WEBVTT
Kind: captions
Language: en
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hello friends welcome to my second lecture
on solution of second order homogenous
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linear differential equation with constant
coefficients we are discussing the solution
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of second order homogenous linear differential
equation with constant coefficients
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that is given by y double dash plus a y dash
plus b equal to zero b y equal to zero the
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characteristic equation here
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is given by m square plus a m plus b equal
to zero let us find the roots of this quadratic
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equation m equal to minus a plus minus under
root a square minus four b divided by two
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so there are two values of m we have
three cases case one distinct roots case two
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complex conjugate roots
and then case three double root
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so the case one when the two roots of
the characteristic equation are distinct we
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have already covered in previous lecture and
the case two where the two roots are complex
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conjugate that also we have covered in
the previous lecture
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so in this lecture we begin with the case
three let us say that the root of the
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quadratic the characteristic equation
has double root that is in the case three
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the discriminant a square minus four b equal
to zero in this case the discriminant a square
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minus four b equal to zero so we have m equal
to minus a y two which is a double root now
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this double root at first gives us one solution
of the second order linear differential
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equation one we can say the one solution
is one solution of the given solution
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one we can write as y one equal to e to the
power minus a by two into x now to obtain
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the other independent solution by two we apply
the method of variation of parameters
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this method of variation of parameters was
given by lagrange by by given by lagrange
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for first order linear differential equation
with first order linear differential equation
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so in this method what we do is let us assume
the other solution by two x to be equal to
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u x into y one x we shall find out a x so
that y two is the other independent solution
00:04:50.389 --> 00:04:54.500
of the given differential equation
so let us assume that y two x is equal to
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u x into y one x is the solution of
equation one then we shall have let us substitute
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by two x so let us by two x gives y two dash
x equal to u dash x into y one x plus u x
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into y one dash x and y two double dash x
equal to u double dash x into y one x plus
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u dash x into y one dash x plus u dash x into
y one dash x plus u x into y one double dash
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x or we can say this u double dash x into
y one x plus two times u dash x into y one
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dash x plus u x into y one dash double dash
x now lets substitute the values of y two
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y two dash y two double dash in equation one
so substituting y two and its derivatives
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in one we have
u double dash x into y one x plus two u dash
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x into y one dash x plus u x y one double
dash x plus a times y dash is y two dash so
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y two dash is equal to u dash x into y one
x plus u x into y one dash x plus b times
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y y two and y two is u into y one
now let us collect the coefficients of the
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derivatives of u so we may write this equation
as u double dash
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the coefficients of u dash x are two y one
dash x
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plus a times y one x
the coefficient of u x is y one double dash
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x
plus a y one dash x plus b y one x equal to
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equal to zero now since y one is a solution
of equation one
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we get y one double dash x plus a y one dash
x plus b y one x equal to zero also y one
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x is equal to e to the power minus a by two
into x so let us differentiate it with respect
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to x it will give me
which is equal to minus e by two y one or
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we can say two y one dash x plus a y one x
equal to zero so substituting these values
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so [substitute/substituting] substituting
these values here it reduces we get u double
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dash x into y one x equal to zero the coefficient
of u dash x is zero from here the coefficient
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of u x is zero from here so u double dash
x into y one x is equal to zero since y one
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x e e to the power minus a y two into x which
is not equal to zero for any x belonging to
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r we have
u double dash x equal to zero
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so when we integrate this twice we get which
implies u x equal to x thus the other solution
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of the equation one is given by
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x into e to the power minus a by two into
x now so this is this is for one solution
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and then we to find the other solution we
applied the method of variation parameters
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where we assumed y two x equal to u x into
y one x now this so the now let us see that
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y one and y two we see that y one over y two
is equal to one over x which is not a constant
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so the y one and y two are independent of
two each other
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so we got two independent solutions of the
equation one and therefore we can write the
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general solution as y x equal to c one plus
c two x into e to the power minus a by two
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into x where c one and c two are arbitrary
constants
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now let us take an example on this first due
to gravity which will at downward will be
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m g so let us take f one equal to m g now
due to the mass m that is due to the gate
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m g the there will be a extension in this
spring
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before it comes to the equilibrium position
so let us say in the equilibrium position
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the force due to gravity that is f one equal
to m g will have downward while the spring
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force the spring force will be lambda s naught
where lam this spring force is
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when lambda a spring modulus and s naught
is the extension r stretch in the spring
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so lets say this is s naught this is spring
force will let in the direction of opposite
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to m g
so in the equilibrium position m g and lambda
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is not will balance each other and we shall
have
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m g equal lambda s naught here we are using
the hooks law which says that the restoring
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force of the spring r we can also call it
as the spring force is proportional to the
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stretch or extension of the spring so here
there is a stretch or extension as not in
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this spring so the spring force are
restoring force is proportional to s naught
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and therefore we write it as equal to lambda
s naught by lambda is constant or proportionality
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we call it as a spring modulus
now let us let us pull the mass m downward
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and then release it should be mass m is pulled
down and then released ok then the there
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will be a vertical motion in the spring the
forces that will be obtained let say at some
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time t during the motion of the spring
y may be displacement from the equilibrium
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position at time t so then the forces that
will be acting on the mass m will be
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the force due to gravity m g downwards
f one equal to m g f two equal to this spring
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force which will be lambda times s plus s
naught plus y because s naught plus y is the
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total extension in the a spring so this lambda
s naught plus y will act in the direction
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opposite to m g the resultant force therefore
is
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since m g is equal to lambda s naught we have
resultant force as minus lambda y now let
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us apply the newtons second law newton by
newtons second law resultant force is is equal
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to m times d square y over d t square so we
have the equation of motion as
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m by double dot equal to minus lambda by r
where y double dot means d square y over d
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t square
now we can also write it as
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so if we denote omega naught let us denote
by omega naught square lambda by m then we
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have the auxiliary equation is let me not
write m ok so alpha square plus omega naught
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square equal to zero and now since this equation
has two complex roots so we have the general
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solution
of equation one as by t equal to c one cos
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omega naught t plus c two sin omega naught
into t now it can also be expressed as c times
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cos omega naught t minus delta where c is
equal to under root c one square plus c two
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square and ten delta is equal to c two by
c one it can be easily checked now if y naught
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is equal to if if we impose the initial conditions
in on this motion as y suppose y zero
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is equal to y naught that means when we release
the mass m at that time at that time t
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equal to zero the displace the displacement
of the mass m from the equilibrium position
00:22:46.220 --> 00:22:58.500
is y naught and the velocity at that instant
is given by y one then we have then we
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can see from here then put t equal to zero
in this we get y zero equal to y naught y
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naught is equal to c cos delta and from y
t equal to c cos omega naught t minus delta
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if you use y dash zero equal to y one you
get y one equal to c omega naught sin delta
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so from these two equations y naught equal
to c cos delta and y one equal to c omega
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naught sin delta we can find the value
of c c square is equal to y naught square
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plus y one square over omega naught square
and delta equal to ten inverse y one over
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y naught omega naught so the two constants
c and delta can be found from the initial
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conditions y zero equal to y naught and y
dash zero equal to y one or we can also put
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c as one over omega naught under root y one
square plus omega naught square y naught square
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until delta equal to this
now we can see here that since y t is equal
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to c cos omega naught t minus delta the the
motion is a harmonic oscillation now let us
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go to damped oscillation so in the case of
dumped oscillation suppose the mass is connected
00:24:23.990 --> 00:24:44.650
to a dash pot
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this is dash pot
00:24:59.040 --> 00:25:05.880
so suppose the mass is connected to a dash
pot then we have to consider the viscous force
00:25:05.880 --> 00:25:13.020
due to the viscosity of the liquid in the
dash pot now the dumping force due to
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the viscous liquid x in the direction opposite
to the instantaneous motion and is proportional
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to the viscosity so the dumping force
is proportional to the viscosity when the
00:25:36.290 --> 00:25:43.630
when it is small so we we have then
we have then we have the third force which
00:25:43.630 --> 00:25:53.700
is f we have taking as f three f three equal
to minus c y dot y dot is the velocity at
00:25:53.700 --> 00:26:00.890
instant t so the force due to the viscosity
of the liquid which is there in the dash pot
00:26:00.890 --> 00:26:06.940
there will be a force which is called dumping
force it x in the direction opposite to the
00:26:06.940 --> 00:26:13.090
instantaneous motion and this c is a dumping
constant this c is a dumping constant
00:26:13.090 --> 00:26:25.630
now lets us see a so in the now the resultant
force will be
00:26:25.630 --> 00:26:32.330
f one plus f two plus f three which is equal
to f one plus f two we have seen is minus
00:26:32.330 --> 00:26:40.580
lambda y so minus lambda y plus minus c y
naught dot thus we get the equation motion
00:26:40.580 --> 00:26:49.040
as n y double dot plus c y dot plus lambda
y equal to zero the auxiliary equation will
00:26:49.040 --> 00:26:54.910
be given by alpha square plus c by m into
alpha plus lambda by m equal to zero the two
00:26:54.910 --> 00:26:59.870
roots of this auxiliary equation are given
by alpha one and alpha two the alpha one is
00:26:59.870 --> 00:27:04.460
minus c by two m plus one by two m under root
c square minus four m lambda and alpha two
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is minus c by two m minus one by two m under
root c square minus four m lambda
00:27:09.880 --> 00:27:17.929
now due to the discriminant c square
minus four m lambda there arise three cases
00:27:17.929 --> 00:27:23.890
the case one is c square is greater than four
m lambda in that case the two roots alpha
00:27:23.890 --> 00:27:30.660
one and alpha two of the auxiliary equation
will be distinct so now we have three cases
00:27:30.660 --> 00:27:36.100
in the case one when c square is greater four
lambda we have the case of over dumping because
00:27:36.100 --> 00:27:40.970
the dumping constant c is high here so in
this case the general solution is given by
00:27:40.970 --> 00:27:45.830
y t equal to c one e to the power minus p
minus q into t plus c two e to the power minus
00:27:45.830 --> 00:27:50.970
p plus q into t now where p is equal to c
by two m and q equal to one over two m under
00:27:50.970 --> 00:27:55.580
root c square minus four m lambda now from
here we can see that p is is strictly greater
00:27:55.580 --> 00:28:02.860
than q and so when t goes to infinity
e to the power minus p minus q into t goes
00:28:02.860 --> 00:28:09.110
to zero as well as e to the power minus p
plus q into t goes to zero so in the equation
00:28:09.110 --> 00:28:15.250
one when y t goes to zero s t goes to infinity
that is the system comes to rest after a sufficiently
00:28:15.250 --> 00:28:21.830
long time in equation two we see that y t
is given by c e to the power minus p t cos
00:28:21.830 --> 00:28:29.490
omega bar t and therefore the motion is a
harmonic oscillation the displacement curve
00:28:29.490 --> 00:28:34.610
y t lies between y e equal to c times e to
the power minus p t and y equal to minus c
00:28:34.610 --> 00:29:06.240
times e to the power minus p t
so this is let us say t x is this is y t these
00:29:06.240 --> 00:29:16.170
are displace these are the curves y t equal
to c e to the power minus p t and y t equal
00:29:16.170 --> 00:29:28.950
to minus
the displacement curve lies between these
00:29:28.950 --> 00:29:38.880
two curves and touches these curves when omega
bar t minus delta is a interior multiple of
00:29:38.880 --> 00:29:54.190
pi moreover the frequency here is frequency
of oscillation is equal to omega bar over
00:29:54.190 --> 00:30:03.080
two pi so when omega bar is more frequency
is more and when let us see when omega
00:30:03.080 --> 00:30:12.380
bar will be more omega bar will be more
when c is smaller when c is smaller here and
00:30:12.380 --> 00:30:22.600
it will be greatest when c tends to zero so
when c tends to zero
00:30:22.600 --> 00:30:37.580
omega bar goes to one over two m under root
four m lambda which is equal to under root
00:30:37.580 --> 00:30:46.010
lambda by m which is equal to omega naught
the omega naught which be found in the case
00:30:46.010 --> 00:30:58.310
of the simple harmonic oscillation
this omega naught y t equal to c cos omega
00:30:58.310 --> 00:31:02.890
naught here omega naught you can see omega
naught is root lambda by m
00:31:02.890 --> 00:31:11.040
now let us discuss the last case in the
equation three you can see we have y y
00:31:11.040 --> 00:31:16.110
t equal to e a plus b t into e to the power
minus p t since e to the power minus p t is
00:31:16.110 --> 00:31:26.161
not equal to zero y t can never be zero except
possibly once when a plus b t is equal to
00:31:26.161 --> 00:31:32.580
zero that is t is is equal to minus a by b
hence the motion can have at most one passage
00:31:32.580 --> 00:31:37.630
through the equilibrium position now if the
initial conditions are such that a and b are
00:31:37.630 --> 00:31:45.350
positive then you can see y t can never be
zero because if a and b are positive then
00:31:45.350 --> 00:31:50.740
y t can never be zero so the displacement
will never be zero and this case is similar
00:31:50.740 --> 00:31:57.540
to the case one in the case one you can see
y t tends to zero as t goes to infinity so
00:31:57.540 --> 00:32:01.350
the system comes to rest after sufficiently
long time
00:32:01.350 --> 00:32:07.539
thanks