Linear Differential Equation of Second Order-Part-1 Video Lecture | Calculus for MAT

welcome to lecture series on
differential equations for undergraduate
students
today's topic is linear differential
equation of second order till now we had
learned about the first order
differential equations and the solutions
now we will go for higher order so we
will start with second order and we will
start with linear differential equations
of second order what are they a general
form is y double dash plus PX y dash
plus QX times y is equal to RX here y is
the unknown function y dash is first
derivative of y and y double dash is
second derivative of y we are seeing
here is that is the highest order
occurring in this equation is the second
order so this is a second order equation
moreover we are finding it out that Y y
dash and y double dash they are
occurring in first degree only are
separately none of the terms contain the
two things simultaneously so this is a
linear equation now right hand side are
X and the coefficients px and QX they
are function of X only so we have this
as a so an equation if it can be written
in this form this is called the linear
differential equation of second order
here if this right hand side is 0 then
we call this a homogeneous equation if
Rx is any function of X we call it
non-homogeneous moreover this px and QX
they are called the coefficients of the
equation let us see some examples of
second-order equation for C homogeneous
linear equation 1 minus X square y
double dash minus 2xy - plus 6y is equal
to 0 we see here is that is why y - and
y double dash they are occurring in
first degree and separately the
coefficients are 1 minus X square - x +
6 they are none of them are containing Y
and the right hand side is 0 so this is
a homogeneous linear equation of course
this is not in a standard form let us
see another example non homogeneous
linear equation y double dash plus 4y is
equal to e to the power minus x times
sine X again we see here that Y and Y
Double Dash they are occurring in first
degree and separately and the right hand
side is a function of X which is not 0
so this is non-homogeneous and linear
let us see some example of nonlinear
differential equations see here the
first example x times y double dash y
plus y dash square plus 2 y - y is equal
to 0 we see here that y double dash and
why they are occurring in the same term
again y dash is having the thickened
degree so we and here again we do have y
dash and why that is occurring in the
same term so this is non linear now let
us see another example y double dash is
equal to square root of y dash square
plus 1 this is not in a rational form so
if I rationalize it I do get y double
dash square minus y dash square minus 1
is equal to 0 we find it here out that
of course this equation is of second
order but both y dash and y double dash
are occurring in the second degree so
this is not a linear equation so both of
these are the examples of nonlinear
equation now we will try to learn about
first about the homogeneous linear
equations so let us first revisit some
of the definitions which we already have
in the first order and in general we'll
revisit in the terms of this homogeneous
linear differential equations of second
order so first definition we will see of
solution a solution of second order
differential equation is a function YX
that has first and second derivatives as
y - x and y double dash X and satisfies
the given differential equation for all
X in a given interval that function we
will call a solution let us see one
example let us take a differential
equation y double dash minus y is equal
to 0 this is a second-order equation we
can see that e to the power x + e to the
power minus x these two are the solution
of this equation why there would be
solution if I substitute y Y - and y
double dash in this equation they should
satisfy the equation so let us see it
one by one first take Y is equal to e to
the power X it says y dash would be also
e to the power X as well y double dash
will also be e to the power X so we are
getting y double dash and why they are
same so we will satisfy the equation y
double dash minus y would be 0 so this
is a solution similarly we can check
with e to the power minus 6 what will be
its derivative minus e to the power
minus X second derivative e to the power
minus X again we are getting the second
derivative and the function they are
same so their difference would give me 0
that is again it is satisfying our
equation so both our solution of this
differential equation moreover we can
see if I take a 2 times e to the power X
and B to the times e to the power minus
X
they are also solution of the same
equation you can check it here moreover
if I take a linear combination of these
two solutions that is a times e to the
power X plus B times e to the power
minus 6 this will also be
solution of this equation let us try to
see this function our equation is y
double dash minus y is equal to zero
what we want to check that a times e to
the power X plus B times e to the power
minus X is a solution so we have to
substitute the function and its
derivative
what's first derivative a times e to the
power X minus b times e to the power
minus 6 second derivative a times e to
the power x plus b times e to the power
minus x again we find out that the
function or its second derivative they
are same so their difference would be
zero that says it is satisfying this
given equation so we have got actually
what we have seen if a function is a
solution of a given differential
equation its constant multiple is also
solution of its this equation moreover
if two functions are solution of a
differential equation the linear
combination was also becoming a solution
of differential equations so this
differential equation was homogeneous
now from here we get our first result
that is the basic property of the
fundamental theorem about the
homogeneous linear differential equation
of second order for a homogeneous linear
differential equation y double dash plus
px y dash plus Q XY is equal to 0 any
linear combination of two solution of
this equation is again a solution or in
other words if y1 and y2 are two
solution of the given equation that is y
double dash plus px y dash plus QX Y is
equal to 0 then c1 y1 plus c2 y2 that is
the linear combination of both solution
will also be a solution where the c1 and
c2 are any arbitrary constants
from here we've just come to the general
solution for a second order homogeneous
linear equation the equation is again
same in the standard form y double dash
plus PX y dash plus QX y is equal to
zero a general solution will be of the
form c1 y1 plus c2 y2 where y1 and y2
are two solutions of this given
homogeneous equation now and c1 and c2
are any constants then more particular
salt initial value problem a initial
value problem will have the equation
that y double dash plus PX y dash plus
QX y is equal to zero and two initial
conditions y x naught is equal to k
naught and y dash at x naught is equal
to k1 if you do remember in first-order
equation we had only one initial
condition now it is a second-order
equation and we will have two initial
condition that is the value of the
function at a particular point and the
derivative of the function at that
particular point the solution of these
two conditions if I use in the general
solution we cannot attain the solution
of this particular problem or this
particular solution of initial value
problem again let us see this by example
again I have used the same equation y
double dash minus y is equal to zero and
the two initial conditions are Y at 0 is
4 and the derivative at 0 is minus 2 see
we had already known that this equation
we had seen that e to the power x and e
to the power minus X was two solutions
so the general solution will be of the
form C 1 times e to the power X plus C 2
times e to the power minus X now we will
use the initial conditions that ad 0
what will be at 0 0 it would be C 1 1
plus C 2 that is C
plus c2 and the condition given is that
y 0 is 4 so we have got the first
equation that c1 plus c2 is equal to 4
second was the derivative at 0 what is
the derivative of this c1 e to the power
X minus C 2 e to the power minus X now
if I put at X is equal to 0 I would get
it again as c1 minus c2 which is given
as minus 2 so what we have got we have
got these two equations in two unknowns
these are linear equations we can solve
it and what will be the solution that c1
is 1 and C 2 is 3 so what is the
particular solution we have got that Y
is equal to e to the power X plus 3
times e to the power minus X that is I
have put the values of c1 and c2 in the
general solution we can check see the
graph of this function also you see this
is the graph of the function e to the
power X plus 3 times e to the power
minus x we see that here at 0 the value
of this function is 4 and the slope of
this at this point if you see the slope
would be this way so that is the
negative minus 2 slow now let us do some
more experimentation in this example let
us say I choose these two as a solution
that is one solution e to the power X we
do know that a times e to the power X is
also a solution so I have chosen here 4
times e to the power X this will also be
a solution so now I have chosen the two
solutions y 1 and y 2 as e to the power
x + 4 e times e to the power X if I am
choosing these two as a solution what
will be the general solution that would
be c1 times e to the power X plus 4 c2
times e to the power X now I would like
to find out the particular solution
using those initial conditions so I
would find out what is y - that would be
C 1 e to the power X plus 4 times C 2 e
to the power X now put the initial
condition at 0 it gives C 1 plus 4 C 2
and y dash at 0 will give me again C 1
plus 4 C 2
patience given up y0 is four so I have
got the first equation C 1 plus 4 C 2 is
equal to 4 and second equation C 1 plus
4 C 2 is equal to minus 2 now you see I
have got two equations which are not
consistent that means I cannot find out
the solution of this I cannot find out C
1 and C 2 which satisfy these two
equations what it says given initial
conditions I cannot find out the values
of C 1 and C 2 what is the problem see
here there is choice of y 1 and y 2 what
is this ratio if I take this ratio this
ratio is coming out to be 1 by 4 a
constant for all X while s earlier what
was our choice was e to the power x + e
to the power minus 6 which was e to the
power 2 X that is it is a function for
all X now here is the point when I am
getting this as a constant I'm not able
to get the particular solution but when
it was not a constant I was getting it
as a the particular solution I was able
to find it out that says is now we
require to modify our definition of
general solution a general solution of
an equation y double dash plus px Y -
plus Q XY is equal to zero is of the
form C 1 Y 1 plus C 2 by 2 with y1 and
y2 not being proportional solution and C
1 and C 2 are arbitrary constants
because we have got this a problem when
y1 and y2 are having ratio as constant
that we are calling a proportional so
let us just define this proportional
solution so this y1 y2 which we are
obtaining is not being proportional they
are called the bases are the
fundamentals
some of given equation so now what we
have got the general solution consists
of linear combination of two solutions
and those two solutions has to form the
basis or they must be the fundamental
system of the given equation now let us
see more definitions about this so
particular solution particular solution
after differential equation y double
dash plus PX y dash plus QX y is equal
to 0 can be obtained if specific values
of c1 and c2 are assigned in the general
solution c1 y1 plus c2 y2 now let us
define proportional we call the two
solutions y1 and y2 as a proportional if
the ratio y1 by Y 2 is a constant the
other term which we are using for being
proportional and not being proportional
that are more standard in mathematics
they are linear independence and linear
dependence so let us first say linear
independence two functions y1 x and y 2x
are called linearly independent if the
linear combination k 1 by 1 plus K 2 y2
is 0 only if K 1 is 0 1 K 2 is equal to
0 that is if the two functions are such
that I cannot obtain this linear
combination to be equal to 0 unless I do
take both the constants to be 0 they are
called linearly independent the other
term is linear dependence the two
functions y1 and y2 are called linearly
dependent if for some K 1 and K 2 not
both 0 this linear combination turns out
to be 0 so in our example if I just see
e to the power x + 4 times e to the
power X when I have taken y1 and y2 as
that then if I choose K 1 as minus 4 and
K 2 as 1 I will get that 0 will
k1 and k2 both are not zero but if I
take e to the power X and e to the power
minus six I will not be able to find out
any constant other than zero so that I
can make this as a zero so we have got
this the definition of linear
independence and linear dependence so
let us reformulate our definition of
basis of the fundamental system a
linearly independent pair of solutions
y1 and y2 for differential equation y
double dash plus PX y dash plus QX y is
equal to zero is called a basis or
fundamental system so the basis of
fundamental system is linearly
independent solutions of the
second-order differential equation so in
our previous example e to the power x
and e to the power minus 6 they form the
basis are they are the fundamental
system now we will like to first we will
discuss how to obtain a basis if one
solution is known this is a very
important one because the one solution
we may be able to guess are we are
knowing from some other method so of
course we would like to know what is the
fundamental system that means we would
be interested in finding out the basis
so the technique which we are going to
learn here is not as the reduction of
order technique see what is this
technique let y1 be a nonzero solution
of given equation y double dash plus bxy
dash plus Q X Y is equal to 0 then the
second solution y2 which is linearly
independent of Phi 1 can be obtained as
UX times y1 X now question is how to get
it what we do is we simply put y2 and
its derivative in the given equation
that is y2 which we are having is UX
times y1 x
and why - - you - y1 + uy 1 - y double
dash of course q double - y 1 + 2 u - y
1 dash plus uy 1 double dash will put
all these in the given equation y double
dash plus PX y dash plus QX y is equal
to 0 so we put 1 by 1 we get y double
dash y 1 plus 2 u - y 1 dash plus u
times y 1 Double Dash plus P times u - y
1 plus uy 1 dash plus Q u y 1 is equal
to 0 now let us collect the terms in you
Double Dash you - and you so the
equation is this now we are collecting
the term the coefficient of U Double
Dash is only y 1 that is the first term
coefficient of U - 2 y 1 dash plus PU y
1 coefficient of U y 1 Double Dash plus
py 1 dash plus qy 1 now you are quitting
it to see you now we see this
coefficient of U this is what is our
given equation y double dash plus py -
plus qy + y 1 is a solution so it
satisfies it should satisfy the given
equation that means this equation should
be 0 because y 1 is the solution of the
given equation that says is what is
remaining to us you double - y 1 + u - 2
y 1 dash plus py 1 is equal to 0 now
this is an equation in U dash and you
double dash let us stay you - as capital
u then you double dash would be capital
u dash now rewrite this given equation
in the terms of this capital u and
capital u dash writing in a standard
form we will get u double dash plus 2 y
1 dash upon y 1 plus P times U is equal
to 0 now you see this is a first order
equation in capital u that is why we are
calling this reduction of order
technique now we have
a differential equation of first order
for which we do know the methods to
solve it so we will solve it by using
the variable separable method that is
for separating the variables and then
integrating it so separating the
variables we get D u upon u is equal to
minus 2y 1 - upon y 1 plus P times DX
now integrating on both the sides we get
log U is equal to minus 2 log Y 1 minus
integral PDX taking antilog U is 1 upon
Y 1 square into e to the power minus
integral PDX now U is nothing but U - so
what will be the small u integral of
capital u with respect to X that is
integral of 1 upon Y 1 square e to the
power minus PD X with respect to X now
we see that this function U this is the
ratio of Y 2 and y 1 we have got the
second solution as Y 2 the first
solution was Y 1 this ratio is U this is
not a constant why we see that you -
that is this capital u this is 1 upon Y
1 square e to the power minus PD X
whatever with this P it may be the 0
even then this exponential function is
never 0 that says is I will never get
this capital u R that is you - as 0
since you - is not 0 you will never be a
constant thus we are getting that y2 and
y1 would be linearly independent let us
see one example
find the general solution of
differential equation 1 minus X square y
double dash minus 2x y dash plus 2y is
equal to 0 using the fact that y is
equal to X is a solution of this
equation I am going to use this just now
which we have learned a technique of
reduction of order we are given that X
is a solution so first we will check
whether it is a solution or not y1 is X
so Y 1 dash would be 1 and Y 1 Double
Dash would be 0 if I substitute this in
the given equation I would get that the
first term is 0 second term will give me
minus 2x third term also give me plus 2x
that is it is 0 it is satisfying so this
is a solution what will be the way to
find out the second solution we will use
the technique UX times y 1 X now my
given equation is 1 minus X square Y
Double Dash minus 2 XY dash plus 2y is
equal to 0
let us rewrite it in standard form Y
Double Dash minus 2 X upon 1 minus X
square y dash plus 2 upon 1 minus X
square Y is equal to 0 what I have done
I have just divided this equation by 1
minus X square to make it in standard
form in standard form the coefficient of
Y Double Dash has to be 1 this says is
that in standard form if I write what
will be the P here is minus 2 X 0.1
minus X square
now the second solution would be UX
times X that is X was the first solution
now we can obtain this u by substituting
this Y - y - - + y - double - in this
given equation but we are not this
leaving up for you I am just going to do
the formulation just now we had obtained
the formula for finding u that is
integral of 1 upon y1 square e to the
power minus P X DX so what is integral
minus px DX this is minus log 1 - X
square hence e to the power minus px DX
would be 1 upon 1 minus X square so what
will be you I am substituting all these
in the formula y 1 is X square so it
should be 1 upon X square 1 minus X
square it is integral with respect to X
integrating of course we have to do
first the partial fractions 1 upon X
square DX plus 1 upon 1 minus X square
DX again one more partial fraction 1
upon X square DX 1 integral second
integral half 1 upon 1 minus X DX
third integral half integral 1 upon 1
plus X DX now integrating these things
we get minus 1 by X plus half log 1 plus
X upon 1 minus X so what will be the
second solution x times UX that would be
minus 1 plus 1/2 x times log 1 plus X
upon 1 minus X thus we have got the
general solution as C 1 times X plus C 2
times minus 1 plus 1/2 times X into log
1 plus X upon 1 minus X so given one
solution X we had obtained the second
solution which is linearly independent
of this solution and we had obtained the
general solution let us try one more
example here find the basis of solution
for differential equation X square y
double dash minus X y dash plus y is
equal to 0 now here we have not been
given any solution so what we will do I
am just going to do is that is guessing
one solution and then obtaining the
second solution is linearly independent
of the one guessed solution the given
equation is this for guessing one what
we just think is that is here if I take
Y as X then I would get here X y dash
would be 1 so this would be minus x and
y double dash would be 0 so this term
would be ruled out and
we'll get that there should satisfy the
solution so let us take y1 is equal to
exam checking as a solution y1 is X Y 1
- would be 1 + y1 double - would be 0
substituting I would get minus X plus X
that is satisfying the given equation so
it is a solution now for second solution
again I am going to use the formulation
so I have to write this equation in the
standard form that is I have to get the
coefficient of Y Double Dash as constant
one so I have divided by X square now
this we could do if X is not 0 that
means this solution which we are going
to obtain that would be only true when X
is not 0 so we are making it in standard
form it gives me PS minus 1 by X now for
you I am just using the formula so
integral PDX is log - log X so I would
get it first solution 1 upon X square e
to the power minus of minus px DX that
is log X this if I rewrite it again I
would get X upon X square DX that is
integral of 1 upon X DX which is log X
so I have got the second solution as x
times log X so what we have got the
bases of solution X and X log X so of
course what I have done I have guessed
here one solution and obtain second
linearly independent solution as X log X
that is giving me the basis now we will
learn how to obtain the general solution
of a second-order differential equation
that is we are not going to do the
guessing job or anything and then
finding out the second solution by the
technique which we have learned now we
will do in general how to solve this one
so for that first we will do the second
order homogeneous Lien's with constant
coefficients
to the a second-order equation is y
double dash plus PX y dash plus Q XY is
equal to 0 here this coefficients px and
QX they are function of X now I want
these coefficients to be constant not a
function of X that is if I change this
px and QX to the constants a and B we
would get the second-order differential
equation with constant coefficient so
the standard form for this would be y
double dash plus a y dash plus B Y is
equal to 0 we have a and B are some
constants how to solve this equation
will again go with the things which we
already know we have done the first
order equations there we know that
linear equations have the solution of
the form e to the power lambda X so we
will try with this kind of function so
what we will try we will try that this
let us see or let us assume that the
solution will be of the form e to the
power lambda X we'll substitute Y y dash
and y double dash in this given equation
what will be Y - that would be lambda
times e to the power lambda X and y
double dash as lambda square times e to
the power lambda X so we are going to
substitute this Y y dash and Y Double
Dash this equations what we are getting
is y double dash lambda square times e
to the power lambda X plus a times
lambda times e to the power lambda X
plus B times e to the power lambda X now
take this e to the power lambda X as
common what we get lambda square plus a
lambda plus B times e to the power
lambda X is equal to 0 now see what we
have started we have started that e to
the power lambda X is a solution if it
is a solution this cannot be a zero
function
and if it is a solution this must
satisfy the equation that says is I must
get this coefficient lambda square plus
a lambda plus B is equal to zero so that
if the power lambda X should be a
solution of this given equation this is
called the characteristic equation that
is this equation which we have often
this is called the characteristic
equation so let us have a formal
definition of this term also the
quadratic equation thus obtained lambda
square plus a lambda plus B is equal to
0 is called the characteristic equation
our auxilary equation of the
differential equation y double dash plus
a y dash plus B y is equal to 0 what we
have seen here that if we are having a
second-order differential equation with
constant coefficients then a
corresponding characteristic equation we
are getting as lambda square plus a
lambda plus B is equal to 0 that is we
are having this a quadratic equation
with first coefficient as same as the
constant one are the coefficients of
this equation and this equation of C
this is called the characteristic
equation now we see that this equation
is a quadratic equation so what will be
the solution of our differential
equation they will depend upon word the
solution of this quadratic equation we
are getting solution of quadratic
equation that we do not know we are also
calling them the roots of this quadratic
equation we do know the roots of this
quadratic equations depend upon the
discriminant function a square minus 4 B
accordingly we get three cases and
accordingly we will get the solutions on
the different cases so let us make those
three cases the case one if my
discriminant function a square minus
four B is positive we will get two real
roots and the second case is if the
discriminant function a square minus 4b
is 0 you will get real root double 1
that is I will get the same roots and
the K is 3 if my a square minus 4b is
less than 0 if we get the complex
conjugate roots I will discuss these
cases one by one on the solution
therefore for the given differential
equation case one let us say 2 distinct
real roots lambda 1 and lambda 2 we will
say that the two solutions we can make
because we have got the two roots
lambda 1 and lambda 2 so we can make the
two solutions 1 e to the power lambda 1
X and second Y 2 as e to the power
lambda 2 X we can see that these two
solutions are linearly independent they
form the basis we can check it by the
proportionality Y 1 upon Y 2 is equal to
e to the power lambda 1 minus lambda 2 X
now this we are getting is lambda 1
minus lambda 2 since they are distinct
real roots so lambda 1 minus lambda 2
this is not 0 that says it will never be
a constant thus they are linearly
independent solution so the general
solution will be of the form C 1 e to
the power lambda 1 X plus C 2 e to the
power lambda 2 x case 1 two distinct
real roots lambda 1 and lambda 2 so we
do have the two solutions y 1 e to the
power lambda 1 x and y 2 as e to the
power lambda 2 X we see that these two
solutions are linearly independent we
can check by the Ray taking the ratio of
these two solutions since Y 1 upon y 2
is e to the power lambda 1 minus lambda
2 X now we do know that lambda
lambda2 are distinct they are not equal
so lambda 1 minus lambda 2 will not be 0
that is this will not be a constant so
they are linearly independent that is
these two solutions form the basis are
the fundamental system hence in this
case the general solution of the given
differential equation will be of the
form C 1 e to the power lambda 1 X plus
C 2 e to the power lambda 2 X let us
discuss the case 2 that real double root
lambda when a square minus 4 B is 0 then
the single root we do get lambda as
minus a by 2 that means we get only one
solution e to the power minus a X by 2
how to get the other solution which is
linearly independent of this we will use
again the technique which we do know
that reduction of order technique so we
will use Y 2 as u times y 1 now our
equation is y double dash plus a y dash
plus B y is equal to 0 this is in the
standard form P is the constant a so we
will get u by the formula 1 upon y 1 is
square e to the power minus P X is a so
a times DX DX now what is integral that
y 1 is we are getting is e to the power
minus a X by 2 so when I substitute this
y1 and this a so what we get e to the
power minus X upon e to the power minus
X that is integral with respect to X
this is constant so it's integral is
just the function X that is we have got
the second solution as X y1 is equal to
x times e to the power minus a X by 2 so
we have got the two solution in this
case first solution is e to the power
minus X by 2 and second solution is x
times e to the power minus a X by 2 so
you will get the general solution
as y times y is equal to c1 times e to
the power minus ax by 2 plus c2 times x
times e to the power minus ax by 2 or we
can write it as C 1 plus C 2 x times e
to the power minus ax by 2 now let us
see the third case complex root when a
square minus 4b is less than 0 the roots
would be complex numbers that is lambda
1 would be minus a by 2 plus 1/2 square
root of a square minus 4 B and the other
root would be minus a by 2 minus 1/2
square root of a square minus 4 B now
since a square minus 4 B is negative so
this square root is an imaginary number
let us say that imaginary number to be I
T so let us say this lambda 1 and lambda
2 we are saying is s plus I T and s
minus I T where s is what this minus a I
by 2 and T is a square root of 4 B minus
a square by 2 since I am using this
imaginary minus 1 I am taking square
root of -1 I am writing it as I then
what will be the two solutions the two
solutions would have fun as e to the
power lambda 1 X that is e to the power
s plus I T X which we could write as e
to the power s X into e to the power I
TX this again we are using with the
demeyers here we can write cos TX plus I
sine T X similarly the second solution e
to the power lambda 2 X that is e to the
power s minus i TX we can get as e to
the power s X into cos TX minus I sine
TX they would be linearly independent
actually we can get these are a little
bit lengthy solutions we can get smaller
ones that is we can get a linear
combination of these two solutions
which are again linearly independent how
to get that one we do is that is we add
these two solutions if we add we do get
is that two times e to the power s X cos
T X and if I divided by 2 I would get e
to the power s X cos T X similarly if I
subtract from second one to the first
one I would get or from first one the
second one I do get to I times sine T X
and of course multiplied with e to the
power s x two times so if I divided by
two I I would get the two solutions as e
to the power minus a by 2 so assist
minus a by 2 minus a by 2 X cos T X and
another is e to the power minus a by 2 X
sine T X we can check again that they
are linearly independent because if I
take the ratio what I will get Y 1 upon
Y - I will get cos T X upon sine T X
that is caught a cotangent T X which is
a function of X not zero are not a
constant thus we are getting that y1 and
y2 they are linearly independent and
what are this T is here is of course
half of square root of 4 B minus a
square so thus we will get the general
solution as C 1 times e to the power
minus a by 2 X cos T X plus C 2 times e
to the power minus a by 2 X sine T X so
we have seen that we have got for a
linear differential equation with
constant coefficient characteristic
equation which has two roots and
depending upon the discriminant function
we are getting three types of solutions
so let us summarize this method how we
are going to solve this equation so to
find the general solution of y double
dash plus a y dash plus B Y is equal to
zero
where a and B are constant first we
write the characteristic equation the
characteristic equation how you are
writing we are just writing is that is
we write a second order are a quadratic
equation in lambda where the
coefficients are same as the
coefficients of given differential
equation that is lambda square plus a
lambda plus B is equal to zero then we
check its roots lambda 1 and lambda 2
what there would be there a minus a plus
square root of a square minus 4 B by 2
and minus a minus square root of a
square minus 4 B by 2 so of course
depending upon the value of a square
minus 4 B we do have different cases so
in case a square minus 4 B is positive
we will get the two roots lambda 1 and
lambda 2 there would be two distinct
real numbers and the general solution
would be c1 e to the power lambda 1 X
plus c2 times e to the power lambda 2 X
where of course my lambda 1 would be
minus a plus square root of a square
minus 4 B by 2 and lambda 2 would be
minus a minus square root of a square
minus 4 B by 2 and depending upon what
is the value of this we would get this
lambda 1 and lambda 2 then if this a
square minus 4 B is 0 that is the second
part would be 0 I would get lambda 1 and
lambda 2 both us minus a by 2 in this
case the two solutions we get as e to
the power minus a by 2 X and X times e
to the power minus a by 2 X or in short
here we are just saying is that is then
the general solution will be of the form
C 1 plus C 2 x times e to the power
minus a by 2 X where a is of course the
coefficient of Y - in the given
differential equation third case was
that if a square minus 4 B is less than
0 we get the roots
1 and lambda 2 as the conjugate pair and
the general solution is of the form C 1
e to the power minus a by 2 X cos beta X
plus C 2 times e to the power minus a by
2 X sine beta X where a is the
coefficient of Y - in the given equation
and beta is nothing but 1/2 times square
root of 4 B minus a square so now let us
to understand are to clear these steps
let us do some examples for solving
second-order linear differential
equations with constant coefficients
first example find the general solution
of y double dash minus 9 y dash plus 20
y is equal to 0 C given differential
equation is y double dash minus 9 y dash
plus 20 y is equal to 0 so corresponding
characteristic equation would be lambda
square minus 9 lambda plus 20 is equal
to 0 to find its root we'll just do the
factorization here so factorization
gives me lambda minus 5 into lambda
minus 4 is equal to 0 thus roots are 4
and 5 they are real and distinct so this
is the first case what will be the
general solution general solution would
be C 1 e to the power 4 X plus C 2 times
e to the power 5 X so what we have got
the general solution of the differential
equation y double dash minus 9 y dash
plus 20 y is equal to 0 the general
solution is C 1 e to the power 4 X plus
C 2 e to the power 5 X let us do one
more example so on the initial value
problem y double dash plus y dash minus
6 Y is equal to 0 with the initial
condition
why add zero is 10 and why - at zero is
zero how we are going to do is the given
differential equation is y double dash
plus y dash minus 6y is equal to zero so
the characteristic equation lambda
square plus lambda minus six is equal to
zero factorize it we get lambda plus
three into lambda minus two is equal to
zero that says the roots are minus three
and two again what we are getting is
real and distinct roots so the general
solution would be C one e to the power
minus three X plus C 2 e to the power 2
X now for the initial problem initial
value problem the particular solution
general solution is C 1 e to the power
minus 3x plus C 2 e to the power 2 X
this gives y dash as minus 3 C 1 e to
the power minus 3x plus 2 C 2 e to the
power 2 X the initial condition is y at
0 is 10 and y dash at 0 is 0 so we will
put X is equal to 0 in the first one we
get is here C 1 plus C 2 this is given
as 10 in the second Y - if I put X is
equal to 0 what I get minus 3 C 1 plus 2
C 2 this is given as 0 now this is
linear equations in two unknowns we can
solve it and what we will get we will
get the solution as C 1 is equal to 4
and C 2 is equal to 6 this is the unique
solution so we will get only single
solution particular solution for the
initial value problem and that is C 1 we
are putting as for C 2 we are putting as
6 we are getting is 4 times e to the
power minus 3x plus 6 times
e to the power 2 X so we have got this
is the particular solution of our
initial value problem see some more
examples so that we can practice all the
way in which all those three cases we
can complete solve the initial value
problem y double dash minus 4 y dash
plus 4 y is equal to 0 with the initial
conditions y at 0 is 2 and y dash at 0
is 1 let us solve it the differential
equation is y double dash minus 4 y dash
plus 4 y is equal to 0 so the
characteristic equation that will be
lambda square minus 4 lambda plus 4 is
equal to 0 we see this is the whole
square of lambda minus 2 that says I do
get the double root lambda is equal to 2
so the one solution I would get e to the
power 2 X what should be the other
solution if you do remember we i have
got x times e to the power 2 x so the
general solution we will get as c1 plus
c2 x times e to the power 2 x now we
have to solve the initial-value problem
so we have to use the initial conditions
for that we find out the YX and its
derivative what will be derivative of
phi XY exist c 1 plus c 2 XE to the
power 2 X so it's derivative is 2 C 1
times e to the power 2 X plus 2 C 2 x
times e to the power 2 X plus C 2 times
e to the power 2 X the derivative of X
is 1 now use at 0 the initial conditions
are given at 0 that is y 0 is at 2 and y
dash 0 is 1 when I put here in the first
one X is equal to 0 I do get the only
term C 1 so we get C 1 is equal to 2
when I put in y dash X is equal to 0 I
do
the two terms 2 C 1 and C 2 this is
giving us 132 C 1 plus C 2 is 1 since C
1 is 2 so C 2 you would be getting as
minus 3 so what the particular solution
we have got now we'll put this value C 1
and C 2 in this given in this general
solution particular solution would be 2
minus 3x e to the power 2 X so we have
got the particular solution of this
initial value problem as 2 minus 3x e to
the power 2 X let us see one more
example give the general solution of
differential equation for y double dash
plus 4 y dash plus 10y is equal to 0 see
how to solve it differential equation is
for y double dash plus 4 y dash plus 10y
is equal to 0 the characteristic
equation would be 4 lambda square plus 4
lambda plus 10 is equal to 0 you see
here we are not having this in a
standard form that is the coefficient of
Y Double Dash is not 1 the
characteristic equation we write with
the same coefficients as they are in the
given differential equation so we get 4
lambda square plus 4 lambda plus 10 to
find out its roots let us use the
formula for finding out the roots that
is minus b plus/minus square root of b
square minus 4ac f1 2 a so this we are
having is minus 4 plus minus B square is
16 minus 4ac that is 160 upon it now you
do get it that is this is negative minus
144 so we will get actually the complex
oods
minus 4 plus minus 12i upon 8r more
simplified manner we are getting is the
first root as minus half
plus 3x2 I and the second route as minus
1/2 minus 3 by 2 I so we are getting the
complex suits they are always coming as
in conjugate pairs what will be our
general solution we see here is my minus
1/2 and you see there is what we have
got minus a by 2 as the coefficient of Y
- the coefficient of Y - will we were
having the equation in the standard form
here the equation is not if I write it
in the standard form that is the
coefficient of Y Double Dash as 1 then I
would get it as a as 1 because I will
divide it by 4 so I will get coefficient
of Y - as 1 so that is why it is minus 1
by 2 and so on so we do get the so the
general solution will be y as e to the
power minus X by 2 c1 times cos 3 by 2 X
plus c2 sine 3 by 2 X R we can write it
as C 1 e to the power minus X by 2 cos 3
by 2 X plus C 2 times e to the power
minus X by 2 sine 3 by 2 X now next
example solve the initial-value problem
y double dash plus 2 y dash plus 2 is
equal to 0 the given initial conditions
are Y at PI by 4 is 2 and derivative of
Y that is y dash at PI by 4 is minus 2
see how to solve it the given
differential equation is y double dash
plus 2 y dash plus 2 is equal to 0 the
characteristic equation would be lambda
square plus 2 lambda plus 2 is equal to
0 its roots would be again we'll find
out from the form this directly minus 2
plus minus square root of 4 minus 8 by 2
again we see this discriminant function
is negative so we will
get the complex roots minus one plus
minus I that is the two roots we are
getting as minus one plus I and minus
one minus I so what will be the general
solution general solution will get C 1 e
to the power minus x cos x plus C 2 e to
the power minus X sine X now for the
particular solution we have to find out
the values of C 1 and C 2 for that we
have to use the initial condition so we
will find out what is the Y - X Y - X is
C 1 e to the power minus x cos x s
derivative would be minus cos x minus
sine X e to power minus X similarly it
would be C 2 minus sine X plus cos XE to
the power minus X now the initial
conditions are being given at PI by 4 so
what will be Y at 5 by 4 I will put the
value of x as PI by 4 I will get e to
the power minus PI by 4 and cos PI by 4
and here again sine PI by 4 we do know
the value of pi by 4 and sine PI by 4
they are 1 by square root 2 so we do get
it 1 by square root 2 e to the power
minus PI by 4 times C 1 plus C 2 this is
given us as a similarly we can find out
y dash at PI by 4 we do get here is that
is bigger sphere getting as minus cos x
plus cos x something and e to the power
minus XS as common so we would be
getting it actually and the coefficients
we would be getting it as actually C 1 e
to the power minus PI by 4 times the
square root 2 now put this in the given
initial conditions they are being given
that y at PI by 4 is 2 and y dash at PI
by 4 is minus 2 so substitute in those
find out the values that is 1 by root 2
e to the power minus PI by 4 times C 1
plus C 2 this is equal to e to the power
minus PI by 4 C 1 plus C 2 is 2 times a
square root 2 similarly the second one
we would get
that C 1 times e to the power minus PI
by 4 is square root 2 so from here we
are getting the value of C 1 and that
will put over here and we'll get the
value of C 2 and both we are getting as
equal to square root 2 times e to the
power minus Pi by 4 so what will be our
particular solution we will put the
values of C 1 and C 2 into the general
solution general solution was e to the
power minus x cos x plus e to the power
minus x sine X so we'll put C 1 and C 2
both are square root 2 e to the power
minus PI by 4 we get the solution as
square root 2 times e to the power minus
PI by 4 e to the power minus x times cos
x plus sine X so we had learned today
about the second order linear
differential equations in that we had
learned that is what we mean by the
solution what we mean by the general
solution what we mean by the particular
solution and if one solution is known
how to find out linearly independent
another solution
so that we can make the general solution
and we had learned that linear
differential equations of second order
with constant coefficients how to find
out the general solution we'll continue
with this kind of equations in the next
lectures also today we are finishing up
our lecture here so thank you
Oh