hello friends welcome to my second lecture on solution of second order homogenous
linear differential equation with constant coefficients we are discussing the solution
of second order homogenous linear differential equation with constant coefficients
that is given by y double dash plus a y dash plus b equal to zero b y equal to zero the
characteristic equation here
is given by m square plus a m plus b equal to zero let us find the roots of this quadratic
equation m equal to minus a plus minus under root a square minus four b divided by two
so there are two values of m we have three cases case one distinct roots case two
complex conjugate roots and then case three double root
so the case one when the two roots of the characteristic equation are distinct we
have already covered in previous lecture and the case two where the two roots are complex
conjugate that also we have covered in the previous lecture
so in this lecture we begin with the case three let us say that the root of the
quadratic the characteristic equation has double root that is in the case three
the discriminant a square minus four b equal to zero in this case the discriminant a square
minus four b equal to zero so we have m equal to minus a y two which is a double root now
this double root at first gives us one solution of the second order linear differential
equation one we can say the one solution is one solution of the given solution
one we can write as y one equal to e to the power minus a by two into x now to obtain
the other independent solution by two we apply the method of variation of parameters
this method of variation of parameters was given by lagrange by by given by lagrange
for first order linear differential equation with first order linear differential equation
so in this method what we do is let us assume the other solution by two x to be equal to
u x into y one x we shall find out a x so that y two is the other independent solution
of the given differential equation so let us assume that y two x is equal to
u x into y one x is the solution of equation one then we shall have let us substitute
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
into y one dash x and y two double dash x equal to u double dash x into y one x plus
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
x or we can say this u double dash x into y one x plus two times u dash x into y one
dash x plus u x into y one dash double dash x now lets substitute the values of y two
y two dash y two double dash in equation one so substituting y two and its derivatives
in one we have u double dash x into y one x plus two u dash
x into y one dash x plus u x y one double dash x plus a times y dash is y two dash so
y two dash is equal to u dash x into y one x plus u x into y one dash x plus b times
y y two and y two is u into y one now let us collect the coefficients of the
derivatives of u so we may write this equation as u double dash
the coefficients of u dash x are two y one dash x
plus a times y one x the coefficient of u x is y one double dash
x plus a y one dash x plus b y one x equal to
equal to zero now since y one is a solution of equation one
we get y one double dash x plus a y one dash x plus b y one x equal to zero also y one
x is equal to e to the power minus a by two into x so let us differentiate it with respect
to x it will give me which is equal to minus e by two y one or
we can say two y one dash x plus a y one x equal to zero so substituting these values
so [substitute/substituting] substituting these values here it reduces we get u double
dash x into y one x equal to zero the coefficient of u dash x is zero from here the coefficient
of u x is zero from here so u double dash x into y one x is equal to zero since y one
x e e to the power minus a y two into x which is not equal to zero for any x belonging to
r we have u double dash x equal to zero
so when we integrate this twice we get which implies u x equal to x thus the other solution
of the equation one is given by
x into e to the power minus a by two into x now so this is this is for one solution
and then we to find the other solution we applied the method of variation parameters
where we assumed y two x equal to u x into y one x now this so the now let us see that
y one and y two we see that y one over y two is equal to one over x which is not a constant
so the y one and y two are independent of two each other
so we got two independent solutions of the equation one and therefore we can write the
general solution as y x equal to c one plus c two x into e to the power minus a by two
into x where c one and c two are arbitrary constants
now let us take an example on this first due to gravity which will at downward will be
m g so let us take f one equal to m g now due to the mass m that is due to the gate
m g the there will be a extension in this spring
before it comes to the equilibrium position so let us say in the equilibrium position
the force due to gravity that is f one equal to m g will have downward while the spring
force the spring force will be lambda s naught where lam this spring force is
when lambda a spring modulus and s naught is the extension r stretch in the spring
so lets say this is s naught this is spring force will let in the direction of opposite
to m g so in the equilibrium position m g and lambda
is not will balance each other and we shall have
m g equal lambda s naught here we are using the hooks law which says that the restoring
force of the spring r we can also call it as the spring force is proportional to the
stretch or extension of the spring so here there is a stretch or extension as not in
this spring so the spring force are restoring force is proportional to s naught
and therefore we write it as equal to lambda s naught by lambda is constant or proportionality
we call it as a spring modulus now let us let us pull the mass m downward
and then release it should be mass m is pulled down and then released ok then the there
will be a vertical motion in the spring the forces that will be obtained let say at some
time t during the motion of the spring y may be displacement from the equilibrium
position at time t so then the forces that will be acting on the mass m will be
the force due to gravity m g downwards f one equal to m g f two equal to this spring
force which will be lambda times s plus s naught plus y because s naught plus y is the
total extension in the a spring so this lambda s naught plus y will act in the direction
opposite to m g the resultant force therefore is
since m g is equal to lambda s naught we have resultant force as minus lambda y now let
us apply the newtons second law newton by newtons second law resultant force is is equal
to m times d square y over d t square so we have the equation of motion as
m by double dot equal to minus lambda by r where y double dot means d square y over d
t square now we can also write it as
so if we denote omega naught let us denote by omega naught square lambda by m then we
have the auxiliary equation is let me not write m ok so alpha square plus omega naught
square equal to zero and now since this equation has two complex roots so we have the general
solution of equation one as by t equal to c one cos
omega naught t plus c two sin omega naught into t now it can also be expressed as c times
cos omega naught t minus delta where c is equal to under root c one square plus c two
square and ten delta is equal to c two by c one it can be easily checked now if y naught
is equal to if if we impose the initial conditions in on this motion as y suppose y zero
is equal to y naught that means when we release the mass m at that time at that time t
equal to zero the displace the displacement of the mass m from the equilibrium position
is y naught and the velocity at that instant is given by y one then we have then we
can see from here then put t equal to zero in this we get y zero equal to y naught y
naught is equal to c cos delta and from y t equal to c cos omega naught t minus delta
if you use y dash zero equal to y one you get y one equal to c omega naught sin delta
so from these two equations y naught equal to c cos delta and y one equal to c omega
naught sin delta we can find the value of c c square is equal to y naught square
plus y one square over omega naught square and delta equal to ten inverse y one over
y naught omega naught so the two constants c and delta can be found from the initial
conditions y zero equal to y naught and y dash zero equal to y one or we can also put
c as one over omega naught under root y one square plus omega naught square y naught square
until delta equal to this now we can see here that since y t is equal
to c cos omega naught t minus delta the the motion is a harmonic oscillation now let us
go to damped oscillation so in the case of dumped oscillation suppose the mass is connected
to a dash pot
this is dash pot
so suppose the mass is connected to a dash pot then we have to consider the viscous force
due to the viscosity of the liquid in the dash pot now the dumping force due to
the viscous liquid x in the direction opposite to the instantaneous motion and is proportional
to the viscosity so the dumping force is proportional to the viscosity when the
when it is small so we we have then we have then we have the third force which
is f we have taking as f three f three equal to minus c y dot y dot is the velocity at
instant t so the force due to the viscosity of the liquid which is there in the dash pot
there will be a force which is called dumping force it x in the direction opposite to the
instantaneous motion and this c is a dumping constant this c is a dumping constant
now lets us see a so in the now the resultant force will be
f one plus f two plus f three which is equal to f one plus f two we have seen is minus
lambda y so minus lambda y plus minus c y naught dot thus we get the equation motion
as n y double dot plus c y dot plus lambda y equal to zero the auxiliary equation will
be given by alpha square plus c by m into alpha plus lambda by m equal to zero the two
roots of this auxiliary equation are given by alpha one and alpha two the alpha one is
minus c by two m plus one by two m under root c square minus four m lambda and alpha two
is minus c by two m minus one by two m under root c square minus four m lambda
now due to the discriminant c square minus four m lambda there arise three cases
the case one is c square is greater than four m lambda in that case the two roots alpha
one and alpha two of the auxiliary equation will be distinct so now we have three cases
in the case one when c square is greater four lambda we have the case of over dumping because
the dumping constant c is high here so in this case the general solution is given by
y t equal to c one e to the power minus p minus q into t plus c two e to the power minus
p plus q into t now where p is equal to c by two m and q equal to one over two m under
root c square minus four m lambda now from here we can see that p is is strictly greater
than q and so when t goes to infinity e to the power minus p minus q into t goes
to zero as well as e to the power minus p plus q into t goes to zero so in the equation
one when y t goes to zero s t goes to infinity that is the system comes to rest after a sufficiently
long time in equation two we see that y t is given by c e to the power minus p t cos
omega bar t and therefore the motion is a harmonic oscillation the displacement curve
y t lies between y e equal to c times e to the power minus p t and y equal to minus c
times e to the power minus p t so this is let us say t x is this is y t these
are displace these are the curves y t equal to c e to the power minus p t and y t equal
to minus the displacement curve lies between these
two curves and touches these curves when omega bar t minus delta is a interior multiple of
pi moreover the frequency here is frequency of oscillation is equal to omega bar over
two pi so when omega bar is more frequency is more and when let us see when omega
bar will be more omega bar will be more when c is smaller when c is smaller here and
it will be greatest when c tends to zero so when c tends to zero
omega bar goes to one over two m under root four m lambda which is equal to under root
lambda by m which is equal to omega naught the omega naught which be found in the case
of the simple harmonic oscillation this omega naught y t equal to c cos omega
naught here omega naught you can see omega naught is root lambda by m
now let us discuss the last case in the equation three you can see we have y y
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
not equal to zero y t can never be zero except possibly once when a plus b t is equal to
zero that is t is is equal to minus a by b hence the motion can have at most one passage
through the equilibrium position now if the initial conditions are such that a and b are
positive then you can see y t can never be zero because if a and b are positive then
y t can never be zero so the displacement will never be zero and this case is similar
to the case one in the case one you can see y t tends to zero as t goes to infinity so
the system comes to rest after sufficiently long time
thanks