will state a result about under which
conditions the solution of initial value
problem does exist and under what
circumstances that solution would be
unique so the result here is existence
and uniqueness theorem for initial value
problem if px and QX are the continuous
functions on some interval I and X
naught is in I for differential equation
y double dash plus PX y dash plus QX y
is equal to 0 and two initial conditions
that is function y at X naught is K
naught and the derivative of Y at X
naught is K 1 then this initial value
problem has a unique solution Y X on
interval I we have learned it that this
solution unique solution of this initial
value problem is called the particular
solution and this particular solution we
can obtain by giving a specific value to
the general constants in the general
solution how we are obtaining those
constant values for this particular
solution what we are doing is we
here putting these initial conditions to
the general solution and then find out
the values for those constants so now we
will learn that
we had learned that general solution for
second-order equations contains two
linearly independent solutions those two
linearly independent solutions we are
obtaining by some methods we are
checking that those solutions which we
have obtained for the second-order
equation whether they are independent or
not for that either we had implied the
basic definition of linear independence
of function all we had learned one more
thing that is those two solutions are
not proportional today we will learn
something more about this linear
independence of solution which is known
as wronskian our Ron C determinant what
we do call the wronskian brown-skinned
for the second order equation y double
dash PX y dash plus QX y is equal to 0
if it has two solutions y1 and y2 then
the wronskian of y1 and y2 is being
given as a determinant whose first row
contains y1 and y2 the two solutions on
the second row contains the derivative
of those two solutions of course we
don't know the e this is equal to y1 y2
- minus y 1 - y - this is called the
wronskian on how you're checking the
linear independence if this wronskian is
not 0 for any value of x not in the
interval i then we say on that interval
i the two solutions y 1 and y 2 are
linearly independent some more
properties of this run skin given two
functions FX and GX that are
differentiable on some interval i the
wronskian of f and g at some point x
naught is not 0 for some X naught in I
then we call that FX and GX are linearly
independent on I moreover if FX and GX
are linearly dependent on I then this
wronskian off if I'm G would be 0 for
all X in I so we see is that is if
is linearly dependent we will get
wronskian of F and G to be zero for all
X and if it is linearly independent I
could find out that it's not some X not
where it is not zero now on special
property which we are using for the
differential equation that is called
Everest theorem what is that if y1 and
y2 are two solutions to the differential
equation y double dash plus PX y dash
plus QX y is equal to 0 then there are
skin of the two solution is at some
point X that is w y1 y2 at X is equal to
w y1 y2 at X naught into e to the power
integral minus 0 to X px DX we can show
this result also so we are going to
prove this the differential equation y
double dash plus py dash plus qy is
equal to 0 we have been given that it
has two solutions y1 and y2 since they
are two solutions so they would satisfy
our differential equation what it says
is that y double dash plus py 1 dash
plus qy 1 is equal to 0 and y double
dash plus py 2 dash plus qy 2 would be 0
now from these two equations if I
eliminate the Chum containing Q that is
I would multiply the first equation by Y
2 on the second equation by y1 and
substract then what we would get we
would get Y 2 Double Dash Y 1 minus y 1
double dash Y 2 plus py 2 dash Y 1 minus
y 1 dash y 2 is equal to 0 now let us
see what we have obtained again come
back to our run skin it is y 1 y 2 that
is y 1 y 2 dash minus y 1 dash Y 2 now
if I differentiate this function because
this is a function of X if I
differentiate with respect to X what I
would get Y 1 dash Y 2 dash plus y 1 y 2
Double Dash minus y 1 dash Y 2 dash
minus y 1 double dot
while that is y1 y2 double dash minus y1
double dash Y to since Y 1 dash Y 2 dash
is getting canceled it out now we see
what the equation we had obtained by
laminating cube we had obtained y 2
double dash y 1 minus y 1 double dash y
2 plus P times y 1 y 2 dash minus y 2 y
1 dash is equal to 0 now we see that the
coefficient of P is nothing but the
wronskian and what we are getting here
the constant term that is nothing but
the derivative of the wronskian now
these are the function of X so we can
write this equation as W dash plus P W
is equal to 0 where W dash is the
derivative of W with respect to X that
says is that we are getting a
differential equation with respect to W
now what we are having is now this is
first order equation we do know how to
solve it the solution of this is W X
naught e to the power minus 0 to X
integral px DX we do know that this
solution we do know and we do know this
is what we have replaced W X naught as
the constant which we are having that we
had replaced with a particular value X
naught how we could get this particular
value if I have been given the value of
W at a particular value it's not so this
is what we are getting as a solution
this we do know that is how what that
Able's theorem says now from here what
we are getting the result that is if W
at some X naught is not 0 I would get
that W X will never be 0 since this
whatever be this part this will never be
0 for any value of x that says is for
linearly independent I require
w X to be not 0 for only a single point
while when they are linearly dependent
they would be 0 for all X now let us see
one example how we are using this
wronskian to find out the linear
independence
find the general solution of y double
dash minus two y dash plus y is equal to
zero this is a linear differential
equation with constant coefficients we
do know how to solve it so solution let
us see what is the characteristic
equation that would be lambda square
minus 2 lambda plus 1 is equal to 0 now
we do know this is nothing but lambda
minus 1 whole square so I would get the
double root lambda is equal to 1 we
already know that the two solutions for
this we are taking as e to the power x
and x times e to the power X now we are
going to check that these two solutions
are linearly independent and we would
employ the method of Bronski
so we'll find out the Transkei
determinants for these two solutions so
what it would be e to the power X that
is the first solution then the second
solution x times e to the power X on in
the second row the derivative of e to
the power X that is e to the power it's
derivative of x times e to the power X
lattice e to the power X plus x times e
to the power x if I calculate what is
the value of this determinant that would
be e to the power 2 X plus X times e to
the power 2 X minus x times e to the
power 2 X that is equal to e to the
power 2 X that is not zero for any value
of x this says is that the two solution
e to the power x and x times e to the
power x they are linearly independent we
know that we had already checked it
using the method of constant proportion
so if I take the ratio of e to the power
x and x times e to the power x that is
coming out to be 1 by x which is not a
constant so this also we do know from
the first method of not proportion that
they are linearly independent so what
will be the general solution of this
given equation that would be C 1 e to
the power X plus C two x times e to the
power X now we had learned about the
linear independence
if I one more method that is called
Ron's key determinant our wronskian now
I will move to the non homogeneous
equations
so first we'll do non-homogeneous second
order linear differential equations what
are they we do know that they are second
order linear differential equations with
right-hand side not to be 0 that is y
double dash plus PX y dash plus Q XY is
equal to rx and I should have that rx
should not be 0 this we do now we'll
learn about that is what the we are
going to solve this so what we says one
more term we are associating with this
that is corresponding homogeneous
equation where I will take the equation
as such but the right-hand side s 0 this
is called the corresponding homogeneous
equation for this non-homogeneous
equation are sometimes it is also
referred as Associated homogeneous
equation so this equation we are
obtaining has the same equation but the
right hand side rx we have made it here
0 what we get one result that suppose
Capital y 1 and capital y 2 are two
solutions of non-homogeneous equation y
double dash plus PX y dash plus QX y is
equal to RX and suppose is small y1 and
y2 are fundamental set of solution of
corresponding homogeneous differential
equation that is y double dash plus PX y
dash plus QX y is equal to 0 here what
we have taken that is y1 and y2 are the
fundamental said that means they are
linearly independent solution of this
homogeneous equation then Capital y 1
minus y 2 is also a solution of this
homogeneous equation and since it is a
solution of the homogenous equation it
can be written as a linear
combination of the fundamental set of
solution that is y1 x and y 2x that is I
could write Capital y 1 minus Capital y
2 c1 times y 1 plus c2 times y 2 again
this is again a simple result which we
can show here by the knowledge whatever
we are knowing so let's prove it
since y 1 x and y 2 X second derivative
would be nothing but the second
derivative of y 1 - second derivative of
Phi 2 and the derivative of y 1 minus y
2 is nothing but the derivative of Y 1
minus derivative of Phi 2 will
substitute this in our homogeneous
equation that is y double dash plus P XY
- plus Q XY is equal to 0 since we want
to show that Y 1 minus y 2 is a solution
of the homogenous equation let us
substitute it what we do get Y 1 Double
Dash minus y 2 Double Dash plus P times
y 1 dash minus y 2 dash plus Q times y 1
minus y 2 now rearrange the terms first
I write the terms corresponding to Y 1
that is y 1 Double Dash plus px times y
1 dash plus Q XY 1 then minus y 2 - y 2
Double Dash plus px times y 2 dash plus
Q XY 2 now the first term which we are
having this is since Capital y 1 is the
solution of non-homogeneous equation
that means this first terms what we are
having that would be that would be equal
to the right hand side of
non-homogeneous equation that is our X
similarly the second term will also be
the non homogeneous solution of
non-homogeneous equation that is it
should also be equal to rx that says
this rx minus rx will get SIL what it
says if I am substituting in this
equation y 1 minus y 2 I am getting
right-hand side equal to 0 that says Y 1
minus y 2 is satisfying this equation
which says that
is a solution now since it is a solution
we can write it as a linear combination
of the fundamental set of solutions that
is I could write it as c1 y1 plus c2 y2
for some c1 and c2 now from here what we
could get we could get one more result
say what we could do let's say let us
consider the two solution y + YP X what
are they they are actually we are taking
this non-homogeneous equation while
double dot plus px slider plus QX y is
equal to rx and let's say Y is a general
solution of this non-homogeneous
equation and YP X is a particular
solution of this non-homogeneous
equation so now what we are getting we
are getting this as capital y1 and this
as capital y to the two solutions are
non-homogeneous equation then by the
previous result we do know that Y minus
y fee would be solution of corresponding
homogenous equation let's say that is
the fundamental set of homogeneous
equation solution of some homogeneous
equation would be Y small Y one and Y 2
that means this would be equal to c1 y1
plus c2 y2 now from here I could write Y
as c1 y1 plus c2 y2 plus YP X so the
first term c1 y1 X plus c2 y2 x is
nothing but the general solution of the
corresponding homogeneous equation and
YP X is a particular solution of the
non-homogeneous equation that says is
the general solution of non-homogeneous
equation YX I can write as y HX plus YP
X where YH X is the solution of
homogeneous equation that is c1 y1 X
plus c2 y2 x this is the general
solution of Associated homogeneous
equation and YP X is a particular
solution of the non-homogeneous equation
thus
we are getting now you see we had
already learn how to find out or that is
these solutions y1 and y2 they are
linearly independent
now this YP which we are saying if I am
keeping this in the general solution
this YP should also be linearly
independent of y1 and y2 so this is
important thing over here now we come to
our main theorem of non-homogeneous
equation this is about general solution
for a second order linear differential
equation which is non-homogeneous that
is y double dash plus PX y dash plus QX
y is equal to RX it's solution general
solution can be given as general
solution of Associated homogeneous
equation plus particular solution of the
non-homogeneous equation that is why X
can be written as y HX plus YP X where
YX is the general solution of
corresponding homogeneous equation y
double dash plus PX y dash plus QX y is
equal to 0 and YP X is a particular
solution of our non-homogeneous equation
now what will be the particle result
about the particular solution the
particular solution we do know can be
obtained from the general solution y HX
plus YP X by giving the specific values
c1 c2 in the wallet X since we do know
that y HX is the general solution of
Associated homogeneous equation y px is
a particular solution so it is not
containing any arbitrary constants the
arbitrary constants are over here now if
I keep this c1 and c2 0 I will get only
YP X that is the particular solution
which we are keeping over here but that
is not the only particular solution we
can obtain some other particular
solutions as well by giving some values
to the c1 and c2 so what will happen to
the initial value problem that is from
there we will obtain this c1 and see
by putting the initial values in the
general solution so the general solution
of one now one more result over here the
general solution of non-homogeneous
equations includes all solution what
this is or says this is if px qyx and rx
are continuous on some open interval I
then every solution of y double dash
plus PX y dash plus QX y is equal to RX
on I is obtained by giving suitable
values to the arbitrary constants in
general solution y HX plus YP X that is
every solution of this non-homogeneous
equation can be obtained through this
general solution it may imply there is
no singular solution now what we have
drawn the practical conclusion out of
these results we have obtained that to
solve a non-homogeneous equation y
double dash plus PX y dash plus QX y is
equal to RX first are an initial value
problem what we do first we do solve the
homogeneous equation y double dash plus
PX y dash plus QX y is equal to 0 and
find a particular solution by PF
non-homogeneous equation and then we do
keep it over there we had learnt the
techniques to solve homogeneous linear
equations of course there we have not
done it for general PX and QX we have
done only for the constants px and QX
that says this now so let us do first
the examples we will do about this
constant then we will learn about
functions as well so we do know if this
constants we do know how to solve this
homogeneous equations but the things
remain is how to find out a particular
solution for this non-homogeneous
equation we'll start with the basics so
let us first find out that is basic
technique to solve the non-homogeneous
equation
it is method to fight YP it doesn't
start with the as I say this with basics
so let us I start with an example find
the general solution of y double dash
plus three y dash plus 2y is equal to 12
times e to the power X we see this is a
linear equation its coefficients are
constants but the right-hand side is not
0 that is it is non-homogeneous so what
we do is first we'll solve ax associated
homogeneous equation what is the
corresponding homogeneous equation y
double dash plus 3 y dash plus 2y is
equal to 0 we do know how to solve it
first we find out the characteristic
equation what is the characteristic
equation lambda is square plus 3 lambda
plus 2 is equal to 0 what are its roots
of course it is lambda plus 2 into
lambda plus 1 is equal to 0 so it has
two roots - what on - 2 they are
distinct real numbers so what will be
the corresponding general solution of
the homogenous equation that will be C 1
e to the power minus X plus C 2 e to the
power minus 2x now the differential
equation was y double dash plus 3y -
plus 2y is equal to 12 times e to the
power X now I have to find out the
particular solution particular solution
means is that is it should be a solution
of this equation now let us see
right-hand side we are having the charm
e to the power X so now if I experiment
with e to the power X we do know it's
derivative as well as double derivative
all our e to the power X so if I am
substituting so we can try like this one
so what we are doing is we are taking
particular solution as a times e to the
power X because I do not know what will
stop satisfy this equation that means it
should if I'm putting e to the power X I
should get equal to 12 times e to the
power X so I have put I have taken that
as it constant because e to the power X
I can obtain by putting e to the power X
but this constant we have to
or cancer I have put here in how to
obtain this yeah now if I am assuming
that this is a solution this must
satisfy this equation so now I will
substitute wipey wipey - and Val P
double - in this equation so what is y P
- a times e to the power x + y P double
- that is again a e times e to the power
it's let us substitute it we would be
getting is a times e to the power X into
1 plus 3 plus 2 is equal to 12 times e
to the power X what is this is that 6 a
times e to the power X is equal to 12
times e to the power X which implies
that my a must be 2 that is the
particular solution should be 2 times e
to the power X we can check that this is
satisfying this equation so this is a
particular solution so now what will be
the general solution of our
non-homogeneous equation that would be
the general solution of homogeneous plus
the particular solution so I could say 1
e to the power minus x plus C 2 e to the
power minus 2x + 2 times e to the power
X so this is a general solution or
homogeneous equation we can check that
if I substitute this function under its
derivatives in this equation for any
value of C 1 and C 2 that is take the
general constant C 1 and C 2 I will
often that is this will satisfy this
equation so this is the general solution
of the non-homogeneous equation so this
is here we have done with an example on
the Basic Instinct we have got that is
right hand side we had seen that it is
containing the chunky to the power X and
we had already experimented from the
beginning of this course with the
function e to the power X so we could do
it now let us learn the methods to find
out why P we will learn here the two
methods to find out this particular
solution the one method method of
undetermined coefficient that is
something like that what we have done in
this example and there is another method
called method of variation of parameter
so first we will learn the method of
undetermined coefficients
what this method says is this method is
first let let us non that is where this
method is applicable this method is
applicable in the equations with
constant coefficients of C we would be
dealing with only linear equations with
constant coefficients and right-hand
side rx that is also a special form that
special form that is either my RFC's of
the form e to the power X or it is a
polynomial or it may be a function of
cosine ax signed something like that
then only this method is applicable now
let us learn what this method is this
method says is that if Rx is of a
special form I would choose that
particular kind of YP how we are doing
it
let me summarize it by a table we said
that is if I do have the rx as K times e
to the power gamma X that as in our
example I was having 12 times e to the
power X that is gamma was 1 and what I
have choose our chose the y PS e to the
power X and a here that says is that if
our X has of the form K times e to the
power gamma X choose y PS say x into the
power gamma x if my Rx is containing the
polynomials that is here I have given
the example of K times X to the power n
then twice of Y P will be a polynomial
of degree n see here we are not taking
only K times X to the power and we are
taking the complete polynomial K naught
K 1 X plus so on K and X to the power n
so if it is a polynomial or if it is
some terms of polynomial in the choice
of Phi P we will always choose the
complete polynomial of degree n then if
the terms in are X R cosine are sign
that is K times cosine Omega X odd K
times cosine sine Omega X we will choose
Y P as a
sine Omega X plus B cosine Omega X again
you see my rx may contain only cosine
our only sign but in the YP we would
choose both sine and cosine functions
now if the Charmander X is having a
multiple of e to the power alpha X and
cosine are sine terms we would be
choosing y fears
e to the power L Phi X is sine Omega X
plus B cosine Omega X that is now what
are the rules to apply this method so
we'll learn some basic rules rules of
this method our first is called the
basic rule basic rule just now as I had
explained you that is for second order
linear differential equation which is
non-homogeneous y double dash plus PX y
dash plus QX y is equal to RX if R
exists one of the function in the first
column of the table just now as I had
explained I would choose corresponding
function YP in the second column and
find the value of undetermined
coefficient by putting YP and its
derivative in the given equation so we
had got that is by using that table we
can choose YP and as in our example what
we have done we had find out the value
of a by putting wipey wipey - and why P
double dash in the equation and then we
have determined what is a so here
whatever the undetermined coefficients
KC k1 K naught and so on we could find
out all of them by putting YP and its
derivatives in the non-homogeneous given
differential equation
I'm solve for the unknown coefficients
that is why it is called the method of
undetermined coefficients now you do
remember I said that the general
solution contains two things one is
general solution of Associated
homogeneous equation y HX and another is
YP X now it would be the general
solution that
I should have YP also linearly
independent of the solutions which we
are having as y1 and y2 of the
homogenous equation that says sometimes
it may happen that is whatever I am
getting as the choice of YP that may
turn out to be one of the solution of
the homogenous equation then of course
because here I am choosing it a constant
time that function that will not be
linearly independent of the solutions of
homogenous one so how to solve this kind
of problem that says is the second rule
as modification rule it says if any term
in the choice for YP is also in the
solution Y H of corresponding
homogeneous equation y double dash plus
px Y - plus Q X Y is equal to zero then
multiply the choice YP by X R by X
square if the solution corresponds to
the double root of the characteristic
equation of the homogeneous equation why
we are doing it self we can understand
this very well since that particular
solution is also coming us in the
solution of homogeneous equation so with
a constant coefficient of course we
cannot get it as linearly independent so
we have to multiply with its now if it
is solution corresponding to the double
root then we do know that it's times
that solution is also coming as the
fundamental solution so I have to
multiply it by X square then there is
one more rule that is called some rule
what it says is I may have R X that is
our X is sum of different functions
given in the column one of the table
then choose YP is the sum of
corresponding functions in the second
column now let us see that is how to use
these rules and how to use this
particular method to find out the
general solution of non-homogeneous
equation that we will see through some
examples
so let us go with the examples first
example let's say solve the
non-homogeneous equation y double dash
plus 4y is equal to 8 X square we see
this is a second order equation constant
coefficients are constants and
right-hand sides of the special form
that is it is the polynomial that is 8
times X to the power 2 so go for the
solution give an equation is y double
dash plus 4y is equal to 8 X square the
corresponding homogenous equation will
be y double dash plus 4y is equal to 0
to solve this we require the
characteristic equation what is the
characteristic equation that is lambda
squared plus 4 is equal to 0 what will
be its root it is lambda square is equal
to minus 4 that means it will have
complex conjugate roots the roots would
be plus minus 2i we do know in this case
what is this L the general solution of
homogeneous equation that is C 1 cosine
2x plus c2 sine 2x now what is my right
hand side that is a Texas square now if
you're not remembering we of course go
back to our table we are having 8 X is 4
that is here n is equal to 2 so what we
would choose we would choose K naught
plus K 1 X plus K 2 X square so let's
see since our X is 8 X square Y P we
would choose K 2 X square plus K 1 X
plus K not now substitute its
derivatives in the equation that is
first find out what is Phi P dash that
is 2 K 2 X plus K 1 YP Double Dash would
be 2 K 2 now substitute this in the
given equation y double dash plus 4 y is
equal to 8 X square what we get we get 2
K 2 plus K 2 square for K 2 X X square
plus 4 K 1 X plus 4 K naught is equal to
8 X square now to find out these
constants will compare the
coefficients of different powers so
equating the coefficients we get the
coefficient of x is square on the right
hand side is a while the coefficient of
x is square on the left hand side is
fourth it was the first equation we are
getting is for K 2 is equal to 8 then we
are getting is the right hand side the
coefficient of x is zero while on the
left hand side the coefficient of x is
for K 1 so I get for K 1 is equal to 0
now right hand side I don't have any
constant term while on the left hand
side I do have the constant term 2 K 2 +
4 K naught that says is 2 K 2 + 4 K
naught is also 0 now from the first
equation I am getting K 2 is equal to 2
second equation gives K 1 as 0 on the
first equation the third equation gives
me that K naught is equal to -1 now
substitute this in our choice of Y P
what I would get I would get Y PS 2 x
squared - 1 so what will be the general
solution of our equation y double dash
plus 4y is equal to 8x square that is c1
cosine 2x plus c2 sine two x that is the
general solution of homogeneous equation
plus this particular solution 2 X square
minus 1
now you can obtain check with this
solution find out its derivative and
double-dare a second derivative
substitute in this equation and check
that is in general if I am taking only
c1 and c2 this will satisfy for all
values of c1 and c2 so if this would be
the interesting exercises for you to do
you can check it that this is a solution
of this non-homogeneous equation now let
us do one more example
yeah one more thing here you would be
worrying that is why I have chosen this
as I said is we will always choose
polynomial you can try one more
interesting exercise you can try only
with KX square don't use the complete
polynomial and see that what the
solution you are getting that will not
be actually
solution of this non-homogeneous
equation is another interesting exercise
which you can try so let us move to the
second example solve the initial-value
problem y double dash plus two y dash
plus y is equal to e to the power minus
x where the initial conditions are that
the function r 0 is minus 1 and the
derivative at 0 is 1 move to the
solution the given differential equation
is y double dash plus 2 y dash plus y is
equal to e to the power minus x see
again we are having the equation has
constant coefficients on right hand side
is of a special form e to the power of
minus x first find out the corresponding
homogeneous equation that is y double
dash plus 2 y dash plus y is equal to 0
for solving this we require its
associated characteristic equation that
is lambda square plus 2 lambda plus 1 is
equal to 0 we see this is the whole
square of lambda plus 1 that is we are
going to get the double root lambdas
minus 1 and minus 1 we do know in
homogeneous equations when we are having
the double root the solution is e to the
power minus X and the other solution we
choose x times e to the power minus x so
the general solution would be c1 plus c2
x times e to the power minus X now right
hand side is e to the power minus X now
you see if I am using our table I would
get right y PS a times e to the power
minus XR C times e to the power minus X
but this e to the power minus X is also
coming us here in the e to the power
minus X is also a solution of my
homogeneous equation moreover this is a
solution corresponding to the double
root that is x times e to power minus x
is also a solution so we have to apply
the modification rule so again I am
repeating this these steps first we see
from our table
I do have e to the power minus X I would
choose C times e to the power minus X
that is gamma is minus 1 since again
this e to the power minus X is coming as
a solution of the homogeneous linear
equation we do apply the modification
rule the modification rule says is
multiply the choice of Phi P by X R by X
square so here my solution e to the
power minus x is corresponding to the
double root so I have to multiply by X
square that says this I should choose my
Y P as a times e to the a times X square
times e to the power minus X find out
what is YP - and why P double - so Y P -
its derivative - ax e to the power minus
6 minus ax s squared e to the power
minus 6 second derivative - e to the
power minus 6 minus 2 ax e to power
minus 6 minus 2a X e to the power minus
X plus X square times e to the power
minus X that is it will be 2 e to the
power minus X minus 4a x times e to
power minus X and plus a X is called
power minus 6 substitute this in given
equation that is Phi Double Dash plus 2y
- plus y is equal to e to power minus 6
left hand side what we would be getting
to a e to the power minus X minus 4a x
times e to power minus X plus ax is
called power minus 6 plus 2 times its
derivative that is we are getting it as
2a X e to the power minus X minus ax is
square 8 4 minus X plus y is 8 X is
square e to power minus 6 this must be
equal to 8 power minus X let's see in
the left hand side here I am having this
minus 4a X e to power minus X this would
be plus 4a X e to power minus X that is
this term is get cancel it out here I am
getting is a times X is square e to the
power minus 6 that is a coefficient of x
is called 4 minus 6 here is a here is
minus 2 is since this 2 and here is one
more a plus a again it is cancelling out
so what is being left I have been left
with 2 e to the
- X is equal to e to power minus X so
comparing the coefficients we get to a
is equal to 1 R is equal to 1/2 so what
will be my wife YP would be 1/2 X square
e to the power minus X so the general
solution would be C 1 plus C 2 X e to
power minus 6 that is the general
solution of homogeneous equation Plus
this particular solution 1/2 X square e
to the power minus X now you can
actually again check with this solution
that you could find it out that
substitute this in this equation and
find out that this is satisfying this
equation and moreover here we are
getting is e to the power minus x x
times e to the power minus x and x is
square e to power minus x all of them
are linearly independent no to find out
the particular solution because this is
an initial value problem which we had
started so by finding out the particular
solution what are the initial conditions
we have been given we have been given
that the function R 0 is minus 1 and its
derivative at 0 is 1 what the general
solution we had got C 1 plus C 2 XE to
the power minus X plus 1/2 X square e to
the power minus X now if I substitute
this function at X as 0 that is I find
out the value of the function at 0 that
is given as minus 1 from here in this
solution if I am substituting X as 0 I
would get only C 1 so it gives C 1 is
equal to minus 1 then find out its
derivative derivative is minus C 1 plus
C 2 X e to power minus X minus 1/2 X is
called 4 minus 6 plus say 2 times e to
the power minus x plus x times e to the
power minus X now again the second
condition is y dash 0 is 1 so again we
are putting X is equal to 0 what I would
be left I would be left minus C 1 plus C
2 which is equal to 1 from the first
result we have got C 1 as minus 1 thus
gives C to us 0 so we have got the
particular solution that is I will put C
is equal to minus one I'm c2 is equal to
zero in our general solution of
homogeneous equation so I would get the
solution of initial value problem as
minus e to the power minus x plus half X
square e to the power minus X we can
check that this is satisfying both
initial conditions that is the function
value of zero is minus 1 and the
function derivative of the function r0
is plus one let us do one more example
over here find the general solution of y
double dash minus three y dash plus sign
y double dash minus three y dash minus
4y is equal to three times e to the
power 2 X plus 2 times sine X minus 8
times e to the power minus X now we see
our right-hand side is actually sum of
many functions which are in our table e
to the power 2 X e to the power minus X
sine X and so on
so it says is that we have to use the
sum rule let us see if we have to use
some other rules also so first we go for
the solution what is the corresponding
homogenous equation y double dash minus
3 y dash minus 4 y is equal to 0 to
solve this we require the characteristic
equation what is the characteristic
equation lambda square minus 3 lambda
minus 4 is equal to 0 we do find it out
that it is nothing but lambda minus 4
into lambda plus 1 the factors so it
will have two roots minus 1 at 4 they
are real and distinct so the general
solution of homogeneous equation will be
C 1 e to the power minus X plus C 2 e to
the power 4 X now we see right hand side
right hand side our X we have 3 times e
to the power 2 X 2 sine X minus 8 times
e to the power minus 6
now so if I'll just go with our table if
you do remember something that is e to
the power gamma X we do take C times e
to the power gamma X so here if I do
take I have to choose one y PS e to the
power minus 6 and that is here again as
a solution corresponding to the one root
that means we have to use the
modification rule that is in this one we
would be going to use the sum rule as
well as the modification rule for the
song rule we require which functions we
have to choose we have here function of
e to the power gamma X and sine X so let
us consult our table e to power gamma X
minus means I have to use constant times
e to the power gamma X and sine Omega X
I have to use the function which is
having a constant time sine Omega X
under constant times cosine Omega X with
plus so let us see what will be our
choice of YP the choice of YP would be a
times e to the power 2 X plus b1 sine X
plus b2 cos x plus J x times e to the
power minus X Y I had chosen here since
e to the power minus X was already a
solution so I have to modify it as ifs
times e to the power minus X so here I
had use a modification rule I'm in this
complete one I had use the sum rule so
we have got this as the choice of YP now
find out its derivative to e to the
power 2 X plus b1 cosine X minus v2 sine
X minus D X e to power minus X plus D
times e to the power minus 6 second
derivative 4a e to the power 2 X minus
b1 sine X derivative of sine X is cosine
X so B 2 cosine X derivative of DX e to
power minus X is plus DX 8 power minus X
minus D to power minus X and the
derivative of this is minus D to the
power minus X now substitute this in the
given equation what was the equation
equation was y double dash minus 3 y
dash minus 4 y is equal to 3 times e to
the power 2 X plus 2 sine X minus 8
times e to the power minus six so
substituting it and of course
calculations you can check it we get
minus six a times e to the power 2 X
minus 5 B 1 minus 3 B 2 sine X minus 5 B
2 plus 3 B 1 cosine X minus 5 d e to the
power minus X this should be equal to
the right hand side that is 3 times e to
the power 2 X plus 2 times sine X minus
8 times e to the power minus X now it
create the coefficients we get the
coefficient of here you see e to the
power 2 X its coefficient here is minus
6a here is 3 so I would be creating this
as this one the coefficient of sine X
here is minus 5 B 1 minus 3 B 2 and here
is plus 2 so I would get 5 B 1 minus 3 B
2 is equal to minus 2 the cosine X term
we don't have over here so though this
fight be 2 plus 3 B 1 would be 0 the
coefficient of e to the power minus x is
minus 5 D here here it is minus 8 so I
would get 5 D is equal to 8 so equating
the coefficient I am getting minus 6 a
is equal to 3 which gives me is equal to
minus 1/2 stick in what we are getting
for cosine and sine 5 B 1 minus 3 B 2 is
equal to minus 2 and 5 B 2 plus 3 B 1 is
equal to 0 solving these equations we
get B 1 as minus 5 by 17 and B 2 as
three by 17 and last one we are getting
5 D is equal to 8 gives me D is equal to
8 by 5 so what will be our particular
solution substitute this a B 1 B 2 and D
minus hot e to the power 2 X minus 5 by
17 sine X plus 3 by 17 person X plus 8
by 5 times X into e to the power minus X
so the general solution would be this is
the general solution of homogeneous
equation C 1 in the power of minus X
plus C 2 e to the power 4 X is this
particular solution that the same thing
I have written
here and of course we get a status since
here it is e to power minus X here I'm
getting is x times e to power minus X so
now this is what is the general solution
of this homogeneous equation where now
we had used all the three rule status we
have seen the basic rule then we have
seen the modification rule we had used
the sum rule also now we will learn the
second method of finding the particular
solution that is method of variation of
parameters this method is more general
it doesn't require any special
conditions on your equation neither it
requires any condition on your right
hand side rx it is more general but a
little bit more complicated let us see
what this method says this status may
have p x QX as any there's no special
assumptions on this how this method is
working or what is this method let's say
non-homogeneous equation y double dash
plus PX y dash plus QX y is equal to RX
where i do have that PX QX & RX are
arbitrary but continuous only that is we
require the condition only that these
coefficients on the right hand side has
to be continuous on and some interval i
we also require to know the basis
solutions of the corresponding
homogenous equation so associated
homogeneous equation y double dash plus
PX y dash plus QX y is equal to 0
let's say that is the fundamental set of
system are the basis of solution is y1
and y2 then how we are finding the
particular solution this would be the
general solution of homogeneous equation
the particular solution we are saying is
another function as u 1 y 1 plus u 2 Phi
2 that is we had obtained corresponding
homogeneous equation its general
solution then for particular solution we
have
chosen you 1 y 1 plus u 2 y 2 where this
u 1 x and u 2 EPSA any arbitrary
functions of x only continuous on
differentiable how to obtain this u 1
and u 2 for this what we are doing is we
are substituting this wipey wipey - and
Y P Double Dash in this given equation
so when we substitute it we do use this
condition also that is y 1 and y 2 is
the solution of corresponding homogenous
equation all those things when we do we
can find out this u 1 u 2 as the
solution of the system of this system
that is U 1 - y 1 plus u 2 - y 2 is
equal to 0 u 1 - y 1 dash plus u 2 - y 2
dash is equal to rx you see here I am
getting the two equations which are
having y 1 y 1 - y 2 and Y 2 dash but
they are both are having u 1 cash and u2
- that is I would get the solution of
this system of linear equations as the u
1 - and you Tudor and then u 1 and u 2
we can obtain us integral of them so
what will be that solution let's see
here
u 1 X is minus integral y 2 x RX upon
wronskian of phi 1 by 2 w of y1 y2 is
the wronskian of the solution y1 and y2
y1 y2 are the general solution are the
fundamental set of solution of
Associated homogeneous equation this
integrated with respect to X and u 2 is
y1 x RX upon wronskian of y1 y2
integrated of 1 with respect to X these
are the two solutions so what I will get
that particular solution we would get so
here this has been written that
wronskian of w y1 y2 is the wronskian of
the fundamental solutions y1 and y2 y
people would get - that is U 1 Y 1 so
minus y1 integral Y 2 RX upon transcon
of y 1 and y 2 plus y 2 times integral y
1 RX upon R on scalar Phi 1 by 2
integrated with respect to X you see
here that we are now there we had London
on skin and we said is we are using it
for showing the linear independence of
the solution now you find it off if they
are linearly independent we do know that
this R on skin cannot be 0 since it
cannot be 0 so in the division it won't
create any problem and we are seeing now
here that is how we are using round skin
to find out the particular solution of
non-homogeneous equation let us try to
learn that how to use this method so let
us learn one with one example find the
general solution of y double dash minus
4 y dash plus 4 y is equal to e to the
power 2 X upon X you see of course we
don't require any special conditions on
the coefficients of the differential
equation but since till now we had
learned only to find out the solution of
linear differential equations with
constant coefficients so in this example
I have taken this coefficients to be
constant but the right hand side is not
of the form in which I could use the
undetermined coefficient that's this
method of undetermined coefficients is
what I am having the right hand side e
to the power 2 X upon X now it is not
coming as the sum of any known functions
because it is the ratio of two functions
which are in that table so I cannot use
that method so I do not have any option
other than to use this method of
variation of parameter so let us try to
solve this first we'll find out
associated homogeneous equation that is
y double dash minus 4 y dash plus 4 y is
equal to 0 to solve it we require the
characteristic equation that will be
lambda square minus 4 lambda plus 4 is
equal to 0 now we see this is nothing
but lambda minus 2 whole square that is
again we are getting the double root
lambda is equal to 2 and 2 what will be
the general solution of the
sar what will be the fundamental set of
solutions the fundamental set of
solutions would be e to the power 2 x
and x times e to the power 2 X now so
what should be the run skin of this
fundamental solutions that is I have got
two solutions e to the power 2 x and x
times e to the power 2 X fur on skin
first row is assets second row the
derivative that is 2 times e to the
power 2 X 2 X times e to the power 2 X
plus e to the power 2 X when we solve
this determinant we get since here we
would be getting is 2 X e to power 4 X
plus e to the power 4 X minus 2 times X
times e to the power 4 X I would get e
to the power 4 X which is not sealed
since these are linearly independent I
am getting this is not 0 for any value
of x so we have got e to the power 4 X
as the run skin so what will be now my u
1 and u 2 rather than using that putting
u 1 y 1 in those ones and finding out
what it is we would be using the
formulas directly Hawaii it is C 1 e to
the power 2 X plus C 2 x times e to the
power 2 X 4 YP our X is e to the power 2
X upon X so now what will be my you won
the formula says minus integral Y to X R
X upon run skin of Phi 1 by 2 integrate
with respect to X substitute this values
of Y to our X and so on y 2 is X e to
the power 2 X our X is e to the power 2
X upon X Ron skin W y 1 y 2 is e to the
power 4 X let us substitute X e to the
power 2 X into e to the power 2 X upon X
upon e to the power 4 x we are getting
is that this X is being canceled out by
this e to the power 4 X has been
cancelled out with this I'm getting this
integral DX that is a constant with
respect to X I would get minus X so you
wanna have got minus X what will be u 2
u 2 is y 1 x RX r on scanner y 1 y 2
integrated with respect to X Y 1 is e to
the power 2 X R X is e to the power 2 X
upon X wronskian is e to the power 4 X
what I am getting this e to power 4 X 4
X is cancelling out I would be getting
integral of 1 by X DX which we do know
is log X so what will be my particular
solution u 1 y 1 plus u 2 y 2 u 1 we
have got minus X and what was y 1 e to
the power 2 X so minus X e to the power
2 X u 2 is x times e to the power 2 X Y
2 we have got log X so we are getting
log x times X e to the power 2 X so what
will be the general solution general
solution would be C 1 e to the power 2 X
plus C 2 x times e to the power 2 X
minus x times e to the power 2 X plus X
log X e to the power 2 X now we can
rewrite it because we are having eetu
the power 2x all the places so we can
rewrite it as C 1 plus C 2 minus 1 plus
log X whole multiplied with X then whole
multiplied with e to the power 2 X so
this is what is now here we have learned
one method which says is method of
variation of parameter how we are Chinee
the particular solution when the right
hand side is not of this special form
and we are not getting that as from the
undetermined coefficients of course we
have learned only about differential
equations which have constant
coefficients now
this is what all about we had learned
about linear differential equations with
constant coefficients homogeneous
equations we had learned how to solve
them we had learned about non
homogeneous equations as well where we
had learn that is all the methods which
we had learnt in the non homogeneous
equations they were applicable both with
whether the coefficients are constant or
the coefficients are not constant they
are general functions they are not
changing but all the examples which we
had done they were having all the
coefficients as constants what since we
had learned in the linear differential
equations only to solve the constant
coefficients that is why in the non
homogenous we could not do any example
where the coefficients are not constants
so next lecture we would learn about the
linear differential equations which have
the coefficients as non constant that is
my px and QX were are functions rather
than the constants so that's all for the
today's lecture thank you for this
differential equations for undergraduate
students today's lecture is in
continuation of linear differential
equations of second order in last
lecture we had learned about these
equations moreover we have learned how
to solve linear differential equation
with constant coefficient of second
order we will learn something more about
these equations as well we will also
learn in today's lecture about non
homogeneous equations so first we will
learn about the existence and uniqueness
of the solution of initial value problem
here we are not going to find out that
is what are these conditions rather we
will state a result about under which
conditions the solution of initial value
problem does exist and under what
circumstances that solution would be
unique so the result here is existence
and uniqueness theorem for initial value
problem if px and QX are the continuous
functions on some interval I and X
naught is in I for differential equation
y double dash plus PX y dash plus QX y
is equal to 0 and two initial conditions
that is function y at X naught is K
naught and the derivative of Y at X
naught is K 1 then this initial value
problem has a unique solution Y X on
interval I we have learned it that this
solution unique solution of this initial
value problem is called the particular
solution and this particular solution we
can obtain by giving a specific value to
the general constants in the general
solution how we are obtaining those
constant values for this particular
solution what we are doing is we
here putting these initial conditions to
the general solution and then find out
the values for those constants so now we
will learn that
we had learned that general solution for
second-order equations contains two
linearly independent solutions those two
linearly independent solutions we are
obtaining by some methods we are
checking that those solutions which we
have obtained for the second-order
equation whether they are independent or
not for that either we had implied the
basic definition of linear independence
of function all we had learned one more
thing that is those two solutions are
not proportional today we will learn
something more about this linear
independence of solution which is known
as wronskian our Ron C determinant what
we do call the wronskian brown-skinned
for the second order equation y double
dash PX y dash plus QX y is equal to 0
if it has two solutions y1 and y2 then
the wronskian of y1 and y2 is being
given as a determinant whose first row
contains y1 and y2 the two solutions on
the second row contains the derivative
of those two solutions of course we
don't know the e this is equal to y1 y2
- minus y 1 - y - this is called the
wronskian on how you're checking the
linear independence if this wronskian is
not 0 for any value of x not in the
interval i then we say on that interval
i the two solutions y 1 and y 2 are
linearly independent some more
properties of this run skin given two
functions FX and GX that are
differentiable on some interval i the
wronskian of f and g at some point x
naught is not 0 for some X naught in I
then we call that FX and GX are linearly
independent on I moreover if FX and GX
are linearly dependent on I then this
wronskian off if I'm G would be 0 for
all X in I so we see is that is if
is linearly dependent we will get
wronskian of F and G to be zero for all
X and if it is linearly independent I
could find out that it's not some X not
where it is not zero now on special
property which we are using for the
differential equation that is called
Everest theorem what is that if y1 and
y2 are two solutions to the differential
equation y double dash plus PX y dash
plus QX y is equal to 0 then there are
skin of the two solution is at some
point X that is w y1 y2 at X is equal to
w y1 y2 at X naught into e to the power
integral minus 0 to X px DX we can show
this result also so we are going to
prove this the differential equation y
double dash plus py dash plus qy is
equal to 0 we have been given that it
has two solutions y1 and y2 since they
are two solutions so they would satisfy
our differential equation what it says
is that y double dash plus py 1 dash
plus qy 1 is equal to 0 and y double
dash plus py 2 dash plus qy 2 would be 0
now from these two equations if I
eliminate the Chum containing Q that is
I would multiply the first equation by Y
2 on the second equation by y1 and
substract then what we would get we
would get Y 2 Double Dash Y 1 minus y 1
double dash Y 2 plus py 2 dash Y 1 minus
y 1 dash y 2 is equal to 0 now let us
see what we have obtained again come
back to our run skin it is y 1 y 2 that
is y 1 y 2 dash minus y 1 dash Y 2 now
if I differentiate this function because
this is a function of X if I
differentiate with respect to X what I
would get Y 1 dash Y 2 dash plus y 1 y 2
Double Dash minus y 1 dash Y 2 dash
minus y 1 double dot
while that is y1 y2 double dash minus y1
double dash Y to since Y 1 dash Y 2 dash
is getting canceled it out now we see
what the equation we had obtained by
laminating cube we had obtained y 2
double dash y 1 minus y 1 double dash y
2 plus P times y 1 y 2 dash minus y 2 y
1 dash is equal to 0 now we see that the
coefficient of P is nothing but the
wronskian and what we are getting here
the constant term that is nothing but
the derivative of the wronskian now
these are the function of X so we can
write this equation as W dash plus P W
is equal to 0 where W dash is the
derivative of W with respect to X that
says is that we are getting a
differential equation with respect to W
now what we are having is now this is
first order equation we do know how to
solve it the solution of this is W X
naught e to the power minus 0 to X
integral px DX we do know that this
solution we do know and we do know this
is what we have replaced W X naught as
the constant which we are having that we
had replaced with a particular value X
naught how we could get this particular
value if I have been given the value of
W at a particular value it's not so this
is what we are getting as a solution
this we do know that is how what that
Able's theorem says now from here what
we are getting the result that is if W
at some X naught is not 0 I would get
that W X will never be 0 since this
whatever be this part this will never be
0 for any value of x that says is for
linearly independent I require
w X to be not 0 for only a single point
while when they are linearly dependent
they would be 0 for all X now let us see
one example how we are using this
wronskian to find out the linear
independence
find the general solution of y double
dash minus two y dash plus y is equal to
zero this is a linear differential
equation with constant coefficients we
do know how to solve it so solution let
us see what is the characteristic
equation that would be lambda square
minus 2 lambda plus 1 is equal to 0 now
we do know this is nothing but lambda
minus 1 whole square so I would get the
double root lambda is equal to 1 we
already know that the two solutions for
this we are taking as e to the power x
and x times e to the power X now we are
going to check that these two solutions
are linearly independent and we would
employ the method of Bronski
so we'll find out the Transkei
determinants for these two solutions so
what it would be e to the power X that
is the first solution then the second
solution x times e to the power X on in
the second row the derivative of e to
the power X that is e to the power it's
derivative of x times e to the power X
lattice e to the power X plus x times e
to the power x if I calculate what is
the value of this determinant that would
be e to the power 2 X plus X times e to
the power 2 X minus x times e to the
power 2 X that is equal to e to the
power 2 X that is not zero for any value
of x this says is that the two solution
e to the power x and x times e to the
power x they are linearly independent we
know that we had already checked it
using the method of constant proportion
so if I take the ratio of e to the power
x and x times e to the power x that is
coming out to be 1 by x which is not a
constant so this also we do know from
the first method of not proportion that
they are linearly independent so what
will be the general solution of this
given equation that would be C 1 e to
the power X plus C two x times e to the
power X now we had learned about the
linear independence
if I one more method that is called
Ron's key determinant our wronskian now
I will move to the non homogeneous
equations
so first we'll do non-homogeneous second
order linear differential equations what
are they we do know that they are second
order linear differential equations with
right-hand side not to be 0 that is y
double dash plus PX y dash plus Q XY is
equal to rx and I should have that rx
should not be 0 this we do now we'll
learn about that is what the we are
going to solve this so what we says one
more term we are associating with this
that is corresponding homogeneous
equation where I will take the equation
as such but the right-hand side s 0 this
is called the corresponding homogeneous
equation for this non-homogeneous
equation are sometimes it is also
referred as Associated homogeneous
equation so this equation we are
obtaining has the same equation but the
right hand side rx we have made it here
0 what we get one result that suppose
Capital y 1 and capital y 2 are two
solutions of non-homogeneous equation y
double dash plus PX y dash plus QX y is
equal to RX and suppose is small y1 and
y2 are fundamental set of solution of
corresponding homogeneous differential
equation that is y double dash plus PX y
dash plus QX y is equal to 0 here what
we have taken that is y1 and y2 are the
fundamental said that means they are
linearly independent solution of this
homogeneous equation then Capital y 1
minus y 2 is also a solution of this
homogeneous equation and since it is a
solution of the homogenous equation it
can be written as a linear
combination of the fundamental set of
solution that is y1 x and y 2x that is I
could write Capital y 1 minus Capital y
2 c1 times y 1 plus c2 times y 2 again
this is again a simple result which we
can show here by the knowledge whatever
we are knowing so let's prove it
since y 1 x and y 2 X second derivative
would be nothing but the second
derivative of y 1 - second derivative of
Phi 2 and the derivative of y 1 minus y
2 is nothing but the derivative of Y 1
minus derivative of Phi 2 will
substitute this in our homogeneous
equation that is y double dash plus P XY
- plus Q XY is equal to 0 since we want
to show that Y 1 minus y 2 is a solution
of the homogenous equation let us
substitute it what we do get Y 1 Double
Dash minus y 2 Double Dash plus P times
y 1 dash minus y 2 dash plus Q times y 1
minus y 2 now rearrange the terms first
I write the terms corresponding to Y 1
that is y 1 Double Dash plus px times y
1 dash plus Q XY 1 then minus y 2 - y 2
Double Dash plus px times y 2 dash plus
Q XY 2 now the first term which we are
having this is since Capital y 1 is the
solution of non-homogeneous equation
that means this first terms what we are
having that would be that would be equal
to the right hand side of
non-homogeneous equation that is our X
similarly the second term will also be
the non homogeneous solution of
non-homogeneous equation that is it
should also be equal to rx that says
this rx minus rx will get SIL what it
says if I am substituting in this
equation y 1 minus y 2 I am getting
right-hand side equal to 0 that says Y 1
minus y 2 is satisfying this equation
which says that
is a solution now since it is a solution
we can write it as a linear combination
of the fundamental set of solutions that
is I could write it as c1 y1 plus c2 y2
for some c1 and c2 now from here what we
could get we could get one more result
say what we could do let's say let us
consider the two solution y + YP X what
are they they are actually we are taking
this non-homogeneous equation while
double dot plus px slider plus QX y is
equal to rx and let's say Y is a general
solution of this non-homogeneous
equation and YP X is a particular
solution of this non-homogeneous
equation so now what we are getting we
are getting this as capital y1 and this
as capital y to the two solutions are
non-homogeneous equation then by the
previous result we do know that Y minus
y fee would be solution of corresponding
homogenous equation let's say that is
the fundamental set of homogeneous
equation solution of some homogeneous
equation would be Y small Y one and Y 2
that means this would be equal to c1 y1
plus c2 y2 now from here I could write Y
as c1 y1 plus c2 y2 plus YP X so the
first term c1 y1 X plus c2 y2 x is
nothing but the general solution of the
corresponding homogeneous equation and
YP X is a particular solution of the
non-homogeneous equation that says is
the general solution of non-homogeneous
equation YX I can write as y HX plus YP
X where YH X is the solution of
homogeneous equation that is c1 y1 X
plus c2 y2 x this is the general
solution of Associated homogeneous
equation and YP X is a particular
solution of the non-homogeneous equation
thus
we are getting now you see we had
already learn how to find out or that is
these solutions y1 and y2 they are
linearly independent
now this YP which we are saying if I am
keeping this in the general solution
this YP should also be linearly
independent of y1 and y2 so this is
important thing over here now we come to
our main theorem of non-homogeneous
equation this is about general solution
for a second order linear differential
equation which is non-homogeneous that
is y double dash plus PX y dash plus QX
y is equal to RX it's solution general
solution can be given as general
solution of Associated homogeneous
equation plus particular solution of the
non-homogeneous equation that is why X
can be written as y HX plus YP X where
YX is the general solution of
corresponding homogeneous equation y
double dash plus PX y dash plus QX y is
equal to 0 and YP X is a particular
solution of our non-homogeneous equation
now what will be the particle result
about the particular solution the
particular solution we do know can be
obtained from the general solution y HX
plus YP X by giving the specific values
c1 c2 in the wallet X since we do know
that y HX is the general solution of
Associated homogeneous equation y px is
a particular solution so it is not
containing any arbitrary constants the
arbitrary constants are over here now if
I keep this c1 and c2 0 I will get only
YP X that is the particular solution
which we are keeping over here but that
is not the only particular solution we
can obtain some other particular
solutions as well by giving some values
to the c1 and c2 so what will happen to
the initial value problem that is from
there we will obtain this c1 and see
by putting the initial values in the
general solution so the general solution
of one now one more result over here the
general solution of non-homogeneous
equations includes all solution what
this is or says this is if px qyx and rx
are continuous on some open interval I
then every solution of y double dash
plus PX y dash plus QX y is equal to RX
on I is obtained by giving suitable
values to the arbitrary constants in
general solution y HX plus YP X that is
every solution of this non-homogeneous
equation can be obtained through this
general solution it may imply there is
no singular solution now what we have
drawn the practical conclusion out of
these results we have obtained that to
solve a non-homogeneous equation y
double dash plus PX y dash plus QX y is
equal to RX first are an initial value
problem what we do first we do solve the
homogeneous equation y double dash plus
PX y dash plus QX y is equal to 0 and
find a particular solution by PF
non-homogeneous equation and then we do
keep it over there we had learnt the
techniques to solve homogeneous linear
equations of course there we have not
done it for general PX and QX we have
done only for the constants px and QX
that says this now so let us do first
the examples we will do about this
constant then we will learn about
functions as well so we do know if this
constants we do know how to solve this
homogeneous equations but the things
remain is how to find out a particular
solution for this non-homogeneous
equation we'll start with the basics so
let us first find out that is basic
technique to solve the non-homogeneous
equation
it is method to fight YP it doesn't
start with the as I say this with basics
so let us I start with an example find
the general solution of y double dash
plus three y dash plus 2y is equal to 12
times e to the power X we see this is a
linear equation its coefficients are
constants but the right-hand side is not
0 that is it is non-homogeneous so what
we do is first we'll solve ax associated
homogeneous equation what is the
corresponding homogeneous equation y
double dash plus 3 y dash plus 2y is
equal to 0 we do know how to solve it
first we find out the characteristic
equation what is the characteristic
equation lambda is square plus 3 lambda
plus 2 is equal to 0 what are its roots
of course it is lambda plus 2 into
lambda plus 1 is equal to 0 so it has
two roots - what on - 2 they are
distinct real numbers so what will be
the corresponding general solution of
the homogenous equation that will be C 1
e to the power minus X plus C 2 e to the
power minus 2x now the differential
equation was y double dash plus 3y -
plus 2y is equal to 12 times e to the
power X now I have to find out the
particular solution particular solution
means is that is it should be a solution
of this equation now let us see
right-hand side we are having the charm
e to the power X so now if I experiment
with e to the power X we do know it's
derivative as well as double derivative
all our e to the power X so if I am
substituting so we can try like this one
so what we are doing is we are taking
particular solution as a times e to the
power X because I do not know what will
stop satisfy this equation that means it
should if I'm putting e to the power X I
should get equal to 12 times e to the
power X so I have put I have taken that
as it constant because e to the power X
I can obtain by putting e to the power X
but this constant we have to
or cancer I have put here in how to
obtain this yeah now if I am assuming
that this is a solution this must
satisfy this equation so now I will
substitute wipey wipey - and Val P
double - in this equation so what is y P
- a times e to the power x + y P double
- that is again a e times e to the power
it's let us substitute it we would be
getting is a times e to the power X into
1 plus 3 plus 2 is equal to 12 times e
to the power X what is this is that 6 a
times e to the power X is equal to 12
times e to the power X which implies
that my a must be 2 that is the
particular solution should be 2 times e
to the power X we can check that this is
satisfying this equation so this is a
particular solution so now what will be
the general solution of our
non-homogeneous equation that would be
the general solution of homogeneous plus
the particular solution so I could say 1
e to the power minus x plus C 2 e to the
power minus 2x + 2 times e to the power
X so this is a general solution or
homogeneous equation we can check that
if I substitute this function under its
derivatives in this equation for any
value of C 1 and C 2 that is take the
general constant C 1 and C 2 I will
often that is this will satisfy this
equation so this is the general solution
of the non-homogeneous equation so this
is here we have done with an example on
the Basic Instinct we have got that is
right hand side we had seen that it is
containing the chunky to the power X and
we had already experimented from the
beginning of this course with the
function e to the power X so we could do
it now let us learn the methods to find
out why P we will learn here the two
methods to find out this particular
solution the one method method of
undetermined coefficient that is
something like that what we have done in
this example and there is another method
called method of variation of parameter
so first we will learn the method of
undetermined coefficients
what this method says is this method is
first let let us non that is where this
method is applicable this method is
applicable in the equations with
constant coefficients of C we would be
dealing with only linear equations with
constant coefficients and right-hand
side rx that is also a special form that
special form that is either my RFC's of
the form e to the power X or it is a
polynomial or it may be a function of
cosine ax signed something like that
then only this method is applicable now
let us learn what this method is this
method says is that if Rx is of a
special form I would choose that
particular kind of YP how we are doing
it
let me summarize it by a table we said
that is if I do have the rx as K times e
to the power gamma X that as in our
example I was having 12 times e to the
power X that is gamma was 1 and what I
have choose our chose the y PS e to the
power X and a here that says is that if
our X has of the form K times e to the
power gamma X choose y PS say x into the
power gamma x if my Rx is containing the
polynomials that is here I have given
the example of K times X to the power n
then twice of Y P will be a polynomial
of degree n see here we are not taking
only K times X to the power and we are
taking the complete polynomial K naught
K 1 X plus so on K and X to the power n
so if it is a polynomial or if it is
some terms of polynomial in the choice
of Phi P we will always choose the
complete polynomial of degree n then if
the terms in are X R cosine are sign
that is K times cosine Omega X odd K
times cosine sine Omega X we will choose
Y P as a
sine Omega X plus B cosine Omega X again
you see my rx may contain only cosine
our only sign but in the YP we would
choose both sine and cosine functions
now if the Charmander X is having a
multiple of e to the power alpha X and
cosine are sine terms we would be
choosing y fears
e to the power L Phi X is sine Omega X
plus B cosine Omega X that is now what
are the rules to apply this method so
we'll learn some basic rules rules of
this method our first is called the
basic rule basic rule just now as I had
explained you that is for second order
linear differential equation which is
non-homogeneous y double dash plus PX y
dash plus QX y is equal to RX if R
exists one of the function in the first
column of the table just now as I had
explained I would choose corresponding
function YP in the second column and
find the value of undetermined
coefficient by putting YP and its
derivative in the given equation so we
had got that is by using that table we
can choose YP and as in our example what
we have done we had find out the value
of a by putting wipey wipey - and why P
double dash in the equation and then we
have determined what is a so here
whatever the undetermined coefficients
KC k1 K naught and so on we could find
out all of them by putting YP and its
derivatives in the non-homogeneous given
differential equation
I'm solve for the unknown coefficients
that is why it is called the method of
undetermined coefficients now you do
remember I said that the general
solution contains two things one is
general solution of Associated
homogeneous equation y HX and another is
YP X now it would be the general
solution that
I should have YP also linearly
independent of the solutions which we
are having as y1 and y2 of the
homogenous equation that says sometimes
it may happen that is whatever I am
getting as the choice of YP that may
turn out to be one of the solution of
the homogenous equation then of course
because here I am choosing it a constant
time that function that will not be
linearly independent of the solutions of
homogenous one so how to solve this kind
of problem that says is the second rule
as modification rule it says if any term
in the choice for YP is also in the
solution Y H of corresponding
homogeneous equation y double dash plus
px Y - plus Q X Y is equal to zero then
multiply the choice YP by X R by X
square if the solution corresponds to
the double root of the characteristic
equation of the homogeneous equation why
we are doing it self we can understand
this very well since that particular
solution is also coming us in the
solution of homogeneous equation so with
a constant coefficient of course we
cannot get it as linearly independent so
we have to multiply with its now if it
is solution corresponding to the double
root then we do know that it's times
that solution is also coming as the
fundamental solution so I have to
multiply it by X square then there is
one more rule that is called some rule
what it says is I may have R X that is
our X is sum of different functions
given in the column one of the table
then choose YP is the sum of
corresponding functions in the second
column now let us see that is how to use
these rules and how to use this
particular method to find out the
general solution of non-homogeneous
equation that we will see through some
examples
so let us go with the examples first
example let's say solve the
non-homogeneous equation y double dash
plus 4y is equal to 8 X square we see
this is a second order equation constant
coefficients are constants and
right-hand sides of the special form
that is it is the polynomial that is 8
times X to the power 2 so go for the
solution give an equation is y double
dash plus 4y is equal to 8 X square the
corresponding homogenous equation will
be y double dash plus 4y is equal to 0
to solve this we require the
characteristic equation what is the
characteristic equation that is lambda
squared plus 4 is equal to 0 what will
be its root it is lambda square is equal
to minus 4 that means it will have
complex conjugate roots the roots would
be plus minus 2i we do know in this case
what is this L the general solution of
homogeneous equation that is C 1 cosine
2x plus c2 sine 2x now what is my right
hand side that is a Texas square now if
you're not remembering we of course go
back to our table we are having 8 X is 4
that is here n is equal to 2 so what we
would choose we would choose K naught
plus K 1 X plus K 2 X square so let's
see since our X is 8 X square Y P we
would choose K 2 X square plus K 1 X
plus K not now substitute its
derivatives in the equation that is
first find out what is Phi P dash that
is 2 K 2 X plus K 1 YP Double Dash would
be 2 K 2 now substitute this in the
given equation y double dash plus 4 y is
equal to 8 X square what we get we get 2
K 2 plus K 2 square for K 2 X X square
plus 4 K 1 X plus 4 K naught is equal to
8 X square now to find out these
constants will compare the
coefficients of different powers so
equating the coefficients we get the
coefficient of x is square on the right
hand side is a while the coefficient of
x is square on the left hand side is
fourth it was the first equation we are
getting is for K 2 is equal to 8 then we
are getting is the right hand side the
coefficient of x is zero while on the
left hand side the coefficient of x is
for K 1 so I get for K 1 is equal to 0
now right hand side I don't have any
constant term while on the left hand
side I do have the constant term 2 K 2 +
4 K naught that says is 2 K 2 + 4 K
naught is also 0 now from the first
equation I am getting K 2 is equal to 2
second equation gives K 1 as 0 on the
first equation the third equation gives
me that K naught is equal to -1 now
substitute this in our choice of Y P
what I would get I would get Y PS 2 x
squared - 1 so what will be the general
solution of our equation y double dash
plus 4y is equal to 8x square that is c1
cosine 2x plus c2 sine two x that is the
general solution of homogeneous equation
plus this particular solution 2 X square
minus 1
now you can obtain check with this
solution find out its derivative and
double-dare a second derivative
substitute in this equation and check
that is in general if I am taking only
c1 and c2 this will satisfy for all
values of c1 and c2 so if this would be
the interesting exercises for you to do
you can check it that this is a solution
of this non-homogeneous equation now let
us do one more example
yeah one more thing here you would be
worrying that is why I have chosen this
as I said is we will always choose
polynomial you can try one more
interesting exercise you can try only
with KX square don't use the complete
polynomial and see that what the
solution you are getting that will not
be actually
solution of this non-homogeneous
equation is another interesting exercise
which you can try so let us move to the
second example solve the initial-value
problem y double dash plus two y dash
plus y is equal to e to the power minus
x where the initial conditions are that
the function r 0 is minus 1 and the
derivative at 0 is 1 move to the
solution the given differential equation
is y double dash plus 2 y dash plus y is
equal to e to the power minus x see
again we are having the equation has
constant coefficients on right hand side
is of a special form e to the power of
minus x first find out the corresponding
homogeneous equation that is y double
dash plus 2 y dash plus y is equal to 0
for solving this we require its
associated characteristic equation that
is lambda square plus 2 lambda plus 1 is
equal to 0 we see this is the whole
square of lambda plus 1 that is we are
going to get the double root lambdas
minus 1 and minus 1 we do know in
homogeneous equations when we are having
the double root the solution is e to the
power minus X and the other solution we
choose x times e to the power minus x so
the general solution would be c1 plus c2
x times e to the power minus X now right
hand side is e to the power minus X now
you see if I am using our table I would
get right y PS a times e to the power
minus XR C times e to the power minus X
but this e to the power minus X is also
coming us here in the e to the power
minus X is also a solution of my
homogeneous equation moreover this is a
solution corresponding to the double
root that is x times e to power minus x
is also a solution so we have to apply
the modification rule so again I am
repeating this these steps first we see
from our table
I do have e to the power minus X I would
choose C times e to the power minus X
that is gamma is minus 1 since again
this e to the power minus X is coming as
a solution of the homogeneous linear
equation we do apply the modification
rule the modification rule says is
multiply the choice of Phi P by X R by X
square so here my solution e to the
power minus x is corresponding to the
double root so I have to multiply by X
square that says this I should choose my
Y P as a times e to the a times X square
times e to the power minus X find out
what is YP - and why P double - so Y P -
its derivative - ax e to the power minus
6 minus ax s squared e to the power
minus 6 second derivative - e to the
power minus 6 minus 2 ax e to power
minus 6 minus 2a X e to the power minus
X plus X square times e to the power
minus X that is it will be 2 e to the
power minus X minus 4a x times e to
power minus X and plus a X is called
power minus 6 substitute this in given
equation that is Phi Double Dash plus 2y
- plus y is equal to e to power minus 6
left hand side what we would be getting
to a e to the power minus X minus 4a x
times e to power minus X plus ax is
called power minus 6 plus 2 times its
derivative that is we are getting it as
2a X e to the power minus X minus ax is
square 8 4 minus X plus y is 8 X is
square e to power minus 6 this must be
equal to 8 power minus X let's see in
the left hand side here I am having this
minus 4a X e to power minus X this would
be plus 4a X e to power minus X that is
this term is get cancel it out here I am
getting is a times X is square e to the
power minus 6 that is a coefficient of x
is called 4 minus 6 here is a here is
minus 2 is since this 2 and here is one
more a plus a again it is cancelling out
so what is being left I have been left
with 2 e to the
- X is equal to e to power minus X so
comparing the coefficients we get to a
is equal to 1 R is equal to 1/2 so what
will be my wife YP would be 1/2 X square
e to the power minus X so the general
solution would be C 1 plus C 2 X e to
power minus 6 that is the general
solution of homogeneous equation Plus
this particular solution 1/2 X square e
to the power minus X now you can
actually again check with this solution
that you could find it out that
substitute this in this equation and
find out that this is satisfying this
equation and moreover here we are
getting is e to the power minus x x
times e to the power minus x and x is
square e to power minus x all of them
are linearly independent no to find out
the particular solution because this is
an initial value problem which we had
started so by finding out the particular
solution what are the initial conditions
we have been given we have been given
that the function R 0 is minus 1 and its
derivative at 0 is 1 what the general
solution we had got C 1 plus C 2 XE to
the power minus X plus 1/2 X square e to
the power minus X now if I substitute
this function at X as 0 that is I find
out the value of the function at 0 that
is given as minus 1 from here in this
solution if I am substituting X as 0 I
would get only C 1 so it gives C 1 is
equal to minus 1 then find out its
derivative derivative is minus C 1 plus
C 2 X e to power minus X minus 1/2 X is
called 4 minus 6 plus say 2 times e to
the power minus x plus x times e to the
power minus X now again the second
condition is y dash 0 is 1 so again we
are putting X is equal to 0 what I would
be left I would be left minus C 1 plus C
2 which is equal to 1 from the first
result we have got C 1 as minus 1 thus
gives C to us 0 so we have got the
particular solution that is I will put C
is equal to minus one I'm c2 is equal to
zero in our general solution of
homogeneous equation so I would get the
solution of initial value problem as
minus e to the power minus x plus half X
square e to the power minus X we can
check that this is satisfying both
initial conditions that is the function
value of zero is minus 1 and the
function derivative of the function r0
is plus one let us do one more example
over here find the general solution of y
double dash minus three y dash plus sign
y double dash minus three y dash minus
4y is equal to three times e to the
power 2 X plus 2 times sine X minus 8
times e to the power minus X now we see
our right-hand side is actually sum of
many functions which are in our table e
to the power 2 X e to the power minus X
sine X and so on
so it says is that we have to use the
sum rule let us see if we have to use
some other rules also so first we go for
the solution what is the corresponding
homogenous equation y double dash minus
3 y dash minus 4 y is equal to 0 to
solve this we require the characteristic
equation what is the characteristic
equation lambda square minus 3 lambda
minus 4 is equal to 0 we do find it out
that it is nothing but lambda minus 4
into lambda plus 1 the factors so it
will have two roots minus 1 at 4 they
are real and distinct so the general
solution of homogeneous equation will be
C 1 e to the power minus X plus C 2 e to
the power 4 X now we see right hand side
right hand side our X we have 3 times e
to the power 2 X 2 sine X minus 8 times
e to the power minus 6
now so if I'll just go with our table if
you do remember something that is e to
the power gamma X we do take C times e
to the power gamma X so here if I do
take I have to choose one y PS e to the
power minus 6 and that is here again as
a solution corresponding to the one root
that means we have to use the
modification rule that is in this one we
would be going to use the sum rule as
well as the modification rule for the
song rule we require which functions we
have to choose we have here function of
e to the power gamma X and sine X so let
us consult our table e to power gamma X
minus means I have to use constant times
e to the power gamma X and sine Omega X
I have to use the function which is
having a constant time sine Omega X
under constant times cosine Omega X with
plus so let us see what will be our
choice of YP the choice of YP would be a
times e to the power 2 X plus b1 sine X
plus b2 cos x plus J x times e to the
power minus X Y I had chosen here since
e to the power minus X was already a
solution so I have to modify it as ifs
times e to the power minus X so here I
had use a modification rule I'm in this
complete one I had use the sum rule so
we have got this as the choice of YP now
find out its derivative to e to the
power 2 X plus b1 cosine X minus v2 sine
X minus D X e to power minus X plus D
times e to the power minus 6 second
derivative 4a e to the power 2 X minus
b1 sine X derivative of sine X is cosine
X so B 2 cosine X derivative of DX e to
power minus X is plus DX 8 power minus X
minus D to power minus X and the
derivative of this is minus D to the
power minus X now substitute this in the
given equation what was the equation
equation was y double dash minus 3 y
dash minus 4 y is equal to 3 times e to
the power 2 X plus 2 sine X minus 8
times e to the power minus six so
substituting it and of course
calculations you can check it we get
minus six a times e to the power 2 X
minus 5 B 1 minus 3 B 2 sine X minus 5 B
2 plus 3 B 1 cosine X minus 5 d e to the
power minus X this should be equal to
the right hand side that is 3 times e to
the power 2 X plus 2 times sine X minus
8 times e to the power minus X now it
create the coefficients we get the
coefficient of here you see e to the
power 2 X its coefficient here is minus
6a here is 3 so I would be creating this
as this one the coefficient of sine X
here is minus 5 B 1 minus 3 B 2 and here
is plus 2 so I would get 5 B 1 minus 3 B
2 is equal to minus 2 the cosine X term
we don't have over here so though this
fight be 2 plus 3 B 1 would be 0 the
coefficient of e to the power minus x is
minus 5 D here here it is minus 8 so I
would get 5 D is equal to 8 so equating
the coefficient I am getting minus 6 a
is equal to 3 which gives me is equal to
minus 1/2 stick in what we are getting
for cosine and sine 5 B 1 minus 3 B 2 is
equal to minus 2 and 5 B 2 plus 3 B 1 is
equal to 0 solving these equations we
get B 1 as minus 5 by 17 and B 2 as
three by 17 and last one we are getting
5 D is equal to 8 gives me D is equal to
8 by 5 so what will be our particular
solution substitute this a B 1 B 2 and D
minus hot e to the power 2 X minus 5 by
17 sine X plus 3 by 17 person X plus 8
by 5 times X into e to the power minus X
so the general solution would be this is
the general solution of homogeneous
equation C 1 in the power of minus X
plus C 2 e to the power 4 X is this
particular solution that the same thing
I have written
here and of course we get a status since
here it is e to power minus X here I'm
getting is x times e to power minus X so
now this is what is the general solution
of this homogeneous equation where now
we had used all the three rule status we
have seen the basic rule then we have
seen the modification rule we had used
the sum rule also now we will learn the
second method of finding the particular
solution that is method of variation of
parameters this method is more general
it doesn't require any special
conditions on your equation neither it
requires any condition on your right
hand side rx it is more general but a
little bit more complicated let us see
what this method says this status may
have p x QX as any there's no special
assumptions on this how this method is
working or what is this method let's say
non-homogeneous equation y double dash
plus PX y dash plus QX y is equal to RX
where i do have that PX QX & RX are
arbitrary but continuous only that is we
require the condition only that these
coefficients on the right hand side has
to be continuous on and some interval i
we also require to know the basis
solutions of the corresponding
homogenous equation so associated
homogeneous equation y double dash plus
PX y dash plus QX y is equal to 0
let's say that is the fundamental set of
system are the basis of solution is y1
and y2 then how we are finding the
particular solution this would be the
general solution of homogeneous equation
the particular solution we are saying is
another function as u 1 y 1 plus u 2 Phi
2 that is we had obtained corresponding
homogeneous equation its general
solution then for particular solution we
have
chosen you 1 y 1 plus u 2 y 2 where this
u 1 x and u 2 EPSA any arbitrary
functions of x only continuous on
differentiable how to obtain this u 1
and u 2 for this what we are doing is we
are substituting this wipey wipey - and
Y P Double Dash in this given equation
so when we substitute it we do use this
condition also that is y 1 and y 2 is
the solution of corresponding homogenous
equation all those things when we do we
can find out this u 1 u 2 as the
solution of the system of this system
that is U 1 - y 1 plus u 2 - y 2 is
equal to 0 u 1 - y 1 dash plus u 2 - y 2
dash is equal to rx you see here I am
getting the two equations which are
having y 1 y 1 - y 2 and Y 2 dash but
they are both are having u 1 cash and u2
- that is I would get the solution of
this system of linear equations as the u
1 - and you Tudor and then u 1 and u 2
we can obtain us integral of them so
what will be that solution let's see
here
u 1 X is minus integral y 2 x RX upon
wronskian of phi 1 by 2 w of y1 y2 is
the wronskian of the solution y1 and y2
y1 y2 are the general solution are the
fundamental set of solution of
Associated homogeneous equation this
integrated with respect to X and u 2 is
y1 x RX upon wronskian of y1 y2
integrated of 1 with respect to X these
are the two solutions so what I will get
that particular solution we would get so
here this has been written that
wronskian of w y1 y2 is the wronskian of
the fundamental solutions y1 and y2 y
people would get - that is U 1 Y 1 so
minus y1 integral Y 2 RX upon transcon
of y 1 and y 2 plus y 2 times integral y
1 RX upon R on scalar Phi 1 by 2
integrated with respect to X you see
here that we are now there we had London
on skin and we said is we are using it
for showing the linear independence of
the solution now you find it off if they
are linearly independent we do know that
this R on skin cannot be 0 since it
cannot be 0 so in the division it won't
create any problem and we are seeing now
here that is how we are using round skin
to find out the particular solution of
non-homogeneous equation let us try to
learn that how to use this method so let
us learn one with one example find the
general solution of y double dash minus
4 y dash plus 4 y is equal to e to the
power 2 X upon X you see of course we
don't require any special conditions on
the coefficients of the differential
equation but since till now we had
learned only to find out the solution of
linear differential equations with
constant coefficients so in this example
I have taken this coefficients to be
constant but the right hand side is not
of the form in which I could use the
undetermined coefficient that's this
method of undetermined coefficients is
what I am having the right hand side e
to the power 2 X upon X now it is not
coming as the sum of any known functions
because it is the ratio of two functions
which are in that table so I cannot use
that method so I do not have any option
other than to use this method of
variation of parameter so let us try to
solve this first we'll find out
associated homogeneous equation that is
y double dash minus 4 y dash plus 4 y is
equal to 0 to solve it we require the
characteristic equation that will be
lambda square minus 4 lambda plus 4 is
equal to 0 now we see this is nothing
but lambda minus 2 whole square that is
again we are getting the double root
lambda is equal to 2 and 2 what will be
the general solution of the
sar what will be the fundamental set of
solutions the fundamental set of
solutions would be e to the power 2 x
and x times e to the power 2 X now so
what should be the run skin of this
fundamental solutions that is I have got
two solutions e to the power 2 x and x
times e to the power 2 X fur on skin
first row is assets second row the
derivative that is 2 times e to the
power 2 X 2 X times e to the power 2 X
plus e to the power 2 X when we solve
this determinant we get since here we
would be getting is 2 X e to power 4 X
plus e to the power 4 X minus 2 times X
times e to the power 4 X I would get e
to the power 4 X which is not sealed
since these are linearly independent I
am getting this is not 0 for any value
of x so we have got e to the power 4 X
as the run skin so what will be now my u
1 and u 2 rather than using that putting
u 1 y 1 in those ones and finding out
what it is we would be using the
formulas directly Hawaii it is C 1 e to
the power 2 X plus C 2 x times e to the
power 2 X 4 YP our X is e to the power 2
X upon X so now what will be my you won
the formula says minus integral Y to X R
X upon run skin of Phi 1 by 2 integrate
with respect to X substitute this values
of Y to our X and so on y 2 is X e to
the power 2 X our X is e to the power 2
X upon X Ron skin W y 1 y 2 is e to the
power 4 X let us substitute X e to the
power 2 X into e to the power 2 X upon X
upon e to the power 4 x we are getting
is that this X is being canceled out by
this e to the power 4 X has been
cancelled out with this I'm getting this
integral DX that is a constant with
respect to X I would get minus X so you
wanna have got minus X what will be u 2
u 2 is y 1 x RX r on scanner y 1 y 2
integrated with respect to X Y 1 is e to
the power 2 X R X is e to the power 2 X
upon X wronskian is e to the power 4 X
what I am getting this e to power 4 X 4
X is cancelling out I would be getting
integral of 1 by X DX which we do know
is log X so what will be my particular
solution u 1 y 1 plus u 2 y 2 u 1 we
have got minus X and what was y 1 e to
the power 2 X so minus X e to the power
2 X u 2 is x times e to the power 2 X Y
2 we have got log X so we are getting
log x times X e to the power 2 X so what
will be the general solution general
solution would be C 1 e to the power 2 X
plus C 2 x times e to the power 2 X
minus x times e to the power 2 X plus X
log X e to the power 2 X now we can
rewrite it because we are having eetu
the power 2x all the places so we can
rewrite it as C 1 plus C 2 minus 1 plus
log X whole multiplied with X then whole
multiplied with e to the power 2 X so
this is what is now here we have learned
one method which says is method of
variation of parameter how we are Chinee
the particular solution when the right
hand side is not of this special form
and we are not getting that as from the
undetermined coefficients of course we
have learned only about differential
equations which have constant
coefficients now
this is what all about we had learned
about linear differential equations with
constant coefficients homogeneous
equations we had learned how to solve
them we had learned about non
homogeneous equations as well where we
had learn that is all the methods which
we had learnt in the non homogeneous
equations they were applicable both with
whether the coefficients are constant or
the coefficients are not constant they
are general functions they are not
changing but all the examples which we
had done they were having all the
coefficients as constants what since we
had learned in the linear differential
equations only to solve the constant
coefficients that is why in the non
homogenous we could not do any example
where the coefficients are not constants
so next lecture we would learn about the
linear differential equations which have
the coefficients as non constant that is
my px and QX were are functions rather
than the constants so that's all for the
today's lecture thank you for this
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Nov 23, 2024 Last updated |
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