WEBVTT
Kind: captions
Language: en
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hello friends welcome to my lecture on method
of undetermined coefficients this method we
00:00:25.010 --> 00:00:33.900
we will be used to solve second order linear
differential equations which are non homogenous
00:00:33.900 --> 00:00:40.710
that they are of the form y double dash plus
a y dash plus b y equal to r x now we have
00:00:40.710 --> 00:00:47.620
already solved the homogenous linear second
order differential equations so let us see
00:00:47.620 --> 00:00:53.580
how we shall solve this second order non homogenous
linear differential equation first of all
00:00:53.580 --> 00:01:11.149
we shall show that if y c x is the general
solution of
00:01:11.149 --> 00:01:30.149
of the homogenous equation y double dash plus
a y dash plus b y equal to zero and y p x
00:01:30.149 --> 00:01:43.560
is a particular solution of
the non homogenous equation of the equation
00:01:43.560 --> 00:02:03.009
one
then we shall see that sh show that y x equal
00:02:03.009 --> 00:02:19.700
to y c x plus y p x is the general solution
of
00:02:19.700 --> 00:02:26.390
equation one so y c x is the general solution
of the homogenous equation y double dash plus
00:02:26.390 --> 00:02:31.860
a by dash plus b y equal to zero means y c
will contain two arbitrary constants and we
00:02:31.860 --> 00:02:38.170
know how to get the solution of this homogenous
equation y p x is a particular solution of
00:02:38.170 --> 00:02:43.100
the equation one means y p x is a solution
of the equation one without any arbitrary
00:02:43.100 --> 00:02:51.310
constants so if we can show that y x equal
to y c x plus y p x is a solution of the equation
00:02:51.310 --> 00:02:58.600
one then y x because it contains two arbitrary
constants will be the general solution of
00:02:58.600 --> 00:03:05.230
equation one y x contains two arbitrary constants
because y c x contains two arbitrary constants
00:03:05.230 --> 00:03:10.760
and y p x is a solution of equation one without
any arbitrary constants
00:03:10.760 --> 00:03:30.700
so so if we just prove that y x is a solution
of equation one
00:03:30.700 --> 00:03:50.890
then y x will be the general solution of one
ok so let us see when you substitute y
00:03:50.890 --> 00:03:56.770
x equal to y c x plus y p x if it satisfies
the equation one it will be a solution of
00:03:56.770 --> 00:04:11.980
equation one so substituting
00:04:11.980 --> 00:04:33.670
in the left hand side of
of one we get y p double dash x plus a y p
00:04:33.670 --> 00:04:47.660
dash y double dash y double dash x plus y
dash x plus b y dash b y equal to y c double
00:04:47.660 --> 00:05:02.810
dash x plus y p double dash x plus a times
y dash x means y c dash x plus y p dash x
00:05:02.810 --> 00:05:47.430
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
00:05:47.430 --> 00:05:53.470
plus b y c x now since y c x is a general
solution of y double dash plus a y dash plus
00:05:53.470 --> 00:06:00.460
b y equal to zero so this so this first
expression is zero then we have now y p x
00:06:00.460 --> 00:06:05.820
is a particular solution of the equation one
therefore y p double dash plus a y p dash
00:06:05.820 --> 00:06:17.240
plus b y p is equal to r x and thus we get
y x is a solution of equation one again i
00:06:17.240 --> 00:06:21.930
repeat y x will therefore be the general solution
of equation one
00:06:21.930 --> 00:06:28.639
now so if we are having a non homogenous second
order linear differential equation with constant
00:06:28.639 --> 00:06:36.090
coefficients then we will now discuss a
method by which we can determine this particular
00:06:36.090 --> 00:06:46.120
solution y p x y c x we know how to determine
so our now method will determine the
00:06:46.120 --> 00:06:53.850
value of y p x so the method that we are going
to discuss is called as the method of [undeter/undetermined]
00:06:53.850 --> 00:07:00.560
undetermined coefficients in this method the
by we we make an educated guess for y p x
00:07:00.560 --> 00:07:07.760
the particular solution y p x of the equation
one which we are looking for so we make a
00:07:07.760 --> 00:07:35.990
guess of the expression for y p x which
depends on
00:07:35.990 --> 00:07:51.169
the form of
r x and the complimentary function this complimentary
00:07:51.169 --> 00:07:57.610
function is nothing but the general solution
of the associated homogenous equation complementary
00:07:57.610 --> 00:08:18.960
function is nothing but the general solution
of
00:08:18.960 --> 00:08:28.180
so i can say y c x so we may so this is not
an arbitrary arbitrary s it is a it
00:08:28.180 --> 00:08:35.180
is called an educated guess because it depends
on the form of r x and also sometimes it depends
00:08:35.180 --> 00:08:38.560
on the form of the complimentary function
y c x
00:08:38.560 --> 00:08:47.850
now this expression for y p x will contain
constants which will be determined by
00:08:47.850 --> 00:08:52.880
substituting y p x in the equation one so
[tha/that] that is why we call this method
00:08:52.880 --> 00:09:00.959
as method of undetermined coefficients now
here we have certain detritions this method
00:09:00.959 --> 00:09:07.360
can be applied to the second order linear
linear linear differential equation which
00:09:07.360 --> 00:09:14.369
are non homogenous when r x is of the following
type it can be a polynomial function it can
00:09:14.369 --> 00:09:21.379
be an exponential function it can be a sin
or cosine function or a sum or products of
00:09:21.379 --> 00:09:27.310
such functions now or you can take linear
combination of the functions which are
00:09:27.310 --> 00:09:34.550
polynomials or exponential functions or
trigonometric functions sin x cosine x so
00:09:34.550 --> 00:09:40.379
p with p you can take r x to be function
of the form p x or you can take p x into e
00:09:40.379 --> 00:09:45.809
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
00:09:45.809 --> 00:09:50.819
and beta x so r x is a function of this type
where n is non negative integer and alpha
00:09:50.819 --> 00:09:56.269
and beta are real numbers then we can apply
this method of undetermined coefficients
00:09:56.269 --> 00:10:03.339
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
00:10:03.339 --> 00:10:09.310
this method cannot be applied so for second
order non homogenous linear differential equations
00:10:09.310 --> 00:10:15.740
where r x is of those types log x six
to power minus one ten x sin inverse x etcetera
00:10:15.740 --> 00:10:22.420
then we will apply the general method which
we shall discuss in the next lecture now so
00:10:22.420 --> 00:10:27.990
in the general method we will be taking care
of all the functions r x which cannot
00:10:27.990 --> 00:10:35.380
be handled by the method of undetermined coefficients
now let us see if you take the constants
00:10:35.380 --> 00:10:41.190
polynomials exponentials sines or cosines
the remarkable property of those functions
00:10:41.190 --> 00:10:48.009
is that the derivatives of the sums and products
of such functions are again sums and products
00:10:48.009 --> 00:10:54.939
of similar kind of functions so in order to
assume the form of y p x ok it is reasonable
00:10:54.939 --> 00:11:00.850
to assume the same form as that of r x
now let us take an example say y double dash
00:11:00.850 --> 00:11:11.709
plus four y dash minus two y equal to two
x square minus three x plus six so here we
00:11:11.709 --> 00:11:22.589
can see r x is a polynomial in x of degree
two now first we shall find y c x so to find
00:11:22.589 --> 00:11:27.519
y c x we have the write the auxiliary equation
m square plus four m minus two equal to zero
00:11:27.519 --> 00:11:40.980
we can find the value of m for this
so this will come out to be minus four plus
00:11:40.980 --> 00:11:55.389
minus
so m has two distinct values minus two
00:11:55.389 --> 00:12:02.809
plus root six and minus two minus root six
and therefore y c x is equal to c one e to
00:12:02.809 --> 00:12:11.570
the power minus two plus root six into x plus
c two e to the power minus two minus root
00:12:11.570 --> 00:12:21.350
six into x
now since r x is equal to two x square
00:12:21.350 --> 00:12:29.040
minus three x plus six which is the polynomial
of degree two let us assume y p x y p x
00:12:29.040 --> 00:12:36.429
to be equal to a x square plus b x plus c
a polynomial in x of degree two where a b
00:12:36.429 --> 00:12:43.299
c are undetermined coefficients we shall determine
the values of these undetermined coefficients
00:12:43.299 --> 00:12:48.649
a b c by substituting this y p in to the in
the given differential equation because y
00:12:48.649 --> 00:12:53.860
p is a particular solution of the given differential
equation so it should satisfy that equation
00:12:53.860 --> 00:13:03.329
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
00:13:03.329 --> 00:13:13.819
this value y p x in the given equation y double
dash so let us find y p dash x equal to two
00:13:13.819 --> 00:13:32.660
a x plus b y p double dash is equal to two
a so then y p double dash plus four y p dash
00:13:32.660 --> 00:14:12.319
minus two y p equal to two x square minus
three x plus six gives us two a now this
00:14:12.319 --> 00:14:17.970
equation is valid for all values of x so it
must be an identity and therefore the coefficient
00:14:17.970 --> 00:14:25.209
of like powers of x must be same so if we
equate the coefficients of like powers
00:14:25.209 --> 00:14:32.949
of x we get the values of the unknown coefficients
a b c you can see that here x square
00:14:32.949 --> 00:14:40.449
coefficient x square is minus two a so minus
two a equal to two which gives us a equal
00:14:40.449 --> 00:14:51.600
to minus one when we equate the coefficient
of x on both sides we get eight a minus two
00:14:51.600 --> 00:15:01.579
b equal to minus three putting the value of
a equal to minus one we get minus eight minus
00:15:01.579 --> 00:15:10.249
two b equal to minus three so this will give
you two b equal to minus five which give you
00:15:10.249 --> 00:15:15.329
b equal to minus five by two and when you
equate the coefficients on both sides we get
00:15:15.329 --> 00:15:26.269
two a plus four b minus two c equal to six
so substituting the values of a and b minus
00:15:26.269 --> 00:15:47.319
one and minus five by two we shall have
we shall have c equal to minus nine and thus
00:15:47.319 --> 00:15:54.519
y p x will be equal to x square minus five
by two x minus nine
00:15:54.519 --> 00:16:00.070
so the general solution of this differential
equation is the sum of the complimentary
00:16:00.070 --> 00:16:04.990
function c one e to the power minus two plus
root six into x c two e to the power minus
00:16:04.990 --> 00:16:14.980
two minus root six into x hm and then minus
x square minus five by two x minus nine now
00:16:14.980 --> 00:16:23.160
let us take another example y double dash
minus y dash plus y equal to sin three x here
00:16:23.160 --> 00:16:28.010
the auxiliary equation is m square minus m
plus one equal to zero you can find the values
00:16:28.010 --> 00:16:33.850
of m they are one plus minus i root three
by two so we can write the complimentary function
00:16:33.850 --> 00:16:38.939
because these are complex roots so e to the
power alpha x that is e to the power half
00:16:38.939 --> 00:16:44.480
x into c one cos theta x beta is root three
by two so c one cos root three by two x plus
00:16:44.480 --> 00:16:52.220
c two sin root three by two x
now let us make an guess for y p x so the
00:16:52.220 --> 00:16:58.919
the natural guess when you look at the form
of r x r x is equal to sin three x so natural
00:16:58.919 --> 00:17:06.140
guess for the y p x would be that
we write a times sin three x but then when
00:17:06.140 --> 00:17:14.360
you differentiate sin three x it produces
three times cos three x and then when you
00:17:14.360 --> 00:17:18.329
differentiated again you get sin three x so
it differentiation of sin three x produces
00:17:18.329 --> 00:17:23.980
both sin three x and cos three x and therefore
what we should do is we should assume y p
00:17:23.980 --> 00:17:30.170
x to be a linear combination of sin three
x and cos three x functions so let us assume
00:17:30.170 --> 00:17:34.531
y p x to be having both the functions sin
three x and cos three x that is a cos three
00:17:34.531 --> 00:17:38.960
x plus b sin three x
now then you substitute this y p x in the
00:17:38.960 --> 00:17:46.780
given differential equation the given differential
equation is y double dash minus y dash plus
00:17:46.780 --> 00:17:57.890
y equal to sin three x so if you take y p
x equal to a cos three x plus b sin three
00:17:57.890 --> 00:18:06.490
x and then put it in the given differential
equation what you will get is unequating
00:18:06.490 --> 00:18:11.930
the coefficients of sin three x and cos three
x both sides we shall get a to be six by seventy
00:18:11.930 --> 00:18:19.280
three and b to be minus sixteen by seventy
three so we can find y p dash x here this
00:18:19.280 --> 00:18:32.080
is minus three a sin three x plus three b
cos three x we differentiate it again should
00:18:32.080 --> 00:18:44.470
will give you minus nine a cos three x plus
nine b minus nine b sin three x now substitute
00:18:44.470 --> 00:18:50.910
the values of y p y p dash y p double dash
in this differential equation and equate the
00:18:50.910 --> 00:18:55.630
coefficient of sin three x and cos three x
both sides we shall get a equal to six by
00:18:55.630 --> 00:19:05.170
seventy three and b equal to minus sixteen
by seventy three so we will get y p x to be
00:19:05.170 --> 00:19:12.810
equal to minus six by seventy three cos x
minus sixteen by seventy three sin three x
00:19:12.810 --> 00:19:18.180
and thus the general solution y x is the
sum of the complementary function this is
00:19:18.180 --> 00:19:24.171
the complimentary function plus the particular
integral the particular solution of the
00:19:24.171 --> 00:19:27.970
equation one is also called as particular
integral so six by seventy three cos three
00:19:27.970 --> 00:19:30.530
x minus sixteen by seventy threes sin three
x
00:19:30.530 --> 00:19:39.010
now now let us look at the this comment here
the as i said in the beginning y p x is
00:19:39.010 --> 00:19:44.670
not an arbitrary it is not guessed arbitraly
it is an educated guess because it depends
00:19:44.670 --> 00:19:50.030
on the form of r x the functions that make
up r x as well as the functions that make
00:19:50.030 --> 00:19:55.710
up the complimentary function which is
the solution of the associated homogenous
00:19:55.710 --> 00:20:01.080
linear differential equation that is y double
dash plus a y dash plus b y equal to zero
00:20:01.080 --> 00:20:06.341
let us see how it depends on the complimentary
function we have seen that the function
00:20:06.341 --> 00:20:11.910
y p x depends on the form of r x in the next
example we shall see how y p x depends on
00:20:11.910 --> 00:20:19.360
the form of the complimentary function
so let us take up this this problem y double
00:20:19.360 --> 00:20:25.920
dash minus five y dash plus four y equal to
a t to the power x here we can write the auxiliary
00:20:25.920 --> 00:20:30.060
equation as m square minus four five m plus
four equal to zero so we get two values of
00:20:30.060 --> 00:20:34.480
m which are four and one they are distinct
values so the complementary function we can
00:20:34.480 --> 00:20:39.200
write easily y c x equal to c one e to the
power x plus c two e to the power four x
00:20:39.200 --> 00:20:45.970
now when we look at the form of r x r x is
eight e to the power x and successive differentiation
00:20:45.970 --> 00:20:51.680
of e to the power x produce e to the power
x so we can assume y p x to be equal to
00:20:51.680 --> 00:20:56.980
a times e to the power x but when you put
y p x equal to eight e to the power x in this
00:20:56.980 --> 00:21:03.700
equation in the given differential equation
y double dash minus five y dash plus four
00:21:03.700 --> 00:21:11.610
y equal to eight e to the power x let us see
what happens y we are taking as a e to the
00:21:11.610 --> 00:21:18.550
power x so let us find y dash it is a e to
the power x let us find y double dash it is
00:21:18.550 --> 00:21:27.930
again a to the power x so y so a e to the
power x minus five a e to the power x plus
00:21:27.930 --> 00:21:36.850
four a e to the power x equal to eight e to
the power x now this is five a e to the power
00:21:36.850 --> 00:21:44.470
x minus five a to the power x so this is zero
equal to eight e to the power x which is
00:21:44.470 --> 00:21:48.730
not correct of course because e to the power
x is never zero so thus we have [mad/made]
00:21:48.730 --> 00:21:53.360
made a wrong guess for y p x so what should
be the form of y p x
00:21:53.360 --> 00:21:59.860
now let us look at the complimentary function
the complimentary function is c one e to the
00:21:59.860 --> 00:22:07.430
power x plus c two e to the power four x and
therefore e to the power x and e to the power
00:22:07.430 --> 00:22:13.150
four x are both solutions of the associated
homogenous linear differential equation of
00:22:13.150 --> 00:22:18.500
second order that is y double dash minus five
y dash plus four y equal to zero and so a
00:22:18.500 --> 00:22:23.770
e to the power x one substituted in the given
differential equation produces zero on the
00:22:23.770 --> 00:22:29.570
left hand side now what should therefore be
a form of y c x this is the question so let
00:22:29.570 --> 00:22:34.540
us recall the case of repeated v l rules of
the homogenous linear differential equation
00:22:34.540 --> 00:22:40.210
with constant coefficients there we see that
when the two roots are eq two roots are
00:22:40.210 --> 00:22:50.310
same one if one m value of m is equal to
m one which is repeated then the two solutions
00:22:50.310 --> 00:22:56.851
independent solutions of the homogenous
equations are homogenous equations are e to
00:22:56.851 --> 00:23:09.700
the power m one x and x times e to the power
m one x so let us see whether if we assume
00:23:09.700 --> 00:23:19.060
y p x equal to a times x e to the power x
do we get the solution of the given differential
00:23:19.060 --> 00:23:23.470
equation
so let us see if we do this then y p dash
00:23:23.470 --> 00:23:45.640
x will be equal to a times
this y p double dash will be equal to a e
00:23:45.640 --> 00:23:50.690
x times x e to the power x plus two e to the
power x now when you put this in the given
00:23:50.690 --> 00:24:01.620
differential equation let us see what do we
get so a times x e to the power x plus two
00:24:01.620 --> 00:24:10.230
e to the power x that is y double y p double
dash minus five y p dash y p dash means a
00:24:10.230 --> 00:24:20.500
into x e to the power x plus e to the power
x and then we have four times y which is a
00:24:20.500 --> 00:24:28.940
x e to the power x equal to eight e to the
power x now let us see what do we get so a
00:24:28.940 --> 00:24:34.630
times x e to the power x four a times x e
to the power x minus five a times x e to the
00:24:34.630 --> 00:24:42.750
power x they cancel and what we get is
two a e to the power x two a e to the power
00:24:42.750 --> 00:24:57.060
x minus five a e to the power x equal to eight
e to the power x so we get minus three
00:24:57.060 --> 00:25:07.830
a e to the power x is never zero for any x
so we we can divide by e to the power x and
00:25:07.830 --> 00:25:19.190
we get the value of a x minus eight y three
and thus y p x we get as minus eight y three
00:25:19.190 --> 00:25:27.500
x e to the power x so the general solution
therefore is given by c one y equal to c one
00:25:27.500 --> 00:25:31.290
e to the power e x plus c two e to the power
four x minus eight by three x e to the power
00:25:31.290 --> 00:25:36.280
x
in this slide we have listed the choice
00:25:36.280 --> 00:25:43.220
of y p x depending on the choice of r x so
if in the first column we have given the form
00:25:43.220 --> 00:25:51.500
of r x in the second column we have given
the choice of y p x provided we have
00:25:51.500 --> 00:25:59.440
these exceptions so suppose r x is of the
form k x to the power n then where k is a
00:25:59.440 --> 00:26:09.340
constant then we shall make the choice
of y p x to be a polynomial in x of degree
00:26:09.340 --> 00:26:14.220
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
00:26:14.220 --> 00:26:20.220
plus k naught this will be the choice of y
p x provided zero is not a root of the auxiliary
00:26:20.220 --> 00:26:25.180
equation if zero is the root of the auxiliary
equation what we have to do we shall see next
00:26:25.180 --> 00:26:32.400
now in the next slide when r x is k e to
the power p x then we shall choose y p x to
00:26:32.400 --> 00:26:38.200
be a e to the power p x provided p is not
a root of the auxiliary equation like in the
00:26:38.200 --> 00:26:48.480
previous example we have seen that r
x here here here the m equal to one m equal
00:26:48.480 --> 00:26:55.210
m equal to was a root of the auxiliary equation
so the choice of a times e to the power x
00:26:55.210 --> 00:27:03.360
did not was not a correct choice so
so this is an exceptional case so when p is
00:27:03.360 --> 00:27:08.090
not root of the auxiliary equation we will
make a choice of our pi p x as a e to the
00:27:08.090 --> 00:27:15.380
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
00:27:15.380 --> 00:27:22.810
as k cos q x plus m sin q x provided i q is
not a root of the auxiliary equation now these
00:27:22.810 --> 00:27:28.290
are the exceptional cases when zero is a root
of the auxiliary equation r p is a root of
00:27:28.290 --> 00:27:32.490
the auxiliary equation or i q is the root
of auxiliary equation what choice of y p x
00:27:32.490 --> 00:27:36.080
we will have to make it it is given in the
next slide
00:27:36.080 --> 00:27:41.520
so here let us see if r x is a [combi/combination]
combination of the functions in column one
00:27:41.520 --> 00:27:48.650
if r x is the combination of functions given
in the column one then the choice of y p x
00:27:48.650 --> 00:27:54.240
should be made by combining the corresponding
choices in column two in the case of exceptions
00:27:54.240 --> 00:28:00.321
as given in column three the choice function
in column two should be multiplied by x to
00:28:00.321 --> 00:28:07.309
the power m let us see now we have seen that
m equal to one p equal to one was a root of
00:28:07.309 --> 00:28:12.380
the auxiliary equation here p equal to one
was a root of the auxiliary equation here
00:28:12.380 --> 00:28:19.760
so so we had to consider and it occurred
once its multiplicity is one so we had to
00:28:19.760 --> 00:28:27.120
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
00:28:27.120 --> 00:28:32.400
x so in the case of exceptions as given
in column three the choice function column
00:28:32.400 --> 00:28:36.180
two should be multiplied by x to the power
m where m is the [multisticity/multiplicity]
00:28:36.180 --> 00:28:39.780
multiplicity of the root of the auxiliary
equation
00:28:39.780 --> 00:28:46.059
now let us take the example y double dash
plus y dash x equal to r x square let us see
00:28:46.059 --> 00:28:56.070
what do we how do we choose r x y p x here
so y double dash plus y dash x equal to x
00:28:56.070 --> 00:29:04.390
square so auxiliary equation here is m square
plus m equal to zero so m equal to zero and
00:29:04.390 --> 00:29:12.920
minus one and therefore complimentary function
y c x is equal to c one e to the power zero
00:29:12.920 --> 00:29:21.020
x plus c two e two the power minus x or we
can say c one plus c two e to the power minus
00:29:21.020 --> 00:29:28.860
x now let us look at the table zero is a root
of the auxiliary equation so this is an exceptional
00:29:28.860 --> 00:29:36.440
case so in this case and zero occurs once
and therefore we have to multiply y x so the
00:29:36.440 --> 00:29:45.000
the y p x because r x is equal to x
square here r x is given to be x square here
00:29:45.000 --> 00:29:52.240
so natural choice of y p x should have been
a x square plus b x plus c but in this exceptional
00:29:52.240 --> 00:30:02.830
case we will multiply a x square plus b x
plus c y x (Refer Time: 30 :00) because zero
00:30:02.830 --> 00:30:07.920
is the root of the auxiliary equation and
it occurs once so we have to multiply the
00:30:07.920 --> 00:30:15.480
choice a x square plus b x plus c y x with
this choice of y p x when you when you
00:30:15.480 --> 00:30:23.540
substitute this y p x into the given differential
equation what you will have so this x cube
00:30:23.540 --> 00:30:34.010
so three a x square plus two b x plus c and
then y p double dash x you will get as six
00:30:34.010 --> 00:30:41.250
a s plus two b so let us put these values
of y p x and the derivatives in the given
00:30:41.250 --> 00:30:52.190
differential equation so six a x plus two
b which is y three double dash plus y dash
00:30:52.190 --> 00:31:03.620
so that means y p dash so three a x square
plus two b x plus c equal to x square so what
00:31:03.620 --> 00:31:12.640
do we notice is that
so we notice that the coefficient of x
00:31:12.640 --> 00:31:17.350
be it is an identity it is true for
all values of x so we should equate the
00:31:17.350 --> 00:31:23.140
corresponding powers of like powers of x so
three a equal to one so we get a equal to
00:31:23.140 --> 00:31:32.750
one by three then we have to look at the
coefficient of x here so six a plus two b
00:31:32.750 --> 00:31:37.450
equal to zero because there is no coefficient
of x there the it is zero and then we have
00:31:37.450 --> 00:31:44.770
two b plus c equal to zero
so substituting a equal to one by three here
00:31:44.770 --> 00:31:52.590
this will give you six a equal to two two
plus two b equal to zero which will give you
00:31:52.590 --> 00:32:01.160
b equal to minus one and b equal to minus
one then gives you c equal to two so thus
00:32:01.160 --> 00:32:13.090
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
00:32:13.090 --> 00:32:21.830
b x square b x square means minus x square
and then we have c x so two x and so the general
00:32:21.830 --> 00:32:30.900
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
00:32:30.900 --> 00:32:42.370
is y c x plus y p x so this is the general
solution of the given differential equation
00:32:42.370 --> 00:32:50.050
we can have one more example y double dash
x plus nine y dash x equal to cos three x
00:32:50.050 --> 00:32:58.590
which will take care of the other exceptional
case where i q this exceptional case prob
00:32:58.590 --> 00:33:05.720
i i q is not a root of the auxiliary equation
so let us discuss now another example y
00:33:05.720 --> 00:33:14.770
double dash x plus nine y x equal to cos three
x in this case we have the auxiliary equation
00:33:14.770 --> 00:33:25.990
as m square plus nine equal to zero so m equal
to plus minus three i now this is an exceptional
00:33:25.990 --> 00:33:33.710
case because the because in the r x here is
cos three x ok and we see that r x is equal
00:33:33.710 --> 00:33:40.280
to cos three x and i three i is a root of
the auxiliary equation so what we see that
00:33:40.280 --> 00:33:49.940
three i occurs once here therefore y p x will
be assumed as x times a cos three x plus b
00:33:49.940 --> 00:34:01.330
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
00:34:01.330 --> 00:34:16.849
cos three x plus b sin three x plus x times
minus three a sin three x plus three b cos
00:34:16.849 --> 00:34:29.190
three x and then y p double dash x will be
equal to minus three a sin three x plus three
00:34:29.190 --> 00:34:50.739
b cos three x plus minus three a sin three
x plus three b cos three x and then x times
00:34:50.739 --> 00:35:00.499
minus nine a cos three x minus nine b sin
three x
00:35:00.499 --> 00:35:06.960
so when you put these values in the given
differential equation let us see what we get
00:35:06.960 --> 00:35:12.220
so y p double dash now this is minus three
sin three x minus three a sin three x so minus
00:35:12.220 --> 00:35:24.480
six a sin three x and we get three b cos three
b cos three x six b cos three x and then
00:35:24.480 --> 00:35:39.269
x times minus nine a cos three x minus nine
b sin three x so this is y p double dash plus
00:35:39.269 --> 00:35:52.839
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
00:35:52.839 --> 00:36:00.930
three x now nine x into a cos three x will
cancel with minus nine a into x cos three
00:36:00.930 --> 00:36:07.489
x nine x b sin three x will cancel with minus
nine b sin three x so this and this cancel
00:36:07.489 --> 00:36:12.930
ok now equating the coefficients of sin three
x and cos three x both sides we get a equal
00:36:12.930 --> 00:36:29.160
to zero and b equal to one by six and therefore
y p x equal to x times a zero b is one by
00:36:29.160 --> 00:36:42.249
six so x by six sin three x and thus general
solution y x is equal to y c x plus y p x
00:36:42.249 --> 00:36:56.809
which is equal to now y c x here is some constant
c one cos three x plus c two sin three x so
00:36:56.809 --> 00:37:10.660
this will be c one cos three x plus c two
sin three x plus x y six sin three x so this
00:37:10.660 --> 00:37:14.640
is how we shall solve this equation now in
my next lecture we shall [describe/discuss]
00:37:14.640 --> 00:37:21.410
discuss the general method for obtaining
the general solution of second order no
00:37:21.410 --> 00:37:25.380
non homogenous linear differential equation
thank you