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Euler-Cauchy Theorem - Engineering Mathematics Video Lecture
FAQs on Euler-Cauchy Theorem - Engineering Mathematics Video Lecture
| 1. What is the Euler-Cauchy theorem in engineering mathematics? |  |
Ans. The Euler-Cauchy theorem, also known as the Euler-Cauchy differential equation, is a theorem that deals with finding solutions to linear homogeneous differential equations of the form ax^2y'' + bxy' + cy = 0, where a, b, and c are constants.
| 2. How is the Euler-Cauchy theorem applied in engineering mathematics? | |
Ans. The Euler-Cauchy theorem is applied in engineering mathematics to solve differential equations that involve second-order derivatives. By converting the given differential equation into the standard form ax^2y'' + bxy' + cy = 0, we can then use the Euler-Cauchy theorem to find the solutions.
| 3. Can the Euler-Cauchy theorem be used for non-homogeneous differential equations? | |
Ans. No, the Euler-Cauchy theorem is specifically applicable to linear homogeneous differential equations. For non-homogeneous differential equations, other methods such as variation of parameters or undetermined coefficients need to be applied.
| 4. Are there any limitations or conditions for using the Euler-Cauchy theorem? | |
Ans. Yes, there are certain conditions for using the Euler-Cauchy theorem. The coefficients a, b, and c in the differential equation must be constants, and the differential equation must be linear and homogeneous. Additionally, the equation must have a positive value of a.
| 5. Can the Euler-Cauchy theorem be used for partial differential equations? | |
Ans. No, the Euler-Cauchy theorem is not applicable to partial differential equations. It is specifically designed to solve ordinary differential equations that involve only one independent variable. For partial differential equations, different techniques and theorems are used.
Text Transcript from Video
welcome to the lecture series on
differential equations for undergraduate
students today's lecture is on other
kocchi equations till now we have
learned how to solve second order linear
differential equations with constant
coefficients that is px and QX were
constant not the function of X of course
we had learned both the methods that is
for right-hand side our X to be 0 and
not to be 0 that is both the cases
homogeneous equations and non
homogeneous equations now we will learn
the method to solve linear differential
equations where the coefficients px and
QX are not constants so first we will
discuss one special kind of equation
which has many engineering applications
that is called Euler Koshy equation what
is an ayakashi equation and I'll akashi
equation is of the form X square y
double dash plus ax y dash plus B Y is
equal to 0 where the coffee do we see
that this is a second order linear
differential equation where the
coefficient of Y Double Dash is X square
coefficient of y dash is a X and
coefficient of Y is B here this a and B
they are constants but the coefficient
of wilders that is ax and coefficient of
Y ax double dash that is X a squad they
are not constant but the function of X
we can change this equation to a linear
differential equation with constant
coefficient with certain substitutions
and that substitution is in the variable
X so let us see that substitution say X
is equal to e to the power T then dy
over DX that is y dash would be nothing
but e to the power minus T dy over DT
and the second derivative d 2 y over DX
2 will be e to the power minus 2t d 2 y
over DT 2 minus e to the power minus 2t
dy over DT now if I substitute this dy
over DX and D 2 y over DX in the dy over
DT and X
as e to the power T over here in this
equation we will get this equation as
gay 2y over DT 2 plus a minus 1 dy over
DT plus B y is equal to 0 now we see
this is the second order linear
differential equation where the
coefficients are constants a minus 1 and
B both are constants so now we do know
that how to solve this equation and we
can get the solution however the
solution of this equation will be a
function of Y as a function of T this
this will be this will be that is Y as a
function of T wireless we want to a
solution Y as a function of X so what we
get solution as Y as a function of T
will substitute there T as log X and
we'll get the solution in the terms of X
there is one alternative method to get
that alternative method to solve a
low-cost equation see my question is X
square y double dash plus XY dash plus B
Y is equal to 0 you see here that the
coefficient of Y - is X and coefficient
of Y Double Dash is X square say if I
put Y as X to the power M we would be
able to we think is that it could
certain satisfaction Y we see that y
dash would be M times X to the power M
minus 1 y double dash would be M into M
minus 1 times X to the power M minus 2
so if Y is equal to X to the power M
this is a solution of this equation
these things must satisfy this equation
so let us substitute it if we are
substituting we would get M into M minus
1 X to the power M plus a times M X to
the power M plus B times X to the power
M is equal to 0 let us rewrite this
equation again as M square plus a minus
1 M plus B times X to the power n is
equal to 0 now if this X to the power M
is a solution of this equation we want
that is this should not
a trivial solution that is this should
not be zero for all X and if this is a
solution it must satisfy this equation
that is this must be 0 this will be 0 if
either X to the power M is 0 or this M s
square plus a minus 1 M plus B is equal
to 0 this cannot be 0 hence this would
be solution only M square plus a minus 1
M plus B is equal to 0 so we would get
that is this would be a solution if it
is satisfied what it says this equation
which we have gotten M this is called
characteristic our auxiliary equation
let us compare this method with the
previous method where we have changed it
to the linear differential equation with
constant coefficient in that one what we
have got that equation we have got d 2y
over DT 2 plus a minus 1 dy over DT plus
B y is equal to 0 to solve this we
require the characteristic equation what
is the characteristic equation
characteristic equation would be M is
square plus a minus 1 M plus B is equal
to 0 now let us compare what just now we
have got the auxiliary equation that was
also M square plus n minus 1 M plus B so
we have got that characteristic equation
for both days.the methods we are getting
the same and after that what we'll get
we'll get the solution as the roots of
this one so here if we are getting we
would get the solution as e to the power
M T kind of things and here we would get
X to the power M kind of thing and here
we have to get the solution in the terms
of X that is e to the power T I have to
substitute US x so again we will get X
to the power M and here also as X to the
power M so that says this both methods
will give me the same kind of solutions
so to see how we get the solution we see
the equation is M square plus a minus 1
M plus B is equal to 0 this is a
quadratic equation in M so of course it
will have two roots of this equation and
the solution would be depending on those
two roots and those two roots are
depending upon what we are getting the
tools so there would be three cases the
two roots M 1 and M 2 and we'll have
three cases the case one when I do have
both these M 1 and M 2 they are distinct
and real the second case would be that
is when this M 1 and M 2 are real but
they are equal that is double real loot
and the third situation will have when
this both M 1 and M 2 are complex
numbers so of course there would be
entire that is the complex conjugate
roots let us discuss the general
solution in each case one by one so the
case 1 distinct real roots when this
equation M is square plus a minus 1 M
plus B is equal to 0 has two distinct
real roots M 1 and M 2 then the two
solutions we would have as X to the
power M 1 and X to the power M 2 these
two solutions would be linearly
independent we can check this linear
independence using the wronskian do you
remember the Roskilde wronskian of the
two solutions Phi 1 by 2 is given as
determinant containing the solutions y 1
and y 2 and their derivatives that is
the determinant X to the power M 1 X to
the power M 2 and the second row M 1 X
to the power M 1 minus 1 and M 2 times X
to the power M 2 minus 1 what will be
the value of this determinant this would
be simply M 2 times X to the power M 1
plus M 2 minus 1 minus M 1 times X to
the power M 1 plus M 2 minus 1 that is M
2 minus M 1 into X to the power M 1 plus
M 2 minus 1 now this X is not 0 and
since M 1 and M 2 are distinct they are
not equal this will not be 0 for M 1 not
equal to M 2 that is when they are just
these two solution X to the power M 1
and X to the power M 2 there would be
linearly independent
hence the general solution would be C 1
X to the power M 1 plus C 2 X to the
power M 2 this we have discussed the
case 1 now go to the case 2 double root
that says is that when we do have this
equation M square plus a minus 1 M plus
B this has same real root that what will
be that root that root would be 1 minus
a by 2 so the one solution though would
get us X to the power M as X to the
power 1 by 2 1 minus C now to get the
second linearly independent solution
from here we will use the technique of
variation of parameter that is we will
take the second solution y2 as u times
y1 and then we'll find out what is my U
so by doing that technique we would get
actually write to us log X times y 1
that is you we have got log X so the
second solution would be log X into X to
the power 1 by 2 1 minus a and the
general solution would did it C 1 plus C
2 log X into X to the power 1 by 2 1
minus a so when this characteristic
equation has double root we do get the
general solution is of the form C 1 plus
C 2 log X X into to the power half 1
minus a now the third case complex
conjugate roots when this equation M is
square plus a minus 1 M plus B is equal
to 0 has complex roots they would have
appear in the pair's
that is let us say once first root is s
plus I T then the second root would be
of the form s minus item if these are
the roots then the general solution are
the two solutions we would get us first
as X to the power is cos T log X and the
second solution as X to the power s sine
T log X again we can find it out that is
this is the similar kind of thing as we
have done with the different
with constant coefficients this will be
linearly independent again you can check
it using the wronskian and the general
solution will be of the form X to the
power s c1 cos t log X plus c2 sine t
log X where this year this is the real
part in the complex root and T is the
imaginary part in the constant in the
conjugate their complex roots so we have
discussed these three cases let us do
some examples to understand this method
and how to solve this i'll akashi
equations let us do the first example
solve the ayakashi equation x is square
y double dash minus 4xy dash plus 6y is
equal to 0 we'll substitute Y as
function of X to the power M so we'll
substitute Y as X to the power M thus y
dash would be M times X to the power n
minus 1 and y double dash would be M
into M minus 1 times X to the power M
minus 2 if I substitute this in this
given equation we get M into M minus 1
times X to the power M minus 4 M times X
to the power M plus 6 times X to the
power M is equal to 0 now rewriting
these terms we get M square minus 5m
plus 6 times X to the power M is equal
to 0 so we will get the auxiliary
equation are the characteristic equation
as M square minus 5m plus 6 is equal to
0 we see the factors would be M minus 2
into M minus 3 so the roots would be M
is equal to 2 and see we do have two
real roots which are not equal that is
real distinct roots so the two solutions
would be y1 is X square Y to us X cube
so the general solution would be of the
form C 1 X square plus C 2 X cube this
is a solution of this differential
equation X square Y Double Dash minus 4
XY - plus 6y is equal to zero we can
check that this would be the solution so
if let's say they're checking this for X
square if I take X square in a solution
what will be y double dash that will be
two so I will get here 2 X square and y
dash would be 2x so I would be getting
minus eight X square plus 6 X's square
so that is satisfying this equation
similarly we can check with X cube R we
can check with this general solution it
has do one more example example to find
the general solution ax X square y
double dash plus XY - minus Phi is equal
to zero again we'll go with the same
kind of function that is y is equal to X
to the power M Y - as M times X to the
power M minus 1 y double dash as M into
M minus 1 X to the power M minus 2
substituting this in this equation and
finding out the auxiliary equation we
would get from here we could see that is
also very equation should be M is square
into a minus 1 so here is 1 so a minus 1
would be 0 so I would be getting the
auxilary equation as mu square minus 1
is equal to 0 of course it has two roots
MS one and one that is double root
EMAS one and one now this is the second
case so what will be my two solutions
the one solution Y would be X and
another solution would be x times log X
we can again check by putting the
solutions that they are solutions and so
the general solution would be C 1 X plus
C 2 X log X now let us discuss one more
example so example 3 solve the
initial-value problem x square y double
dash plus 4 x times y dash plus 9 Y is
equal to 0 with initial conditions y at
1 is equal to 2 and y dash at 1 is equal
to 0 so first we'll solve this equation
again by the same method that is y is
equal to X to the power M y dash as M
times X to the power M minus 1 y double
dash as M times M into my M
- 1 into X to the power n minus 2
substituting this we would get M into M
minus 1 plus M plus 9 is equal into X to
the power M is equal to 0 so we will get
the auxilary equation as M square plus 9
is equal to 0 now we see that this will
have complex so that is M square is
equal to minus 9 so I would have the
roots as plus minus 3i now if I compare
it with our original one that is s plus
I T so here the real part is 0 and the T
is 3 and we are having the two roots
plus 3i and
minus 3i so what will be our two
solutions our two linearly independent
solutions would be cause three logics
and sign three log X so the general
solution you would get as Y X plus c1
cos 3 log X plus a2 x sine 3 log X to
get the particular solution for the
initial value problem what are the
initial values initial values are given
that the function at value 1 is 2 and
the derivative at 1 is 0 so we get what
is Y - X from here that would be minus C
1 sine 3 log X into 3 by X plus C 2 cos
3 log X into 3 by X now if the first
initial condition is that X is equal to
1 so here if I put X is equal to 1 what
I will get log ik 1 would be 0 so I
would be getting here cos 0 and sine 0
cos 0 would be 1 and sine 0 would be 0
so what would be getting y at 1 we would
be getting is C 1 and that is given as
equal to 2 so this is what we have got
from the first initial condition in the
second initial condition we do have Y -
act 1 Y - at one that is here if I put X
is equal to 1 we will get this term to
be 0 and this Chango would be getting us
1 and this we would be getting as 3 so
what we would be getting y dash at 1 as
3 c2 which is given us to be 0 which
says is that c2 would be 0 so we have
got the two constant values that is c1
as 2 and C 2 as 0 so what will be my
particular solution here we'll
substitute the values of C 1 and C
- so the solution of initial value
problem would be 2 x cos 3 log X now you
can check that you can put this solution
in the given equation this will satisfy
the equation moreover this solution will
also satisfy these two initial
conditions so thus we have discussed the
Euler Cauchy equation of second order we
have learned that its solution is
depending on the roots of its auxilary
are the equation and that there would be
arising three cases now let us do the
practical life application of these
equations who will now discuss some
application linear differential equation
with constant coefficients have many
engineering applications here we will
discuss want such an application which
is the second order linear differential
equation in mechanical vibration here
what we are discussing very basic
mechanical system that is called a mass
on an elastic spring this is motion of
basic mechanical system this is mass on
an elastic spring and this is called
mass spring system what is this system
let us first understand this system we
do have any spring this isn't
unstretched spring if I hang a mass of
size M on this one you see this is
unstretched esprit I heard this is stick
to here on the one side and on this side
I have attached a mass of size this will
stretch this string so let us say this
is stretching is s naught now what I
will do is that is after this is
stretching this is at the static one so
if I pull this mass little bit downward
and then release it that is I pull this
mass downwards if I am pulling it again
this string will get more stretched 1
and then if I release
what will happen this will go up and
then it will go under the motion like
this zone that is where we are having
this that once I had pulled it down and
then left it it will go in a motion we
want to discuss or we want to find out
this basic motion of this spring so we
would see that is how we are going to
model this system model this system is
we will find out what are the
differential equations which are
governing this motion let us see how we
are going to do so we are going to
setting up this model for this mass spec
system here we have taken and spring
bitches come with it which would be
going to be the whether it will be
compost or it would be stretch it will
resist it that is this simple spring it
is a child attached with one hook on the
one side on the second side we had put a
massive size in know that will give it
is a stretching and this is stretching
it says that is this stretch is s not
here this is in this this is not in the
motion once I had put this mass over
there this will just hang on this
position I am taking is the initial
position this initial position and then
this spring has been this mass has been
pulled down if it has been pulled down
it will again is stretch it and then
when we are releasing it will go to the
motion so now will make certain basic
assumptions for setting up this motion
the first setting up is that we take
that motion should be strictly vertical
that is we are not talking about this
kind of motion
we will talk about only a strictly
vertical motion we are taking that
direction towards the down side as
positive and direction towards the
upside as the negative this is the
static condition that is from here
initially we are starting that is when
the mass has been
and that is this initial stretcher is
not we have taken and then we are
pulling it down and then releasing it so
after time T wherever this thing is that
is suppose it is here then this is the
displacement from the initial condition
Y is equal to 0-2 at Y T at time T know
how this system is being governed let us
see this system would be governed by
Newton's second law what is the Newton's
second law that says its mass into
acceleration is equal to force what is
the mass mass has been given us M
acceleration since we have seen in the
last one that is Y is the displacement
we do know if Y is the displacement at
time T is y T then velocity would be Y -
T that is the derivative of Y with
respect to T at time T that would be the
velocity and acceleration would be the
second derivative y double dash T so we
have got the mass we have got the
acceleration now what is the force let
us see the force the force which is
governing this complete system let us
see when this is unstretched spring when
a mass M is being attached it has been
stretched when it has been stretched it
says this there is a force acting
upwards say if not which is balancing it
so this is this force F naught would be
on the upward size and this is not
moving that is it is balancing the
weight of this mass what will be the
weight of this must by the Hookes law
gas force F naught would be actually
minus case not that is this is against
that is whatever there stretching this K
here is called the spring constant and
we are taking this negative size aside
because this is on the upward side this
is balancing the weight weight is M into
G now the system is in this static
equilibrium that it is not moving that
says this via
having this if not plus W that is minus
KS not plus mg this should be zero this
is not giving any motion this force is
zero now when it goes under the motion
that says is when we are pulling it and
then releasing it when we are pulling it
one more for that is it is being
stressed again and what is this is
stretching one this is stretching from
this initial condition why we have taken
zero now this is stretching extra
stretching is up right that will
generate another force f1 and what this
would be again by Hookes law this would
be minus KY again the sign would be -
since this forces upwards and the K is
the spring constant and why is this
stretching at the time T this is called
the restoring force moreover every
system does has a damper if the damping
is not there that is there should be
some force which is governing this
motion then this motion will go on
forever we will learn this little bit
after let's say that damping force is f2
this will again be in the upward
direction this damping force is going to
govern the velocity so that it would
reduce it so a very good approximation
of this f2 is - CY - that is a rider
that is the velocity and - see this see
this is called the damping constant and
again the sinus - since this would force
will also act upon on the upward side so
now the resulting force what would be
this see we are also taking as positive
this K is also positive the resultant
force would be f1 plus f2 so the total
force would be minus KY - CY - now we
are ready to get the model of this
system that is by second love Newton's
second law mass into acceleration is
equal to force masses M acceleration is
y double dash the force we have find it
out its - CY -
- KY that is we have got a differential
equation my double dash plus CY - plus
KY is equal to zero this is a second
order linear differential equations with
constant coefficients since all these
are constants so this is governing this
massive spring system this equation and
this is a homogenous one so we just says
that is what we do get how to get the
solution of this and how this motion we
are trying to understand so first we
will discuss the case of undamped system
undamped system that is there is no
damping is present that is the damping
constant C would be 0 that is what we
would get our equation our equation
would be M y double dash plus KY is
equal to 0 that is the term involving CY
- since C is 0 that is not present there
now here M and K both are positive so we
would get what will be our
characteristic equation for this linear
differential equation that is M lambda
square plus K is equal to 0 its loot
would be of course since both K and a
more positive its route would be complex
1 and those roots would be minus square
root of K by M I plus and minus so the
general solution would be of the form a
cos T times square root K bar m plus B
times sine K times the square root of K
by him t here I am taking as the time T
that is I'm having the equation in Y and
T this is the general solution now let
us see that is what this general
solution is let's see in the graph this
is called harmonic oscillation Y you see
here here we are having is that is for
different values of a and B so let us
say a is equal to 1 and B is equal to 0
but here I have taken em as 4 and K s 3
that is a spring constant as three and
masses 4 then what we are having is that
is this
motion first this black line you see
this is just going like this one now you
see what we are having is it's starting
the motion is being started that is
displacement is going on it goes up at a
distance then it goes down with the same
distance then it goes up again with the
same distance and you see the time it is
taking between the cycles that is from
here to here here to here here to here
the cycle is also same and it is also
taking that the displacement that is the
distance it is covering our displacement
it is making that is also equal that is
that is what I said is when we are not
having a damper the motion will go on
forever so this will go on forever with
equal magnitude similarly if I take my a
to be 0 and B is to be 1 then you see is
that again I am getting this harmonic
motion distinct kind of line again you
are having is that the cycle is same the
magnitude is also saying whether upwards
are the downwards and this will go on
similarly for the different values if a
and B I have taken as 1 we are getting
is that is the magnitude is differing
here and but cycle is there different
cycle different magnitude but it is
going on
similarly with other values as is 2 and
B as 3 we are getting it again that is
the magnitude this or getting higher one
on the cyclist again in the similar
manner smaller one and then if I had a
s4 and B as minus 5 we have got that the
magnitude is again little bit higher on
the cyclist something is a little bit
smaller and we are getting that this
will go on this is never going to down
that is the motion will go on forever so
let us see another example over here
here I have changed that mass to B 3 and
the spring constant case 5 that is it's
more resistant ISTE again you see the
similar kind of pattern we are having
again I have taken this a B as the
similar kind of thing that is a is equal
to 1 B is equal to 0 that is this first
one we are getting is this
this motion then if is equal to zero and
B is equal to one we are getting this
motion and so on we are having again we
are having as this motion is going on
with the equal cycle in one motion and
with the equal magnitude whether upwards
whether upwards are downwards that is
what we are calling it as harmonic
oscillation and without damping this
motion will go on forever forever time
let us see the damped system that is
when the damping is present so first we
see what is the equation and the
solution and then we'll discuss this
motion so the differential equation
would be my double dash plus CY dash
plus KY is equal to zero here this M is
the mass C is the damping constant and K
is the spring constant this equation
will have the characteristic equation as
lambda square plus C by M lambda plus K
by M is equal to zero since this is it
is a quadratic equation it will have two
roots lambda 1 and lambda 2 and those
roots would be of the form minus C by 2
M plus minus 1 by 2 m square root of C
square minus 4 M K that is we will have
one daughters minus C by 2 m plus 1 upon
2 m is square root of C square minus 4 M
K another note we would have minus C by
2 m minus 1 by 2 m square root of C
square minus 4 M K let us rewrite it as
in the simplified manner let us say this
first thing as the Alpha C by 2 m and
the second part that is beta as 1 upon 2
m square root of C square minus M K then
we would have the two roots lambda 1 as
minus alpha plus beta and lambda 2 s
minus alpha minus beta now you see in
the beta we are having this term square
root of C square minus 4 M K that is we
are finding out the square root of some
quantity where the C m and K all they
are constant so of course we will have
our roots will be depending upon what is
this
inside the square root that is C square
minus four and give whether it is
positive quantity it's zero quantity or
its negative quantity accordingly you
have different roots so we'll have
actually the three cases the case one is
when C square my is greater than 4mk in
that case we will have two different
root distinct real roots lambda1 and
lambda2 this case is called over damping
then second case when C square is equal
to 4 Mk will have double real roots this
is called critical damping well known
all these terms why we are calling them
and the third case when C square is less
than 4mk that is its imaginary part beta
would be imaginary and that is complex
conjugate roots this we would call under
damping let us discuss the solution in
each case one by one so general solution
in case one case one is when C square is
greater than four ng that is we would
have two roots as minus alpha minus beta
and minus alpha plus beta so the general
solution we would have YT as C 1 e to
the power minus alpha minus beta T plus
C 2 e to the power minus alpha plus beta
T here if I this alpha we do know is
that C by 2 m and beta is 1 by 2 MC
square minus 4 in case where m is the
mass C is the damping coefficient and K
is the spring constant now for different
values of C m and K we will get so that
the C square minus 4 M K is positive we
will get the different roots and if I do
give some initial conditions so that we
get some particular solution so we'll
discuss this case one by one here what I
have taken I have taken some values of C
m and K such that C square minus 4 m
case positive an initial condition what
I have taken is that is y at 0 is 4 that
is it is positive initial displacement
then we do have the another initial
condition that
would be governing why - that is why -
whether it is I have defined some values
and I'm classifying them into three
parts that is my velocity initial
velocity y - is positive negative R 0
let us see first case Y 1 this y 1 we
are having is here as positive velocity
initial that's why - 1 is positive I
have taken someone so in that case what
how this motion will go on initially it
has been displaced that is pulled
downwards for at 4 then it has been
released it has generated and with the
damping it has generated a velocity
positive velocity so we are getting it
as that is it's going little bit up and
then it is dying down because of damping
if initial velocity is 0 then it is
started dying down from the initially it
has not gone up when the initial
velocity is negative this initial
velocity is upwards then we are having
route could have the two cases one is
that is it is going down like this one
and then it is dying down to the
displacement is coming to be 0 or it may
happen that comes to 0 and then it goes
to the negative one that is we are
having is this motion upwards but again
it will come up and it will go to the 0
so if the damping is present ultimately
will have that our motion would stop and
after some time we would go to the 0
displacement now if I take initial
displacement to be the negative that is
if y 0 is negative so again I have taken
a case when y 0 is minus 4 again with
the second initial condition about the
velocity again I have divided positive
negative and 0 now if the displacement
is negative and my velocity is also our
velocity is also negative that is the
last case Y for this case so you see is
that is it is first going down little
bit and then coming up and it will
ultimately reach to the 0 that is the
displacement would be 0 if the initial
velocity
zero it is starting with the negative
displacement and the displacement is
going up and that is it's reducing and
it will certainly reach to somewhere
zero if initial velocity is positive we
could have again the two cases one case
is that is it has a start a little bit
that is the displacement has gone very
up like this one that is reduced
certainly and then the reduction power
is little bit less and then we are
moving over here we may have this in
this positive case that is it may go to
the 0 and then it crosses this line then
it goes through to the positive
displacement and then it is dying down
so again we have got that when damping
is present we will not have this
harmonic kind of oscillation that the
motion goes on forever rather we would
be having is that after some time the
motion will die down and their
displacement would be zero now let us
discuss the second case that case 2
where C 2 is equal to 4 M K that is beta
would be 0 and we will have double root
that is both lambda 1 and lambda 2 they
are equal to minus alpha in this case we
do know that the solution would be of
the form C 1 plus C 2 T times e to the
power minus alpha T alpha is of course
our C by 2m
in this case if I take some values of C
and M such that C C square is 4 M K
again I have taken initial condition
that my initial displacement Y is
positive and that is 2 again with the
second initial condition about the
velocity we have divided it into
positive negative R 0 so when the
displacement is positive and velocity is
also positive this is the second case
where I am having C square is equal to 4
M K that is we are having double real
root again it is going little bit up and
then it is coming down so of course
after sometime it will come down when
initial velocity is 0 it is starting
from here it is going down very slowly
but it's going down then when the
initial velocity is negative it is going
down first it goes down very fast then
it is going down slowly and it will go
to the
zero it may have another case where it
is going down to the zero and then it is
going down to the negative displacement
then it will come up and it will go to
the zero that is again in this case also
the motion will die down that is the
displacement would be zero after
sometimes the motion will stop
let us discuss them discuss the third
case case three is when C square is
great less than four m'kay that is the
beta is pure imaginary let us say beta I
Omega star where this Omega star is 1
upon 2 m square root of 4 MK minus C
square that is this thing is positive so
Omega star would be positive in this
case the general solution will be of the
form e to the power minus L 4 T a cos
Omega star T plus B sine Omega star T by
the trigonometry we do know that this
thing we can rewrite as C times e to the
power minus alpha T cosine of Omega star
t minus delta we have this C is such
that C square is a square plus B Square
and this Delta is such that tan of Delta
is B upon a now we see this is the
general solution that is this is telling
us the displacement at time T this is
having eetu the power minus alpha T and
cosine of something now the cosine of
any from this cosine function we do know
this lies between minus 1 and plus 1 so
what will have this motion will lie
between minus C to the power minus alpha
T and plus C e to the power minus alpha
T let us see what kind this motion would
be look like you see here this is the
line C to the power minus l4 T and this
is the line minus C e to the power minus
alpha T again I have taken this some C
and M such that C square is less than 4
Mk you see this motion is actually this
one
is going first it's starting with here
initially and then it is going down then
it is coming up and like this you see
each time the magnitude of the motion
you are getting is getting lesser and
lesser and afterwards it is going to be
died down it's coming it to zero so this
is what is the motion when we do have
the underdone and you see is that is
when we had over damping we had got the
motions which are just going down when
we had critical damping again the
motions were just going down and here
what we are having is that motion is
going down and up and down and up and
then they are dying down that is they
are is talking so this is what is called
the under damping so we have discussed
the three cases of this differential
equation which is governing this motion
of basic mechanical system that is
called the mass spring system now it may
happen that is this motion is being
governed by some force that is we are
calling forced oscillation our resonance
what is this this says is that this
motion is being governed by if force are
a force is affecting the motion that
means till now what I was having that is
this was the equation of the motion
which was equal to zero that is this was
balanced by balanced equation now what
we are having is that this motion is
governed by one force called our team
that is now we are having
non-homogeneous equation my double dash
plus CY dash plus KY is equal to R T now
what is R T which is governing this
motion this R T is called input and what
will be the solution of this equation
why that would be called output this is
we are giving is that is sometimes we
say is that this is this is what is the
input and R this is what the force is
being applied on this output we are
calling the response to this force which
is governing this motion so this is also
called the response for example it
take this my Artie that is the force
which is governing the motion is of the
form F naught cos Omega T this Omega is
something different so f naught is a
constant and the forces of this form you
see I have taken this special form that
is cosine function if you do remember we
have to solve this non-homogeneous
equation where this both F naught and
Omega positive then what will have the
equation our differential equation would
be my double dash plus CY dash plus KY
is equal to if not cos Omega T now this
is a non-homogeneous equation it has
solution which is having the solution of
homogeneous equation and a particular
solution which is governed with this
right hand side that is called the
particular solution and then the general
solution we do have as the homogeneous
solution plus the particular solution so
we do have this differential equation
its general solution is of the form Y H
plus YP where Y it is the general
solution of Associated homogeneous
equation M y double dash plus CY dash
plus KY is equal to zero and YP is the
particular solution now we'll
concentrate on this white people if you
do see that is what we are having here
is that our RT is of the form if not cos
Omega T that is we are having a special
form so we'll use this method of
undetermined coefficient do you remember
that method we are having one table
which is telling us how what is the
choice of the particular solution YP
that table is this one so here what we
are having is we are having a constant
times cos Omega so my choice of
particular solution should be of a times
sine Omega X plus B times cos Omega X it
is so we'll have our particular solution
as of the form a cos Omega T plus sine B
sine Omega T if this satisfies the
equation then it will be a solution so
view how to find out
YP - as minus Omega s in Omega T plus
Omega B cos Omega t + YP Double Dash
would be minus Omega square a cos Omega
t minus Omega square B sine Omega T will
substitute this in the given equation
the equation is my double dash plus CY
dash plus KY is equal to F naught cos
Omega T when I am substituting those
wipey wipey - and by P Double Dash who
would get the equation of the form K
minus M Omega square a plus Omega C B
and this whole is multiple of cos Omega
T and then we do have a sine Omega T and
it's coefficient as minus Omega C a plus
K times plus K minus M Omega square
times B which is equal to F naught cos
Omega T to determine this a and B we
would equate these coefficients so what
it gives us that is the coefficient of
cos Omega T on the right side is f
naught and on the left side is this one
so we get the first equation as K minus
M Omega square a plus Omega C B is equal
to F naught and we don't have any term
of sine Omega T on the right side so we
will get the second equation as minus
Omega C a plus K minus M Omega square B
is equal to 0 these are the two
algebraic equations in two unknowns a
and B so we can solve it and we get the
solution so we will get what is the
particular solution homogenous equation
will give me the solution of a by H and
we will get the general solution now we
will discuss one special case rather
which is called damped force oscillation
R which is more generally known in the
mechanical engineering as vibration what
is this one
this governing differential equation
would be my double dash plus CY dash
plus KY is equal to F naught cos Omega T
I am just discussing with this special
one that is f naught cos Omega T
associated homogeneous equation would be
my double dash plus CY dash plus KY is
equal to zero it's solution let us say
suppose that this equation has a complex
OOTS
that is it's a characteristic equation
has complex roots then the solution
would be e to the power minus l4 t a cos
Omega T plus B sine Omega T where this
alpha would be C by 2 m and Omega would
be 1 upon 2 a misfire Ruta for MK minus
C square now this is what we are writing
and then the particular solution we
would be general solution would be of
the form Y H plus YP this general
solution is also known in this vibration
case as the transient solution and this
particular solution YP this is known as
a steady-state solution
why this is called transient solution
and this is a steady-state solution this
we will learn a little later on and we
would actually learn these things with
the help of an example this example we
will do in the next lecture where we
would really model or trial in the
vibration model and we will see that is
how the solution is coming at wow this
vibration is taking place and why we
call one thing as transient solution
this general solution and this
particular solution has a steady-state
solution how this is a response to the
input function all these things we will
see in the help of example and that
example will go with the next lecture
today we have learned a special kind of
equation called Euler Cauchy equations
this is second order linear differential
equations where the coefficients were
not constant but the function of X we
had learned homogeneous Cauchy Euler
equations how to solve them we had
learned some application of linear
differential equation with constant
coefficient we had learned one special
application in the simple mechanical
system called the mass spring system and
we had learned when this there is no
force is governing the motion both the
cases damp is that when there is no
damping and when there is damping who is
present so we have seen the damped
oscillations and we have seen the
undamped motion that is called harmonic
oscillation now next lecture we will
learn one more system mechanical
vibration thank you
differential equations for undergraduate
students today's lecture is on other
kocchi equations till now we have
learned how to solve second order linear
differential equations with constant
coefficients that is px and QX were
constant not the function of X of course
we had learned both the methods that is
for right-hand side our X to be 0 and
not to be 0 that is both the cases
homogeneous equations and non
homogeneous equations now we will learn
the method to solve linear differential
equations where the coefficients px and
QX are not constants so first we will
discuss one special kind of equation
which has many engineering applications
that is called Euler Koshy equation what
is an ayakashi equation and I'll akashi
equation is of the form X square y
double dash plus ax y dash plus B Y is
equal to 0 where the coffee do we see
that this is a second order linear
differential equation where the
coefficient of Y Double Dash is X square
coefficient of y dash is a X and
coefficient of Y is B here this a and B
they are constants but the coefficient
of wilders that is ax and coefficient of
Y ax double dash that is X a squad they
are not constant but the function of X
we can change this equation to a linear
differential equation with constant
coefficient with certain substitutions
and that substitution is in the variable
X so let us see that substitution say X
is equal to e to the power T then dy
over DX that is y dash would be nothing
but e to the power minus T dy over DT
and the second derivative d 2 y over DX
2 will be e to the power minus 2t d 2 y
over DT 2 minus e to the power minus 2t
dy over DT now if I substitute this dy
over DX and D 2 y over DX in the dy over
DT and X
as e to the power T over here in this
equation we will get this equation as
gay 2y over DT 2 plus a minus 1 dy over
DT plus B y is equal to 0 now we see
this is the second order linear
differential equation where the
coefficients are constants a minus 1 and
B both are constants so now we do know
that how to solve this equation and we
can get the solution however the
solution of this equation will be a
function of Y as a function of T this
this will be this will be that is Y as a
function of T wireless we want to a
solution Y as a function of X so what we
get solution as Y as a function of T
will substitute there T as log X and
we'll get the solution in the terms of X
there is one alternative method to get
that alternative method to solve a
low-cost equation see my question is X
square y double dash plus XY dash plus B
Y is equal to 0 you see here that the
coefficient of Y - is X and coefficient
of Y Double Dash is X square say if I
put Y as X to the power M we would be
able to we think is that it could
certain satisfaction Y we see that y
dash would be M times X to the power M
minus 1 y double dash would be M into M
minus 1 times X to the power M minus 2
so if Y is equal to X to the power M
this is a solution of this equation
these things must satisfy this equation
so let us substitute it if we are
substituting we would get M into M minus
1 X to the power M plus a times M X to
the power M plus B times X to the power
M is equal to 0 let us rewrite this
equation again as M square plus a minus
1 M plus B times X to the power n is
equal to 0 now if this X to the power M
is a solution of this equation we want
that is this should not
a trivial solution that is this should
not be zero for all X and if this is a
solution it must satisfy this equation
that is this must be 0 this will be 0 if
either X to the power M is 0 or this M s
square plus a minus 1 M plus B is equal
to 0 this cannot be 0 hence this would
be solution only M square plus a minus 1
M plus B is equal to 0 so we would get
that is this would be a solution if it
is satisfied what it says this equation
which we have gotten M this is called
characteristic our auxiliary equation
let us compare this method with the
previous method where we have changed it
to the linear differential equation with
constant coefficient in that one what we
have got that equation we have got d 2y
over DT 2 plus a minus 1 dy over DT plus
B y is equal to 0 to solve this we
require the characteristic equation what
is the characteristic equation
characteristic equation would be M is
square plus a minus 1 M plus B is equal
to 0 now let us compare what just now we
have got the auxiliary equation that was
also M square plus n minus 1 M plus B so
we have got that characteristic equation
for both days.the methods we are getting
the same and after that what we'll get
we'll get the solution as the roots of
this one so here if we are getting we
would get the solution as e to the power
M T kind of things and here we would get
X to the power M kind of thing and here
we have to get the solution in the terms
of X that is e to the power T I have to
substitute US x so again we will get X
to the power M and here also as X to the
power M so that says this both methods
will give me the same kind of solutions
so to see how we get the solution we see
the equation is M square plus a minus 1
M plus B is equal to 0 this is a
quadratic equation in M so of course it
will have two roots of this equation and
the solution would be depending on those
two roots and those two roots are
depending upon what we are getting the
tools so there would be three cases the
two roots M 1 and M 2 and we'll have
three cases the case one when I do have
both these M 1 and M 2 they are distinct
and real the second case would be that
is when this M 1 and M 2 are real but
they are equal that is double real loot
and the third situation will have when
this both M 1 and M 2 are complex
numbers so of course there would be
entire that is the complex conjugate
roots let us discuss the general
solution in each case one by one so the
case 1 distinct real roots when this
equation M is square plus a minus 1 M
plus B is equal to 0 has two distinct
real roots M 1 and M 2 then the two
solutions we would have as X to the
power M 1 and X to the power M 2 these
two solutions would be linearly
independent we can check this linear
independence using the wronskian do you
remember the Roskilde wronskian of the
two solutions Phi 1 by 2 is given as
determinant containing the solutions y 1
and y 2 and their derivatives that is
the determinant X to the power M 1 X to
the power M 2 and the second row M 1 X
to the power M 1 minus 1 and M 2 times X
to the power M 2 minus 1 what will be
the value of this determinant this would
be simply M 2 times X to the power M 1
plus M 2 minus 1 minus M 1 times X to
the power M 1 plus M 2 minus 1 that is M
2 minus M 1 into X to the power M 1 plus
M 2 minus 1 now this X is not 0 and
since M 1 and M 2 are distinct they are
not equal this will not be 0 for M 1 not
equal to M 2 that is when they are just
these two solution X to the power M 1
and X to the power M 2 there would be
linearly independent
hence the general solution would be C 1
X to the power M 1 plus C 2 X to the
power M 2 this we have discussed the
case 1 now go to the case 2 double root
that says is that when we do have this
equation M square plus a minus 1 M plus
B this has same real root that what will
be that root that root would be 1 minus
a by 2 so the one solution though would
get us X to the power M as X to the
power 1 by 2 1 minus C now to get the
second linearly independent solution
from here we will use the technique of
variation of parameter that is we will
take the second solution y2 as u times
y1 and then we'll find out what is my U
so by doing that technique we would get
actually write to us log X times y 1
that is you we have got log X so the
second solution would be log X into X to
the power 1 by 2 1 minus a and the
general solution would did it C 1 plus C
2 log X into X to the power 1 by 2 1
minus a so when this characteristic
equation has double root we do get the
general solution is of the form C 1 plus
C 2 log X X into to the power half 1
minus a now the third case complex
conjugate roots when this equation M is
square plus a minus 1 M plus B is equal
to 0 has complex roots they would have
appear in the pair's
that is let us say once first root is s
plus I T then the second root would be
of the form s minus item if these are
the roots then the general solution are
the two solutions we would get us first
as X to the power is cos T log X and the
second solution as X to the power s sine
T log X again we can find it out that is
this is the similar kind of thing as we
have done with the different
with constant coefficients this will be
linearly independent again you can check
it using the wronskian and the general
solution will be of the form X to the
power s c1 cos t log X plus c2 sine t
log X where this year this is the real
part in the complex root and T is the
imaginary part in the constant in the
conjugate their complex roots so we have
discussed these three cases let us do
some examples to understand this method
and how to solve this i'll akashi
equations let us do the first example
solve the ayakashi equation x is square
y double dash minus 4xy dash plus 6y is
equal to 0 we'll substitute Y as
function of X to the power M so we'll
substitute Y as X to the power M thus y
dash would be M times X to the power n
minus 1 and y double dash would be M
into M minus 1 times X to the power M
minus 2 if I substitute this in this
given equation we get M into M minus 1
times X to the power M minus 4 M times X
to the power M plus 6 times X to the
power M is equal to 0 now rewriting
these terms we get M square minus 5m
plus 6 times X to the power M is equal
to 0 so we will get the auxiliary
equation are the characteristic equation
as M square minus 5m plus 6 is equal to
0 we see the factors would be M minus 2
into M minus 3 so the roots would be M
is equal to 2 and see we do have two
real roots which are not equal that is
real distinct roots so the two solutions
would be y1 is X square Y to us X cube
so the general solution would be of the
form C 1 X square plus C 2 X cube this
is a solution of this differential
equation X square Y Double Dash minus 4
XY - plus 6y is equal to zero we can
check that this would be the solution so
if let's say they're checking this for X
square if I take X square in a solution
what will be y double dash that will be
two so I will get here 2 X square and y
dash would be 2x so I would be getting
minus eight X square plus 6 X's square
so that is satisfying this equation
similarly we can check with X cube R we
can check with this general solution it
has do one more example example to find
the general solution ax X square y
double dash plus XY - minus Phi is equal
to zero again we'll go with the same
kind of function that is y is equal to X
to the power M Y - as M times X to the
power M minus 1 y double dash as M into
M minus 1 X to the power M minus 2
substituting this in this equation and
finding out the auxiliary equation we
would get from here we could see that is
also very equation should be M is square
into a minus 1 so here is 1 so a minus 1
would be 0 so I would be getting the
auxilary equation as mu square minus 1
is equal to 0 of course it has two roots
MS one and one that is double root
EMAS one and one now this is the second
case so what will be my two solutions
the one solution Y would be X and
another solution would be x times log X
we can again check by putting the
solutions that they are solutions and so
the general solution would be C 1 X plus
C 2 X log X now let us discuss one more
example so example 3 solve the
initial-value problem x square y double
dash plus 4 x times y dash plus 9 Y is
equal to 0 with initial conditions y at
1 is equal to 2 and y dash at 1 is equal
to 0 so first we'll solve this equation
again by the same method that is y is
equal to X to the power M y dash as M
times X to the power M minus 1 y double
dash as M times M into my M
- 1 into X to the power n minus 2
substituting this we would get M into M
minus 1 plus M plus 9 is equal into X to
the power M is equal to 0 so we will get
the auxilary equation as M square plus 9
is equal to 0 now we see that this will
have complex so that is M square is
equal to minus 9 so I would have the
roots as plus minus 3i now if I compare
it with our original one that is s plus
I T so here the real part is 0 and the T
is 3 and we are having the two roots
plus 3i and
minus 3i so what will be our two
solutions our two linearly independent
solutions would be cause three logics
and sign three log X so the general
solution you would get as Y X plus c1
cos 3 log X plus a2 x sine 3 log X to
get the particular solution for the
initial value problem what are the
initial values initial values are given
that the function at value 1 is 2 and
the derivative at 1 is 0 so we get what
is Y - X from here that would be minus C
1 sine 3 log X into 3 by X plus C 2 cos
3 log X into 3 by X now if the first
initial condition is that X is equal to
1 so here if I put X is equal to 1 what
I will get log ik 1 would be 0 so I
would be getting here cos 0 and sine 0
cos 0 would be 1 and sine 0 would be 0
so what would be getting y at 1 we would
be getting is C 1 and that is given as
equal to 2 so this is what we have got
from the first initial condition in the
second initial condition we do have Y -
act 1 Y - at one that is here if I put X
is equal to 1 we will get this term to
be 0 and this Chango would be getting us
1 and this we would be getting as 3 so
what we would be getting y dash at 1 as
3 c2 which is given us to be 0 which
says is that c2 would be 0 so we have
got the two constant values that is c1
as 2 and C 2 as 0 so what will be my
particular solution here we'll
substitute the values of C 1 and C
- so the solution of initial value
problem would be 2 x cos 3 log X now you
can check that you can put this solution
in the given equation this will satisfy
the equation moreover this solution will
also satisfy these two initial
conditions so thus we have discussed the
Euler Cauchy equation of second order we
have learned that its solution is
depending on the roots of its auxilary
are the equation and that there would be
arising three cases now let us do the
practical life application of these
equations who will now discuss some
application linear differential equation
with constant coefficients have many
engineering applications here we will
discuss want such an application which
is the second order linear differential
equation in mechanical vibration here
what we are discussing very basic
mechanical system that is called a mass
on an elastic spring this is motion of
basic mechanical system this is mass on
an elastic spring and this is called
mass spring system what is this system
let us first understand this system we
do have any spring this isn't
unstretched spring if I hang a mass of
size M on this one you see this is
unstretched esprit I heard this is stick
to here on the one side and on this side
I have attached a mass of size this will
stretch this string so let us say this
is stretching is s naught now what I
will do is that is after this is
stretching this is at the static one so
if I pull this mass little bit downward
and then release it that is I pull this
mass downwards if I am pulling it again
this string will get more stretched 1
and then if I release
what will happen this will go up and
then it will go under the motion like
this zone that is where we are having
this that once I had pulled it down and
then left it it will go in a motion we
want to discuss or we want to find out
this basic motion of this spring so we
would see that is how we are going to
model this system model this system is
we will find out what are the
differential equations which are
governing this motion let us see how we
are going to do so we are going to
setting up this model for this mass spec
system here we have taken and spring
bitches come with it which would be
going to be the whether it will be
compost or it would be stretch it will
resist it that is this simple spring it
is a child attached with one hook on the
one side on the second side we had put a
massive size in know that will give it
is a stretching and this is stretching
it says that is this stretch is s not
here this is in this this is not in the
motion once I had put this mass over
there this will just hang on this
position I am taking is the initial
position this initial position and then
this spring has been this mass has been
pulled down if it has been pulled down
it will again is stretch it and then
when we are releasing it will go to the
motion so now will make certain basic
assumptions for setting up this motion
the first setting up is that we take
that motion should be strictly vertical
that is we are not talking about this
kind of motion
we will talk about only a strictly
vertical motion we are taking that
direction towards the down side as
positive and direction towards the
upside as the negative this is the
static condition that is from here
initially we are starting that is when
the mass has been
and that is this initial stretcher is
not we have taken and then we are
pulling it down and then releasing it so
after time T wherever this thing is that
is suppose it is here then this is the
displacement from the initial condition
Y is equal to 0-2 at Y T at time T know
how this system is being governed let us
see this system would be governed by
Newton's second law what is the Newton's
second law that says its mass into
acceleration is equal to force what is
the mass mass has been given us M
acceleration since we have seen in the
last one that is Y is the displacement
we do know if Y is the displacement at
time T is y T then velocity would be Y -
T that is the derivative of Y with
respect to T at time T that would be the
velocity and acceleration would be the
second derivative y double dash T so we
have got the mass we have got the
acceleration now what is the force let
us see the force the force which is
governing this complete system let us
see when this is unstretched spring when
a mass M is being attached it has been
stretched when it has been stretched it
says this there is a force acting
upwards say if not which is balancing it
so this is this force F naught would be
on the upward size and this is not
moving that is it is balancing the
weight of this mass what will be the
weight of this must by the Hookes law
gas force F naught would be actually
minus case not that is this is against
that is whatever there stretching this K
here is called the spring constant and
we are taking this negative size aside
because this is on the upward side this
is balancing the weight weight is M into
G now the system is in this static
equilibrium that it is not moving that
says this via
having this if not plus W that is minus
KS not plus mg this should be zero this
is not giving any motion this force is
zero now when it goes under the motion
that says is when we are pulling it and
then releasing it when we are pulling it
one more for that is it is being
stressed again and what is this is
stretching one this is stretching from
this initial condition why we have taken
zero now this is stretching extra
stretching is up right that will
generate another force f1 and what this
would be again by Hookes law this would
be minus KY again the sign would be -
since this forces upwards and the K is
the spring constant and why is this
stretching at the time T this is called
the restoring force moreover every
system does has a damper if the damping
is not there that is there should be
some force which is governing this
motion then this motion will go on
forever we will learn this little bit
after let's say that damping force is f2
this will again be in the upward
direction this damping force is going to
govern the velocity so that it would
reduce it so a very good approximation
of this f2 is - CY - that is a rider
that is the velocity and - see this see
this is called the damping constant and
again the sinus - since this would force
will also act upon on the upward side so
now the resulting force what would be
this see we are also taking as positive
this K is also positive the resultant
force would be f1 plus f2 so the total
force would be minus KY - CY - now we
are ready to get the model of this
system that is by second love Newton's
second law mass into acceleration is
equal to force masses M acceleration is
y double dash the force we have find it
out its - CY -
- KY that is we have got a differential
equation my double dash plus CY - plus
KY is equal to zero this is a second
order linear differential equations with
constant coefficients since all these
are constants so this is governing this
massive spring system this equation and
this is a homogenous one so we just says
that is what we do get how to get the
solution of this and how this motion we
are trying to understand so first we
will discuss the case of undamped system
undamped system that is there is no
damping is present that is the damping
constant C would be 0 that is what we
would get our equation our equation
would be M y double dash plus KY is
equal to 0 that is the term involving CY
- since C is 0 that is not present there
now here M and K both are positive so we
would get what will be our
characteristic equation for this linear
differential equation that is M lambda
square plus K is equal to 0 its loot
would be of course since both K and a
more positive its route would be complex
1 and those roots would be minus square
root of K by M I plus and minus so the
general solution would be of the form a
cos T times square root K bar m plus B
times sine K times the square root of K
by him t here I am taking as the time T
that is I'm having the equation in Y and
T this is the general solution now let
us see that is what this general
solution is let's see in the graph this
is called harmonic oscillation Y you see
here here we are having is that is for
different values of a and B so let us
say a is equal to 1 and B is equal to 0
but here I have taken em as 4 and K s 3
that is a spring constant as three and
masses 4 then what we are having is that
is this
motion first this black line you see
this is just going like this one now you
see what we are having is it's starting
the motion is being started that is
displacement is going on it goes up at a
distance then it goes down with the same
distance then it goes up again with the
same distance and you see the time it is
taking between the cycles that is from
here to here here to here here to here
the cycle is also same and it is also
taking that the displacement that is the
distance it is covering our displacement
it is making that is also equal that is
that is what I said is when we are not
having a damper the motion will go on
forever so this will go on forever with
equal magnitude similarly if I take my a
to be 0 and B is to be 1 then you see is
that again I am getting this harmonic
motion distinct kind of line again you
are having is that the cycle is same the
magnitude is also saying whether upwards
are the downwards and this will go on
similarly for the different values if a
and B I have taken as 1 we are getting
is that is the magnitude is differing
here and but cycle is there different
cycle different magnitude but it is
going on
similarly with other values as is 2 and
B as 3 we are getting it again that is
the magnitude this or getting higher one
on the cyclist again in the similar
manner smaller one and then if I had a
s4 and B as minus 5 we have got that the
magnitude is again little bit higher on
the cyclist something is a little bit
smaller and we are getting that this
will go on this is never going to down
that is the motion will go on forever so
let us see another example over here
here I have changed that mass to B 3 and
the spring constant case 5 that is it's
more resistant ISTE again you see the
similar kind of pattern we are having
again I have taken this a B as the
similar kind of thing that is a is equal
to 1 B is equal to 0 that is this first
one we are getting is this
this motion then if is equal to zero and
B is equal to one we are getting this
motion and so on we are having again we
are having as this motion is going on
with the equal cycle in one motion and
with the equal magnitude whether upwards
whether upwards are downwards that is
what we are calling it as harmonic
oscillation and without damping this
motion will go on forever forever time
let us see the damped system that is
when the damping is present so first we
see what is the equation and the
solution and then we'll discuss this
motion so the differential equation
would be my double dash plus CY dash
plus KY is equal to zero here this M is
the mass C is the damping constant and K
is the spring constant this equation
will have the characteristic equation as
lambda square plus C by M lambda plus K
by M is equal to zero since this is it
is a quadratic equation it will have two
roots lambda 1 and lambda 2 and those
roots would be of the form minus C by 2
M plus minus 1 by 2 m square root of C
square minus 4 M K that is we will have
one daughters minus C by 2 m plus 1 upon
2 m is square root of C square minus 4 M
K another note we would have minus C by
2 m minus 1 by 2 m square root of C
square minus 4 M K let us rewrite it as
in the simplified manner let us say this
first thing as the Alpha C by 2 m and
the second part that is beta as 1 upon 2
m square root of C square minus M K then
we would have the two roots lambda 1 as
minus alpha plus beta and lambda 2 s
minus alpha minus beta now you see in
the beta we are having this term square
root of C square minus 4 M K that is we
are finding out the square root of some
quantity where the C m and K all they
are constant so of course we will have
our roots will be depending upon what is
this
inside the square root that is C square
minus four and give whether it is
positive quantity it's zero quantity or
its negative quantity accordingly you
have different roots so we'll have
actually the three cases the case one is
when C square my is greater than 4mk in
that case we will have two different
root distinct real roots lambda1 and
lambda2 this case is called over damping
then second case when C square is equal
to 4 Mk will have double real roots this
is called critical damping well known
all these terms why we are calling them
and the third case when C square is less
than 4mk that is its imaginary part beta
would be imaginary and that is complex
conjugate roots this we would call under
damping let us discuss the solution in
each case one by one so general solution
in case one case one is when C square is
greater than four ng that is we would
have two roots as minus alpha minus beta
and minus alpha plus beta so the general
solution we would have YT as C 1 e to
the power minus alpha minus beta T plus
C 2 e to the power minus alpha plus beta
T here if I this alpha we do know is
that C by 2 m and beta is 1 by 2 MC
square minus 4 in case where m is the
mass C is the damping coefficient and K
is the spring constant now for different
values of C m and K we will get so that
the C square minus 4 M K is positive we
will get the different roots and if I do
give some initial conditions so that we
get some particular solution so we'll
discuss this case one by one here what I
have taken I have taken some values of C
m and K such that C square minus 4 m
case positive an initial condition what
I have taken is that is y at 0 is 4 that
is it is positive initial displacement
then we do have the another initial
condition that
would be governing why - that is why -
whether it is I have defined some values
and I'm classifying them into three
parts that is my velocity initial
velocity y - is positive negative R 0
let us see first case Y 1 this y 1 we
are having is here as positive velocity
initial that's why - 1 is positive I
have taken someone so in that case what
how this motion will go on initially it
has been displaced that is pulled
downwards for at 4 then it has been
released it has generated and with the
damping it has generated a velocity
positive velocity so we are getting it
as that is it's going little bit up and
then it is dying down because of damping
if initial velocity is 0 then it is
started dying down from the initially it
has not gone up when the initial
velocity is negative this initial
velocity is upwards then we are having
route could have the two cases one is
that is it is going down like this one
and then it is dying down to the
displacement is coming to be 0 or it may
happen that comes to 0 and then it goes
to the negative one that is we are
having is this motion upwards but again
it will come up and it will go to the 0
so if the damping is present ultimately
will have that our motion would stop and
after some time we would go to the 0
displacement now if I take initial
displacement to be the negative that is
if y 0 is negative so again I have taken
a case when y 0 is minus 4 again with
the second initial condition about the
velocity again I have divided positive
negative and 0 now if the displacement
is negative and my velocity is also our
velocity is also negative that is the
last case Y for this case so you see is
that is it is first going down little
bit and then coming up and it will
ultimately reach to the 0 that is the
displacement would be 0 if the initial
velocity
zero it is starting with the negative
displacement and the displacement is
going up and that is it's reducing and
it will certainly reach to somewhere
zero if initial velocity is positive we
could have again the two cases one case
is that is it has a start a little bit
that is the displacement has gone very
up like this one that is reduced
certainly and then the reduction power
is little bit less and then we are
moving over here we may have this in
this positive case that is it may go to
the 0 and then it crosses this line then
it goes through to the positive
displacement and then it is dying down
so again we have got that when damping
is present we will not have this
harmonic kind of oscillation that the
motion goes on forever rather we would
be having is that after some time the
motion will die down and their
displacement would be zero now let us
discuss the second case that case 2
where C 2 is equal to 4 M K that is beta
would be 0 and we will have double root
that is both lambda 1 and lambda 2 they
are equal to minus alpha in this case we
do know that the solution would be of
the form C 1 plus C 2 T times e to the
power minus alpha T alpha is of course
our C by 2m
in this case if I take some values of C
and M such that C C square is 4 M K
again I have taken initial condition
that my initial displacement Y is
positive and that is 2 again with the
second initial condition about the
velocity we have divided it into
positive negative R 0 so when the
displacement is positive and velocity is
also positive this is the second case
where I am having C square is equal to 4
M K that is we are having double real
root again it is going little bit up and
then it is coming down so of course
after sometime it will come down when
initial velocity is 0 it is starting
from here it is going down very slowly
but it's going down then when the
initial velocity is negative it is going
down first it goes down very fast then
it is going down slowly and it will go
to the
zero it may have another case where it
is going down to the zero and then it is
going down to the negative displacement
then it will come up and it will go to
the zero that is again in this case also
the motion will die down that is the
displacement would be zero after
sometimes the motion will stop
let us discuss them discuss the third
case case three is when C square is
great less than four m'kay that is the
beta is pure imaginary let us say beta I
Omega star where this Omega star is 1
upon 2 m square root of 4 MK minus C
square that is this thing is positive so
Omega star would be positive in this
case the general solution will be of the
form e to the power minus L 4 T a cos
Omega star T plus B sine Omega star T by
the trigonometry we do know that this
thing we can rewrite as C times e to the
power minus alpha T cosine of Omega star
t minus delta we have this C is such
that C square is a square plus B Square
and this Delta is such that tan of Delta
is B upon a now we see this is the
general solution that is this is telling
us the displacement at time T this is
having eetu the power minus alpha T and
cosine of something now the cosine of
any from this cosine function we do know
this lies between minus 1 and plus 1 so
what will have this motion will lie
between minus C to the power minus alpha
T and plus C e to the power minus alpha
T let us see what kind this motion would
be look like you see here this is the
line C to the power minus l4 T and this
is the line minus C e to the power minus
alpha T again I have taken this some C
and M such that C square is less than 4
Mk you see this motion is actually this
one
is going first it's starting with here
initially and then it is going down then
it is coming up and like this you see
each time the magnitude of the motion
you are getting is getting lesser and
lesser and afterwards it is going to be
died down it's coming it to zero so this
is what is the motion when we do have
the underdone and you see is that is
when we had over damping we had got the
motions which are just going down when
we had critical damping again the
motions were just going down and here
what we are having is that motion is
going down and up and down and up and
then they are dying down that is they
are is talking so this is what is called
the under damping so we have discussed
the three cases of this differential
equation which is governing this motion
of basic mechanical system that is
called the mass spring system now it may
happen that is this motion is being
governed by some force that is we are
calling forced oscillation our resonance
what is this this says is that this
motion is being governed by if force are
a force is affecting the motion that
means till now what I was having that is
this was the equation of the motion
which was equal to zero that is this was
balanced by balanced equation now what
we are having is that this motion is
governed by one force called our team
that is now we are having
non-homogeneous equation my double dash
plus CY dash plus KY is equal to R T now
what is R T which is governing this
motion this R T is called input and what
will be the solution of this equation
why that would be called output this is
we are giving is that is sometimes we
say is that this is this is what is the
input and R this is what the force is
being applied on this output we are
calling the response to this force which
is governing this motion so this is also
called the response for example it
take this my Artie that is the force
which is governing the motion is of the
form F naught cos Omega T this Omega is
something different so f naught is a
constant and the forces of this form you
see I have taken this special form that
is cosine function if you do remember we
have to solve this non-homogeneous
equation where this both F naught and
Omega positive then what will have the
equation our differential equation would
be my double dash plus CY dash plus KY
is equal to if not cos Omega T now this
is a non-homogeneous equation it has
solution which is having the solution of
homogeneous equation and a particular
solution which is governed with this
right hand side that is called the
particular solution and then the general
solution we do have as the homogeneous
solution plus the particular solution so
we do have this differential equation
its general solution is of the form Y H
plus YP where Y it is the general
solution of Associated homogeneous
equation M y double dash plus CY dash
plus KY is equal to zero and YP is the
particular solution now we'll
concentrate on this white people if you
do see that is what we are having here
is that our RT is of the form if not cos
Omega T that is we are having a special
form so we'll use this method of
undetermined coefficient do you remember
that method we are having one table
which is telling us how what is the
choice of the particular solution YP
that table is this one so here what we
are having is we are having a constant
times cos Omega so my choice of
particular solution should be of a times
sine Omega X plus B times cos Omega X it
is so we'll have our particular solution
as of the form a cos Omega T plus sine B
sine Omega T if this satisfies the
equation then it will be a solution so
view how to find out
YP - as minus Omega s in Omega T plus
Omega B cos Omega t + YP Double Dash
would be minus Omega square a cos Omega
t minus Omega square B sine Omega T will
substitute this in the given equation
the equation is my double dash plus CY
dash plus KY is equal to F naught cos
Omega T when I am substituting those
wipey wipey - and by P Double Dash who
would get the equation of the form K
minus M Omega square a plus Omega C B
and this whole is multiple of cos Omega
T and then we do have a sine Omega T and
it's coefficient as minus Omega C a plus
K times plus K minus M Omega square
times B which is equal to F naught cos
Omega T to determine this a and B we
would equate these coefficients so what
it gives us that is the coefficient of
cos Omega T on the right side is f
naught and on the left side is this one
so we get the first equation as K minus
M Omega square a plus Omega C B is equal
to F naught and we don't have any term
of sine Omega T on the right side so we
will get the second equation as minus
Omega C a plus K minus M Omega square B
is equal to 0 these are the two
algebraic equations in two unknowns a
and B so we can solve it and we get the
solution so we will get what is the
particular solution homogenous equation
will give me the solution of a by H and
we will get the general solution now we
will discuss one special case rather
which is called damped force oscillation
R which is more generally known in the
mechanical engineering as vibration what
is this one
this governing differential equation
would be my double dash plus CY dash
plus KY is equal to F naught cos Omega T
I am just discussing with this special
one that is f naught cos Omega T
associated homogeneous equation would be
my double dash plus CY dash plus KY is
equal to zero it's solution let us say
suppose that this equation has a complex
OOTS
that is it's a characteristic equation
has complex roots then the solution
would be e to the power minus l4 t a cos
Omega T plus B sine Omega T where this
alpha would be C by 2 m and Omega would
be 1 upon 2 a misfire Ruta for MK minus
C square now this is what we are writing
and then the particular solution we
would be general solution would be of
the form Y H plus YP this general
solution is also known in this vibration
case as the transient solution and this
particular solution YP this is known as
a steady-state solution
why this is called transient solution
and this is a steady-state solution this
we will learn a little later on and we
would actually learn these things with
the help of an example this example we
will do in the next lecture where we
would really model or trial in the
vibration model and we will see that is
how the solution is coming at wow this
vibration is taking place and why we
call one thing as transient solution
this general solution and this
particular solution has a steady-state
solution how this is a response to the
input function all these things we will
see in the help of example and that
example will go with the next lecture
today we have learned a special kind of
equation called Euler Cauchy equations
this is second order linear differential
equations where the coefficients were
not constant but the function of X we
had learned homogeneous Cauchy Euler
equations how to solve them we had
learned some application of linear
differential equation with constant
coefficient we had learned one special
application in the simple mechanical
system called the mass spring system and
we had learned when this there is no
force is governing the motion both the
cases damp is that when there is no
damping and when there is damping who is
present so we have seen the damped
oscillations and we have seen the
undamped motion that is called harmonic
oscillation now next lecture we will
learn one more system mechanical
vibration thank you
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