WEBVTT
Kind: captions
Language: en
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hello friends welcome in this lecture i
will discuss solution of second order homogenous
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linear differential equation with constant
coefficients this topic will be covered
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in two lectures we will cover in
this lecture and the next lecture now let
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us define a homogenous linear differential
equation of second order with constant coefficients
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this such an equation is given by y double
dash x plus a y dash x plus b y x equal to
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zero where a and b are some real constants
and x is any real number belonging to r
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now let us look at this we shall be finding
solution of this second order homogenous
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linear differential equation by with the help
of this solution of first order linear differential
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equation with constant coefficients so let
us look at this equation y [da/dash] dash
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plus k y x equal to zero we know that it possess
a exponential solution y dash x plus k y x
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equal to zero you can write as or we can write
it as
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now integrating both sides
so we can write l n mod y equal to minus k
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x plus some constant c one or we can write
y equal to e to the power minus k x into a
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constant c so y equal to c times e to the
power minus k x here c is an arbitrary constant
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now the question arises whether exponential
solutions exists for homogenous linear higher
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order differential equation or not now so
we will will see that the solutions of higher
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order homogenous linear differential equations
are also either exponential functions or
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they are constructed out of exponential function
let us substitute y equal to e to the power
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x e to the power m x in the second order homogenous
linear differential equation with constant
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coefficients
y double dash plus a y dash plus b y equal
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to zero this our second order homogenous linear
differential equation with constant coefficients
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a and b so let us put y equal to e to the
power m x in this
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we get y dash equal to m times e to the power
m x y double dash equal to m square e to the
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power m x so substituting the values of y
y dash and y double dash in the equation y
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double dash plus a y dash plus b y equal to
zero we get m square plus a m plus b into
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e to the power m x equal to zero now since
e to the power m x is not zero is never zero
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for any x belonging to r we have m square
plus a m plus b equal to zero which is a quadratic
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equation of second order in m this equation
is called as the characteristic equation or
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auxiliary equation so by obtaining the roots
of this equation say the roots are m one and
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m two we can say that we get solutions
y equal to e to the power m one x and e to
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the power m two x of the given differential
equation
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now there there arise the following values
of m one and m two so this gives you m one
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equal to minus a plus under root a square
minus four b y two and m two equal to
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now whether m one and m two are distinct or
they are complex conjugate or they are
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equal to dependents on the discriminant and
[discri/discriminant] discriminant of the
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given second order equation m square plus
a m plus b equal to zero so there arise three
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cases we shall get two distinct real roots
provided the discriminant a square minus four
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b is greater than zero two complex conjugate
roots provided a square minus b four b is
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less than zero and a real double root when
a square minus four b equal to zero so we
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shall discuss these three cases one by one
in the case of two distinct real roots
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we find two solutions corresponding to
the two values of m one and m m so y one
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equal to e to the power m one x and y two
equal to e to the power m two x now y one
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and y two are linearly independent because
y one upon y two is equal to the e to the
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power m one minus m two into x since m one
is not equal to m two e to the power m one
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minus m two into x is not a constant and therefore
y one and y two are linearly independent so
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in this case the general solutions of the
equation is y equal to c one e to the power
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m one x plus c two e to the power m two x
for example let us consider let us consider
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y double dash plus y dash minus two y equal
to zero
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so the auxiliary equation will be m square
plus m minus two equal to zero and we can
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then see that m equal to two m square plus
m minus two equal to zero so we can write
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m square plus two m minus m minus two equal
to zero this will give you
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the roots m equal to minus two and one which
are distinct roots and so the general solution
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of the given equation will be
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y equal to c one e to the power minus two
x plus c two e to the power x
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now let us consider the case of two complex
conjugate roots so lets say m one is equal
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to alpha plus i beta and m two equal to alpha
minus i beta be two complex conjugate roots
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of the auxiliary equation m square plus a
m plus b equal to zero where alpha and beta
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are real numbers and beta is not equal to
zero then the general solution will be
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since in applications [be/we] work with
real solutions therefore instead of complex
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exponentials here let us try to find the real
solutions of the homogenous second order
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linear differential equation so let us recall
the eulers formula the eulers formula tells
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us that e to the power i theta is cos theta
plus i sin theta so e to the power alpha plus
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i beta into x can be written as let it be
y one and y two be e to the power alpha minus
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i beta into x then we can write it as
so adding y one and y two we get
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and also y one minus y two we have
now we have already seen that in the case
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of a homogenous linear differential equation
of nth order if y one y two y k are solutions
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of that equation then c one y one plus c two
y two and so on c k y k is also a solution
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so y one plus y two by two is a solution of
the equation y double dash plus a y dash
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plus b y equal to zero this is our second
order differential equation so if y one and
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y two are solutions of this then y one plus
y two by two is also a solution and y one
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minus y two by two i is also a solution so
e to the power alpha x cos beta x is a solution
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of this differential equation and e power
alpha x into sin beta x is also a solution
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of this differential equation further more
we note that e to the power alpha x cos beta
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x divided by e to the power alpha x sin beta
x is not a constant so e to the power alpha
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x cos beta x and e to the power alpha x sin
beta x are two linearly independent solutions
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of this differential equation
so since e to the power alpha x cos beta x
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upon e to the power alpha x sin beta x is
not a constant
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the general solution in this case will be
y equal to c one cos beta x plus c two
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plus sin beta x into e to the power alpha
x so in this case of two complex conjugated
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gate roots m one equal to alpha plus i beta
and m two equal to alpha minus i beta the
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general solution of the differential equation
y double dash plus a y dash plus b y equal
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to zero will be written as e to the power
alpha x into c one cos beta x plus c two sin
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beta x
now we go over to the an example on this
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so this is eulers formula the general solution
is given by this one y equal to which we have
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already seen c one and c two are two arbitrary
constants let us now take up the problem
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y double dash minus two b this is an initial
value problem we are given a homogenous second
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order linear differential equation with constant
coefficients along with two initial conditions
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at x equal to zero y is given as four and
at x equal to zero the derivative of y is
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given as one so y double dash minus two y
dash plus ten y equal to zero so these two
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initial conditions will determine an unique
solution of the homogenous linear differential
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equation of second order so here auxiliary
be auxiliary equation is m square minus two
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m plus ten equal to zero so we can find m
so this is two plus minus six i at divided
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by two which is equal to one plus minus three
i
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so if you compare it with a alpha plus minus
i beta alpha is one and beta is three so we
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can write the general solution
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y equal to e to the power x into c one cos
three x plus c two sin three x where c one
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and c two are two arbitrary constants now
with the with the given initial conditions
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we shall determine the values of c one and
c two so let us first take y zero equal to
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four when y zero equal to four we get putting
x equal to zero here we get four equal to
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c one because when x is equal to zero cos
three x is one and sin three x is equal to
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zero and e to the power zero is equal to one
so we have got the value of c one now let
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us use second initial condition to determine
the value of c two so we will have to put
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c one equal to four here so we can say
thus y equal to four cos three x plus c two
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sin three x into e to the power x
let us differentiate it with respect to x
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what do we get y dash equal to minus twelve
sin three x plus three c two
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the derivative of e to the power x is e to
the power
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now let us put x equal to zero in this
so x equal to zero when you put sin
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three x will be zero cos three x will be one
so we get three c two this is one here when
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you put x equal to zero this is four this
is zero so three c two plus four now y dash
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zero is given equal to one so we get c
two equal to minus one and thus we have
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y equal to corresponding to the two initial
conditions we have got the values of c two
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arbitrary constants c one and c two so this
is e to the power x times four cos three x
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so this how we find the solutions of this
initial value problem
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thank you