hello friends welcome to my lecture on method of undetermined coefficients this method we
we will be used to solve second order linear differential equations which are non homogenous
that they are of the form y double dash plus a y dash plus b y equal to r x now we have
already solved the homogenous linear second order differential equations so let us see
how we shall solve this second order non homogenous linear differential equation first of all
we shall show that if y c x is the general solution of
of the homogenous equation y double dash plus a y dash plus b y equal to zero and y p x
is a particular solution of the non homogenous equation of the equation
one then we shall see that sh show that y x equal
to y c x plus y p x is the general solution of
equation one so y c x is the general solution of the homogenous equation y double dash plus
a by dash plus b y equal to zero means y c will contain two arbitrary constants and we
know how to get the solution of this homogenous equation y p x is a particular solution of
the equation one means y p x is a solution of the equation one without any arbitrary
constants so if we can show that y x equal to y c x plus y p x is a solution of the equation
one then y x because it contains two arbitrary constants will be the general solution of
equation one y x contains two arbitrary constants because y c x contains two arbitrary constants
and y p x is a solution of equation one without any arbitrary constants
so so if we just prove that y x is a solution of equation one
then y x will be the general solution of one ok so let us see when you substitute y
x equal to y c x plus y p x if it satisfies the equation one it will be a solution of
equation one so substituting
in the left hand side of of one we get y p double dash x plus a y p
dash y double dash y double dash x plus y dash x plus b y dash b y equal to y c double
dash x plus y p double dash x plus a times y dash x means y c dash x plus y p dash x
plus b times y c x plus y p x we can write it as y c double dash x plus a y c dash x
plus b y c x now since y c x is a general solution of y double dash plus a y dash plus
b y equal to zero so this so this first expression is zero then we have now y p x
is a particular solution of the equation one therefore y p double dash plus a y p dash
plus b y p is equal to r x and thus we get y x is a solution of equation one again i
repeat y x will therefore be the general solution of equation one
now so if we are having a non homogenous second order linear differential equation with constant
coefficients then we will now discuss a method by which we can determine this particular
solution y p x y c x we know how to determine so our now method will determine the
value of y p x so the method that we are going to discuss is called as the method of [undeter/undetermined]
undetermined coefficients in this method the by we we make an educated guess for y p x
the particular solution y p x of the equation one which we are looking for so we make a
guess of the expression for y p x which depends on
the form of r x and the complimentary function this complimentary
function is nothing but the general solution of the associated homogenous equation complementary
function is nothing but the general solution of
so i can say y c x so we may so this is not an arbitrary arbitrary s it is a it
is called an educated guess because it depends on the form of r x and also sometimes it depends
on the form of the complimentary function y c x
now this expression for y p x will contain constants which will be determined by
substituting y p x in the equation one so [tha/that] that is why we call this method
as method of undetermined coefficients now here we have certain detritions this method
can be applied to the second order linear linear linear differential equation which
are non homogenous when r x is of the following type it can be a polynomial function it can
be an exponential function it can be a sin or cosine function or a sum or products of
such functions now or you can take linear combination of the functions which are
polynomials or exponential functions or trigonometric functions sin x cosine x so
p with p you can take r x to be function of the form p x or you can take p x into e
to the power alpha x r p x e to the power x cos beta x r p x e to the power alpha x
and beta x so r x is a function of this type where n is non negative integer and alpha
and beta are real numbers then we can apply this method of undetermined coefficients
now if r x is of the type log x r x to the power minus one or ten x r sin inverse x then
this method cannot be applied so for second order non homogenous linear differential equations
where r x is of those types log x six to power minus one ten x sin inverse x etcetera
then we will apply the general method which we shall discuss in the next lecture now so
in the general method we will be taking care of all the functions r x which cannot
be handled by the method of undetermined coefficients now let us see if you take the constants
polynomials exponentials sines or cosines the remarkable property of those functions
is that the derivatives of the sums and products of such functions are again sums and products
of similar kind of functions so in order to assume the form of y p x ok it is reasonable
to assume the same form as that of r x now let us take an example say y double dash
plus four y dash minus two y equal to two x square minus three x plus six so here we
can see r x is a polynomial in x of degree two now first we shall find y c x so to find
y c x we have the write the auxiliary equation m square plus four m minus two equal to zero
we can find the value of m for this so this will come out to be minus four plus
minus so m has two distinct values minus two
plus root six and minus two minus root six and therefore y c x is equal to c one e to
the power minus two plus root six into x plus c two e to the power minus two minus root
six into x now since r x is equal to two x square
minus three x plus six which is the polynomial of degree two let us assume y p x y p x
to be equal to a x square plus b x plus c a polynomial in x of degree two where a b
c are undetermined coefficients we shall determine the values of these undetermined coefficients
a b c by substituting this y p in to the in the given differential equation because y
p is a particular solution of the given differential equation so it should satisfy that equation
so let us put y p equal to y p x equal to a x square p plus b x plus c let us put
this value y p x in the given equation y double dash so let us find y p dash x equal to two
a x plus b y p double dash is equal to two a so then y p double dash plus four y p dash
minus two y p equal to two x square minus three x plus six gives us two a now this
equation is valid for all values of x so it must be an identity and therefore the coefficient
of like powers of x must be same so if we equate the coefficients of like powers
of x we get the values of the unknown coefficients a b c you can see that here x square
coefficient x square is minus two a so minus two a equal to two which gives us a equal
to minus one when we equate the coefficient of x on both sides we get eight a minus two
b equal to minus three putting the value of a equal to minus one we get minus eight minus
two b equal to minus three so this will give you two b equal to minus five which give you
b equal to minus five by two and when you equate the coefficients on both sides we get
two a plus four b minus two c equal to six so substituting the values of a and b minus
one and minus five by two we shall have we shall have c equal to minus nine and thus
y p x will be equal to x square minus five by two x minus nine
so the general solution of this differential equation is the sum of the complimentary
function c one e to the power minus two plus root six into x c two e to the power minus
two minus root six into x hm and then minus x square minus five by two x minus nine now
let us take another example y double dash minus y dash plus y equal to sin three x here
the auxiliary equation is m square minus m plus one equal to zero you can find the values
of m they are one plus minus i root three by two so we can write the complimentary function
because these are complex roots so e to the power alpha x that is e to the power half
x into c one cos theta x beta is root three by two so c one cos root three by two x plus
c two sin root three by two x now let us make an guess for y p x so the
the natural guess when you look at the form of r x r x is equal to sin three x so natural
guess for the y p x would be that we write a times sin three x but then when
you differentiate sin three x it produces three times cos three x and then when you
differentiated again you get sin three x so it differentiation of sin three x produces
both sin three x and cos three x and therefore what we should do is we should assume y p
x to be a linear combination of sin three x and cos three x functions so let us assume
y p x to be having both the functions sin three x and cos three x that is a cos three
x plus b sin three x now then you substitute this y p x in the
given differential equation the given differential equation is y double dash minus y dash plus
y equal to sin three x so if you take y p x equal to a cos three x plus b sin three
x and then put it in the given differential equation what you will get is unequating
the coefficients of sin three x and cos three x both sides we shall get a to be six by seventy
three and b to be minus sixteen by seventy three so we can find y p dash x here this
is minus three a sin three x plus three b cos three x we differentiate it again should
will give you minus nine a cos three x plus nine b minus nine b sin three x now substitute
the values of y p y p dash y p double dash in this differential equation and equate the
coefficient of sin three x and cos three x both sides we shall get a equal to six by
seventy three and b equal to minus sixteen by seventy three so we will get y p x to be
equal to minus six by seventy three cos x minus sixteen by seventy three sin three x
and thus the general solution y x is the sum of the complementary function this is
the complimentary function plus the particular integral the particular solution of the
equation one is also called as particular integral so six by seventy three cos three
x minus sixteen by seventy threes sin three x
now now let us look at the this comment here the as i said in the beginning y p x is
not an arbitrary it is not guessed arbitraly it is an educated guess because it depends
on the form of r x the functions that make up r x as well as the functions that make
up the complimentary function which is the solution of the associated homogenous
linear differential equation that is y double dash plus a y dash plus b y equal to zero
let us see how it depends on the complimentary function we have seen that the function
y p x depends on the form of r x in the next example we shall see how y p x depends on
the form of the complimentary function so let us take up this this problem y double
dash minus five y dash plus four y equal to a t to the power x here we can write the auxiliary
equation as m square minus four five m plus four equal to zero so we get two values of
m which are four and one they are distinct values so the complementary function we can
write easily y c x equal to c one e to the power x plus c two e to the power four x
now when we look at the form of r x r x is eight e to the power x and successive differentiation
of e to the power x produce e to the power x so we can assume y p x to be equal to
a times e to the power x but when you put y p x equal to eight e to the power x in this
equation in the given differential equation y double dash minus five y dash plus four
y equal to eight e to the power x let us see what happens y we are taking as a e to the
power x so let us find y dash it is a e to the power x let us find y double dash it is
again a to the power x so y so a e to the power x minus five a e to the power x plus
four a e to the power x equal to eight e to the power x now this is five a e to the power
x minus five a to the power x so this is zero equal to eight e to the power x which is
not correct of course because e to the power x is never zero so thus we have [mad/made]
made a wrong guess for y p x so what should be the form of y p x
now let us look at the complimentary function the complimentary function is c one e to the
power x plus c two e to the power four x and therefore e to the power x and e to the power
four x are both solutions of the associated homogenous linear differential equation of
second order that is y double dash minus five y dash plus four y equal to zero and so a
e to the power x one substituted in the given differential equation produces zero on the
left hand side now what should therefore be a form of y c x this is the question so let
us recall the case of repeated v l rules of the homogenous linear differential equation
with constant coefficients there we see that when the two roots are eq two roots are
same one if one m value of m is equal to m one which is repeated then the two solutions
independent solutions of the homogenous equations are homogenous equations are e to
the power m one x and x times e to the power m one x so let us see whether if we assume
y p x equal to a times x e to the power x do we get the solution of the given differential
equation so let us see if we do this then y p dash
x will be equal to a times this y p double dash will be equal to a e
x times x e to the power x plus two e to the power x now when you put this in the given
differential equation let us see what do we get so a times x e to the power x plus two
e to the power x that is y double y p double dash minus five y p dash y p dash means a
into x e to the power x plus e to the power x and then we have four times y which is a
x e to the power x equal to eight e to the power x now let us see what do we get so a
times x e to the power x four a times x e to the power x minus five a times x e to the
power x they cancel and what we get is two a e to the power x two a e to the power
x minus five a e to the power x equal to eight e to the power x so we get minus three
a e to the power x is never zero for any x so we we can divide by e to the power x and
we get the value of a x minus eight y three and thus y p x we get as minus eight y three
x e to the power x so the general solution therefore is given by c one y equal to c one
e to the power e x plus c two e to the power four x minus eight by three x e to the power
x in this slide we have listed the choice
of y p x depending on the choice of r x so if in the first column we have given the form
of r x in the second column we have given the choice of y p x provided we have
these exceptions so suppose r x is of the form k x to the power n then where k is a
constant then we shall make the choice of y p x to be a polynomial in x of degree
n that is k n x to the power n plus kn minus one x to the n minus one plus one k one x
plus k naught this will be the choice of y p x provided zero is not a root of the auxiliary
equation if zero is the root of the auxiliary equation what we have to do we shall see next
now in the next slide when r x is k e to the power p x then we shall choose y p x to
be a e to the power p x provided p is not a root of the auxiliary equation like in the
previous example we have seen that r x here here here the m equal to one m equal
m equal to was a root of the auxiliary equation so the choice of a times e to the power x
did not was not a correct choice so so this is an exceptional case so when p is
not root of the auxiliary equation we will make a choice of our pi p x as a e to the
power p x if r x is k cos k q x or k sin q x then the choice of y p x will be taken
as k cos q x plus m sin q x provided i q is not a root of the auxiliary equation now these
are the exceptional cases when zero is a root of the auxiliary equation r p is a root of
the auxiliary equation or i q is the root of auxiliary equation what choice of y p x
we will have to make it it is given in the next slide
so here let us see if r x is a [combi/combination] combination of the functions in column one
if r x is the combination of functions given in the column one then the choice of y p x
should be made by combining the corresponding choices in column two in the case of exceptions
as given in column three the choice function in column two should be multiplied by x to
the power m let us see now we have seen that m equal to one p equal to one was a root of
the auxiliary equation here p equal to one was a root of the auxiliary equation here
so so we had to consider and it occurred once its multiplicity is one so we had to
multiply e to the power x by x we had to assume y p x equal to a times x into e to the power
x so in the case of exceptions as given in column three the choice function column
two should be multiplied by x to the power m where m is the [multisticity/multiplicity]
multiplicity of the root of the auxiliary equation
now let us take the example y double dash plus y dash x equal to r x square let us see
what do we how do we choose r x y p x here so y double dash plus y dash x equal to x
square so auxiliary equation here is m square plus m equal to zero so m equal to zero and
minus one and therefore complimentary function y c x is equal to c one e to the power zero
x plus c two e two the power minus x or we can say c one plus c two e to the power minus
x now let us look at the table zero is a root of the auxiliary equation so this is an exceptional
case so in this case and zero occurs once and therefore we have to multiply y x so the
the y p x because r x is equal to x square here r x is given to be x square here
so natural choice of y p x should have been a x square plus b x plus c but in this exceptional
case we will multiply a x square plus b x plus c y x (Refer Time: 30 :00) because zero
is the root of the auxiliary equation and it occurs once so we have to multiply the
choice a x square plus b x plus c y x with this choice of y p x when you when you
substitute this y p x into the given differential equation what you will have so this x cube
so three a x square plus two b x plus c and then y p double dash x you will get as six
a s plus two b so let us put these values of y p x and the derivatives in the given
differential equation so six a x plus two b which is y three double dash plus y dash
so that means y p dash so three a x square plus two b x plus c equal to x square so what
do we notice is that so we notice that the coefficient of x
be it is an identity it is true for all values of x so we should equate the
corresponding powers of like powers of x so three a equal to one so we get a equal to
one by three then we have to look at the coefficient of x here so six a plus two b
equal to zero because there is no coefficient of x there the it is zero and then we have
two b plus c equal to zero so substituting a equal to one by three here
this will give you six a equal to two two plus two b equal to zero which will give you
b equal to minus one and b equal to minus one then gives you c equal to two so thus
y p x is equal to a a x cube a is equal to one by three so one by three x cube and then
b x square b x square means minus x square and then we have c x so two x and so the general
solution is y x equal to y c x y c x is c one plus c two e to the power minus x this
is y c x plus y p x so this is the general solution of the given differential equation
we can have one more example y double dash x plus nine y dash x equal to cos three x
which will take care of the other exceptional case where i q this exceptional case prob
i i q is not a root of the auxiliary equation so let us discuss now another example y
double dash x plus nine y x equal to cos three x in this case we have the auxiliary equation
as m square plus nine equal to zero so m equal to plus minus three i now this is an exceptional
case because the because in the r x here is cos three x ok and we see that r x is equal
to cos three x and i three i is a root of the auxiliary equation so what we see that
three i occurs once here therefore y p x will be assumed as x times a cos three x plus b
sin three x so if you assume y p x to be equal to this then y p dash x will be equal to a
cos three x plus b sin three x plus x times minus three a sin three x plus three b cos
three x and then y p double dash x will be equal to minus three a sin three x plus three
b cos three x plus minus three a sin three x plus three b cos three x and then x times
minus nine a cos three x minus nine b sin three x
so when you put these values in the given differential equation let us see what we get
so y p double dash now this is minus three sin three x minus three a sin three x so minus
six a sin three x and we get three b cos three b cos three x six b cos three x and then
x times minus nine a cos three x minus nine b sin three x so this is y p double dash plus
nine times y a y p x so y p x is x times a cos three x plus b sin three x equal to cos
three x now nine x into a cos three x will cancel with minus nine a into x cos three
x nine x b sin three x will cancel with minus nine b sin three x so this and this cancel
ok now equating the coefficients of sin three x and cos three x both sides we get a equal
to zero and b equal to one by six and therefore y p x equal to x times a zero b is one by
six so x by six sin three x and thus general solution y x is equal to y c x plus y p x
which is equal to now y c x here is some constant c one cos three x plus c two sin three x so
this will be c one cos three x plus c two sin three x plus x y six sin three x so this
is how we shall solve this equation now in my next lecture we shall [describe/discuss]
discuss the general method for obtaining the general solution of second order no
non homogenous linear differential equation thank you