WEBVTT
Kind: captions
Language: en
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hello friends welcome to my second lecture
on methods for finding particular integral
00:00:22.930 --> 00:00:29.180
for second order linear differential equations
with constant coefficients in my last lecture
00:00:29.180 --> 00:00:46.120
we were trying to find the particular integral
in the case of the differential equation
00:00:46.120 --> 00:00:53.570
d square plus one y equal to cos square one
by two x and we couldnt find it because we
00:00:53.570 --> 00:01:08.320
have to write cos square half x in the form
of one plus cos x by two and when we write
00:01:08.320 --> 00:01:26.200
it in this form what happens is that
00:01:26.200 --> 00:01:32.720
we have to find the particular integral for
one by d square plus one cos x but one by
00:01:32.720 --> 00:01:38.320
d square plus one cos x cannot be found from
the formula one by d square plus a square
00:01:38.320 --> 00:01:55.820
cos a x equal to one over sorry one over
f d square equal to write one over phi d square
00:01:55.820 --> 00:02:10.299
equal to cos a x equal to one over phi minus
a square cos a x we cannot find the of
00:02:10.299 --> 00:02:15.560
particular integral in this case because when
we replace d square by minus a square that
00:02:15.560 --> 00:02:22.400
is d square by minus one square r minus one
this d square plus one on replacing d square
00:02:22.400 --> 00:02:29.560
minus a square becomes zero so phi minus a
square should not be zero to apply this formula
00:02:29.560 --> 00:02:35.540
and therefore let us now discuss how to find
the particular integral in the case of
00:02:35.540 --> 00:02:40.329
one over d square plus c square cos c x
00:02:40.329 --> 00:02:45.560
so let us discuss how to find the particular
integral in the case of one over f d cos c
00:02:45.560 --> 00:02:51.799
x or one over f d sin c x which are where
in the exceptional case we have f d equal
00:02:51.799 --> 00:02:59.190
to d square plus c square and we see that
phi d square when c square d square is replaced
00:02:59.190 --> 00:03:03.980
by minus c square becomes zero so we have
not able to apply the formula one over phi
00:03:03.980 --> 00:03:10.340
d square cos cx equal to one over phi minus
c square cos c x we will prove that one
00:03:10.340 --> 00:03:16.070
over phi d square cos c x which is one over
d square plus c square cos c x gives us x
00:03:16.070 --> 00:03:21.769
sin c x over two c ok
so let us see how we do this we can write
00:03:21.769 --> 00:03:27.889
one over d square plus c square as one
over two i c one over d minus i c minus one
00:03:27.889 --> 00:03:33.320
over d plus i c operating on cos c x that
is we break one over d square plus c square
00:03:33.320 --> 00:03:41.030
in partial fractions now let us recall that
we had proved that one over d minus alpha
00:03:41.030 --> 00:03:50.650
when operates on q or you can say r x then
we had e to the power minus alpha x integral
00:03:50.650 --> 00:03:58.799
e to the power e to the power alpha x e to
the power minus alpha x into q d x
00:03:58.799 --> 00:04:04.440
so let us apply this formula here and to by
using this formula one over d minus i c cos
00:04:04.440 --> 00:04:10.479
c x will be e to the power i c x integral
e to the power minus i c x cos c x d x now
00:04:10.479 --> 00:04:30.590
then what we will have is e to the power
minus i c x cos c x we can write as e to the
00:04:30.590 --> 00:04:40.310
power i c x plus e to the power minus i c
x divided by two by the eulers formula we
00:04:40.310 --> 00:04:43.639
know that [e/ it] for i theta is cos theta
plus i sin theta and e to the power minus
00:04:43.639 --> 00:04:48.230
i theta is cos theta minus i sin theta so
from there cos theta is e to the power i theta
00:04:48.230 --> 00:04:54.520
plus e to the power minus i theta by two
now this is equal to e to the power i c x
00:04:54.520 --> 00:05:13.319
integral this will be one plus e to the power
minus two i c x d x ok let us integrate it
00:05:13.319 --> 00:05:22.500
and we get then integral of one is x and then
e to the power minus two i c x divided by
00:05:22.500 --> 00:05:30.030
two i c
similarly we can write the value of one over
00:05:30.030 --> 00:06:09.889
d plus i c cos c x
this is equal to
00:06:09.889 --> 00:06:17.289
we get this now let us subtract from the
value of one over d minus i c cos c x the
00:06:17.289 --> 00:06:25.490
value of one over d plus i c cos c x and then
divide by two i c ok hence one over d square
00:06:25.490 --> 00:06:38.819
plus c square cos c x is equal to one over
two i c from one over d minus i c cos c x
00:06:38.819 --> 00:06:43.410
we subtract the value of one over d plus i
c cos c x and so this will give you x times
00:06:43.410 --> 00:06:53.599
cos e to the power i c x minus e to the power
minus i c x divided by two and when we then
00:06:53.599 --> 00:07:03.169
we will have minus sin here two two becomes
four i c and we get e to the power minus i
00:07:03.169 --> 00:07:14.259
c x here we get e to the power plus i c x
so e to the power i c x plus e to the power
00:07:14.259 --> 00:07:24.599
minus i c x and this is equal to one over
two i c x times now e to the power i theta
00:07:24.599 --> 00:07:33.289
minus e to the power minus i theta by two
is i sin theta so we get i times sin c x and
00:07:33.289 --> 00:07:38.729
then here we have minus one over four i c
e to the power i theta plus e to the power
00:07:38.729 --> 00:07:57.560
minus i theta is two cos theta so two cos
c x and thus we have x sin c x by two c
00:07:57.560 --> 00:08:07.999
and we have i this two will cancel with that
two here so here cos c x divided by four c
00:08:07.999 --> 00:08:22.699
square now now our differential equation
is of the type our differential equation is
00:08:22.699 --> 00:08:36.070
d square plus a square y equal to cos c x
d square plus c square y equal to cos c x
00:08:36.070 --> 00:08:50.920
so auxiliary equation will be m square plus
c square equal to zero so complementary function
00:08:50.920 --> 00:09:04.199
y h x will be equal to a cos c x plus b sin
c x
00:09:04.199 --> 00:09:09.100
and then when we will operate now we will
operate one over d c square on cos c x to
00:09:09.100 --> 00:09:15.449
find the particular integral we get x sin
c x over two c plus cos c x over four c square
00:09:15.449 --> 00:09:22.100
so this term cos c x over four c square
can be observed in the term a cos c x occurring
00:09:22.100 --> 00:09:27.680
in the complementary function and therefore
when we write the result for one over d
00:09:27.680 --> 00:09:37.579
square plus c square cos c x we just retain
x sin c x by two c so we say that since
00:09:37.579 --> 00:09:57.090
of cos c x over four c square can be observed
in y h x we have one over d square plus c
00:09:57.090 --> 00:10:12.089
square cos c x equal to x sin c x by two c
and similarly we can prove similarly we
00:10:12.089 --> 00:10:18.199
can show one over d square plus c square sin
c x equal to minus x by two a cos c x
00:10:18.199 --> 00:10:24.390
now with these formulas we can then find the
particular integral easily in the case where
00:10:24.390 --> 00:10:29.980
phi minus a square becomes zero for example
one o d let us look at the equation d square
00:10:29.980 --> 00:10:39.790
plus one y equal to cos square one by two
x so particular integral y p x is equal to
00:10:39.790 --> 00:10:45.550
one over d square plus one operating on cos
square x one by two which we can write as
00:10:45.550 --> 00:10:54.580
one plus cos x by two which is equal to one
by two one over d square plus one operating
00:10:54.580 --> 00:11:05.220
on one and then we have one over d square
plus one operating on cos x one can be regarded
00:11:05.220 --> 00:11:12.100
as e to the power zero x so this is one over
two d over d is replaced by a zero so we have
00:11:12.100 --> 00:11:20.140
one and then one over d square plus c square
cos c x gives x sin c x over two c so c
00:11:20.140 --> 00:11:28.760
is equal to one here so we have x sin x divided
by two by using the formula which we have
00:11:28.760 --> 00:11:33.670
just proved and similarly we can solve
example two
00:11:33.670 --> 00:11:38.339
in the case of example two when we find
the particular integrals sin square x by two
00:11:38.339 --> 00:11:45.430
is written as one minus cos x so over two
and then we apply the formula same formula
00:11:45.430 --> 00:11:50.839
one over d square plus c square cos c x equal
to x sin x by two c to obtain the particular
00:11:50.839 --> 00:11:53.990
integral
now let us look at another formula which is
00:11:53.990 --> 00:12:05.230
given by one over f d e to the power a x into
v where v real function of x and a is a
00:12:05.230 --> 00:12:11.360
is some constant so this when one over
f d operates on e to the power a x into v
00:12:11.360 --> 00:12:20.440
what we get is e to the power a x into one
upon f d plus a operating on v so this
00:12:20.440 --> 00:12:25.759
formula is very useful in the case earlier
we have considered one over f d operating
00:12:25.759 --> 00:12:34.231
on e to the power alpha x this formula gives
us one over f alpha e to the power alpha x
00:12:34.231 --> 00:12:42.970
provided f alpha is non zero but if if alpha
if f alpha becomes zero then we cannot apply
00:12:42.970 --> 00:12:48.120
this formula so the case where this f alpha
become zero can be handled by the formula
00:12:48.120 --> 00:12:55.540
which we are now going to prove
so let us see how we prove this result
00:12:55.540 --> 00:13:01.380
when you find the derivative of the function
e to the power a x into v you get e to the
00:13:01.380 --> 00:13:07.709
power a x into derivative of v plus a times
e to the power a x derive[ative] into v and
00:13:07.709 --> 00:13:15.149
this can be expressed as e to the power a
x d plus a operating on v and then you again
00:13:15.149 --> 00:13:20.329
differentiate that is you find the second
derivative of e to the power a x into v what
00:13:20.329 --> 00:13:38.710
you get is if you find d square e to the
power a x into v then this is d of and derivate
00:13:38.710 --> 00:13:49.170
of e to the power a x into v we have found
as e to the power a x d plus a operating
00:13:49.170 --> 00:13:59.199
on v let us operate by d so derivate of e
to the power a x is a times e to the power
00:13:59.199 --> 00:14:12.089
a x
then d plus a
00:14:12.089 --> 00:14:24.839
operating on v then e to the power a x d operating
on d plus a v now this can be expressed as
00:14:24.839 --> 00:14:37.569
e to the power a x then we have a d v a square
d so we can write it as d plus a whole square
00:14:37.569 --> 00:14:45.490
v d d is d square v and then a d v and then
a d v becomes two a d v and then we have a
00:14:45.490 --> 00:14:52.279
square v so a to the power a x d plus a whole
square we get and thus we if we apply f d
00:14:52.279 --> 00:15:01.449
operator if we apply suppose f d is equal
to d square plus alpha d plus beta ok
00:15:01.449 --> 00:15:13.620
let f d v equal to d square plus alpha d plus
beta then f d operating on
00:15:13.620 --> 00:15:21.620
e to the power a x into b will give you d
square plus alpha d plus beta operating on
00:15:21.620 --> 00:15:32.649
this and so what you will get actually d square
plus alpha d plus beta operating on e to the
00:15:32.649 --> 00:15:44.769
power a x into v so this will give you e to
the power a x e to the power a x d plus
00:15:44.769 --> 00:16:04.340
a whole square operating on v then alpha times
d plus a e to the power a x e to the power
00:16:04.340 --> 00:16:14.860
a x d plus a v and then e to the power a x
beta b which is equal to e to the power a
00:16:14.860 --> 00:16:23.459
x f d plus a operating on v so this is
what we get
00:16:23.459 --> 00:16:29.759
and so thus ultimately we get f d operating
on e to the power a x into v it is equal to
00:16:29.759 --> 00:16:35.740
e to the power a x f d plus a into v so this
suggests us this formula one over f d operating
00:16:35.740 --> 00:16:40.300
on e to the power a x into v equal to e to
the power a x one over f d plus a into v now
00:16:40.300 --> 00:16:57.910
let us see how we get this so [vocalised-noise]
we we have to prove one over f d e to the
00:16:57.910 --> 00:17:14.250
power a x into v equal to e to the power a
x one over f d plus a operating on v now
00:17:14.250 --> 00:17:26.100
what we do is let us operate by f d on both
sides so operating in by f d
00:17:26.100 --> 00:17:31.480
f d in one over f d are inverse operators
so we get e to the power a x into v equal
00:17:31.480 --> 00:17:46.220
to f d operating on e to the power a x one
over f d plus a v now let us look at the
00:17:46.220 --> 00:17:58.980
right hand side
00:17:58.980 --> 00:18:04.030
so let us assume that one over f d plus a
operating on v is equal to v one ok let us
00:18:04.030 --> 00:18:16.630
assume that one over f d plus a operating
on v is equal to v one so we shall get f d
00:18:16.630 --> 00:18:24.710
right hand side right hand side here will
be equal to f d operating on e to the power
00:18:24.710 --> 00:18:33.250
a x into v one and when f d operates on e
to the power a x into v what we had got
00:18:33.250 --> 00:18:43.840
earlier that if this is e to the power a x
f d plus a v one we have proved this earlier
00:18:43.840 --> 00:18:50.850
now this implies that f d plus a when you
operate on both sides you get f d plus a v
00:18:50.850 --> 00:18:59.240
one equal to v so this will be equal to e
to the power a x into v and this is what we
00:18:59.240 --> 00:19:04.990
had to prove this is the left hand side so
left right hand side left hand side are same
00:19:04.990 --> 00:19:11.030
and therefor the formula one over f d e
to the power a x into v equal to e to the
00:19:11.030 --> 00:19:14.520
power a x one over f d plus a operating on
v is true
00:19:14.520 --> 00:19:21.560
now let us apply this formula to this eq[uation]
to solve this equation d square plus four
00:19:21.560 --> 00:19:29.840
d plus four which we can write as d plus two
whole square y equal to e to the power two
00:19:29.840 --> 00:19:36.070
x minus e to the power minus two x i will
just find particular integral here complementary
00:19:36.070 --> 00:19:43.670
function is easy to find so this is one over
d plus two whole square e to the power two
00:19:43.670 --> 00:19:52.150
x minus e to the power minus two x now this
is equal to when one over d plus two operator
00:19:52.150 --> 00:19:57.450
a square operates on e to the power two x
we get two plus two four square so we get
00:19:57.450 --> 00:20:02.250
one by sixteen into e to the power two x minus
Refer Time: 20:00) now when one over d plus
00:20:02.250 --> 00:20:07.950
two whole square operates on e to the power
minus two x on replacing d y minus to d
00:20:07.950 --> 00:20:13.790
plus two become zero so we will apply this
formula so we will let us see how we get this
00:20:13.790 --> 00:20:18.323
one over d plus two whole square v operates
separately on e to the power let us operate
00:20:18.323 --> 00:20:24.650
on e to the power minus two x so this is one
over d plus two one over d plus two operating
00:20:24.650 --> 00:20:34.970
on e to the power minus two x we can we
can assume here one so one over d plus two
00:20:34.970 --> 00:20:39.910
operating on e to the power minus two x into
one so v we are taking as one so this is equal
00:20:39.910 --> 00:20:52.070
to one over d plus two e to the power minus
two x d will be replaced by d minus two and
00:20:52.070 --> 00:21:01.290
this cancels we get one over d plus two e
to the power minus two x one over d is d means
00:21:01.290 --> 00:21:05.810
derivative one over d means inverse of the
derivative that is integral integral of one
00:21:05.810 --> 00:21:10.770
is x
now again we will apply the the formula which
00:21:10.770 --> 00:21:16.050
we have just proved two two e to the power
minus two x into x let us take v equal to
00:21:16.050 --> 00:21:21.840
two x so then we will have e to the power
minus two x d will be replaced by d minus
00:21:21.840 --> 00:21:31.180
two now one over d is again integral of
we have to make integral of x so that is x
00:21:31.180 --> 00:21:39.050
square by two o x square by two e to the power
minus two x and thus the particular integral
00:21:39.050 --> 00:21:48.310
is so this is how we can make use of this
formula in the case where under placing
00:21:48.310 --> 00:21:56.820
d by a in the expression one by f d e to the
power a x f a becomes zero now let us
00:21:56.820 --> 00:22:03.000
find another result
so now we discuss how to find the particular
00:22:03.000 --> 00:22:10.300
integral in case r x is of we form x to
the power m where m is a positive integer
00:22:10.300 --> 00:22:20.080
so when when we have to find but particular
integral of this type where m is positive
00:22:20.080 --> 00:22:24.660
integer what we do is in order to evaluate
one over f d x to the power m we extend f
00:22:24.660 --> 00:22:33.350
d in ascending powers of d and then operate
on x to the power m since the derivative
00:22:33.350 --> 00:22:40.040
m plus one of the derivative of x to the power
m is zero an[d] all derivatives more than
00:22:40.040 --> 00:22:45.610
one plus are also zero while writing the
expression in the ascending powers of d we
00:22:45.610 --> 00:22:52.290
need to write the derivate of the powers
of v only up to m rest of the terms will
00:22:52.290 --> 00:22:59.160
be zero so we can evaluate particular
integral by writing in the ascending powers
00:22:59.160 --> 00:23:07.750
of d
let us see we how apply this formula suppose
00:23:07.750 --> 00:23:17.720
we have the equation d square minus d minus
six y equal to one plus x square so this is
00:23:17.720 --> 00:23:25.470
let say lets find find particular integral
here so one over d square minus d minus six
00:23:25.470 --> 00:23:31.780
operating on one plus x square one can be
taken it again by writing e to the power zero
00:23:31.780 --> 00:23:39.120
x so one over d square d square minus d minus
six e to the power zero x and then one over
00:23:39.120 --> 00:23:46.970
d square minus d minus six x square so this
is give by replacing d by zero we get minus
00:23:46.970 --> 00:23:54.480
one over six and this is ok and one over d
square minus d minus six when operates on
00:23:54.480 --> 00:24:00.230
x to the power two we have to expand it in
ascending powers of d so that we can do like
00:24:00.230 --> 00:24:14.590
this one over minus one over six one plus
d minus d square divided by six raise to the
00:24:14.590 --> 00:24:21.170
power minus one
now we will expand it like the expand one
00:24:21.170 --> 00:24:35.400
plus x to the power minus one so this one
will be
00:24:35.400 --> 00:24:46.590
one plus x to the power minus one is one minus
x plus x square
00:24:46.590 --> 00:24:52.570
minus x cube now x cube i am not writing because
x cube in the x cube when you do d minus d
00:24:52.570 --> 00:24:57.660
square to the by six to the power cube will
minimum power of d will be three but we have
00:24:57.660 --> 00:25:02.170
x square here third derivative x square is
zero so we need not write those terms we can
00:25:02.170 --> 00:25:09.340
stop here and then ok
so this operates on this operates
00:25:09.340 --> 00:25:20.550
on x square and so what we will get is minus
one by six minus one by six and then x square
00:25:20.550 --> 00:25:26.370
and then d minus d square when operates on
x square d or in square gives two x d square
00:25:26.370 --> 00:25:37.990
on x square gives two so two x minus two divided
by six and then we have d minus d square square
00:25:37.990 --> 00:25:44.310
one we find we get d square then we
get d cube then we get d to the power four
00:25:44.310 --> 00:25:53.100
so d square one operates on x square gives
you two so two by thirty six all other terms
00:25:53.100 --> 00:26:00.320
will contribute zero because when they operate
on x square we what we get is zero so this
00:26:00.320 --> 00:26:04.630
is the particular integral in this case so
when one over f d operates on x to the power
00:26:04.630 --> 00:26:10.840
m where m is an integer we have to extend
one over f d in ascending powers of d
00:26:10.840 --> 00:26:18.630
now i will tell you how this expression is
justified suppose we have one over one plus
00:26:18.630 --> 00:26:27.010
d operate[n] on x to the power m and we are
writing it as one minus d plus d square minus
00:26:27.010 --> 00:26:37.130
d cube and so on we are expanding one over
one plus d as one minus d plus d square minus
00:26:37.130 --> 00:26:42.220
d cube and so on like we expand one plus x
to the power m minus one so how this expansion
00:26:42.220 --> 00:26:47.550
is valid now what you do is to prove this
you operate by one plus d on both sides
00:26:47.550 --> 00:27:15.910
so operating by one plus d
we will have this ok now one plus d and one
00:27:15.910 --> 00:27:20.430
over one plus d are inverse of operators so
we will have x to the power m and in the right
00:27:20.430 --> 00:27:25.960
side let us operate one plus d on one minus
d plus d square minus d cube what you get
00:27:25.960 --> 00:27:33.100
is when you operate by one you get one minus
d plus d square minus d cube and so on and
00:27:33.100 --> 00:27:40.740
when you operate by d you get plus d minus
d square plus d cube minus and so on x to
00:27:40.740 --> 00:27:47.840
the power m so these terms they go on cancelling
and what we get is x to the power m so both
00:27:47.840 --> 00:27:53.780
sides are equal and therefore one over
one plus d if we write as one minus d plus
00:27:53.780 --> 00:28:00.620
d square minus d cube like we expand bv binomial
expansion this just extension is valid
00:28:00.620 --> 00:28:07.690
justified so then we have to find one
over f d operating on x to the power m we
00:28:07.690 --> 00:28:11.030
have to expand one over f d in ascending
powers of d
00:28:11.030 --> 00:28:12.910
thank you