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