hello friends welcome to my second lecture on methods for finding particular integral
for second order linear differential equations with constant coefficients in my last lecture
we were trying to find the particular integral in the case of the differential equation
d square plus one y equal to cos square one by two x and we couldnt find it because we
have to write cos square half x in the form of one plus cos x by two and when we write
it in this form what happens is that
we have to find the particular integral for one by d square plus one cos x but one by
d square plus one cos x cannot be found from the formula one by d square plus a square
cos a x equal to one over sorry one over f d square equal to write one over phi d square
equal to cos a x equal to one over phi minus a square cos a x we cannot find the of
particular integral in this case because when we replace d square by minus a square that
is d square by minus one square r minus one this d square plus one on replacing d square
minus a square becomes zero so phi minus a square should not be zero to apply this formula
and therefore let us now discuss how to find the particular integral in the case of
one over d square plus c square cos c x
so let us discuss how to find the particular integral in the case of one over f d cos c
x or one over f d sin c x which are where in the exceptional case we have f d equal
to d square plus c square and we see that phi d square when c square d square is replaced
by minus c square becomes zero so we have not able to apply the formula one over phi
d square cos cx equal to one over phi minus c square cos c x we will prove that one
over phi d square cos c x which is one over d square plus c square cos c x gives us x
sin c x over two c ok so let us see how we do this we can write
one over d square plus c square as one over two i c one over d minus i c minus one
over d plus i c operating on cos c x that is we break one over d square plus c square
in partial fractions now let us recall that we had proved that one over d minus alpha
when operates on q or you can say r x then we had e to the power minus alpha x integral
e to the power e to the power alpha x e to the power minus alpha x into q d x
so let us apply this formula here and to by using this formula one over d minus i c cos
c x will be e to the power i c x integral e to the power minus i c x cos c x d x now
then what we will have is e to the power minus i c x cos c x we can write as e to the
power i c x plus e to the power minus i c x divided by two by the eulers formula we
know that [e/ it] for i theta is cos theta plus i sin theta and e to the power minus
i theta is cos theta minus i sin theta so from there cos theta is e to the power i theta
plus e to the power minus i theta by two now this is equal to e to the power i c x
integral this will be one plus e to the power minus two i c x d x ok let us integrate it
and we get then integral of one is x and then e to the power minus two i c x divided by
two i c similarly we can write the value of one over
d plus i c cos c x this is equal to
we get this now let us subtract from the value of one over d minus i c cos c x the
value of one over d plus i c cos c x and then divide by two i c ok hence one over d square
plus c square cos c x is equal to one over two i c from one over d minus i c cos c x
we subtract the value of one over d plus i c cos c x and so this will give you x times
cos e to the power i c x minus e to the power minus i c x divided by two and when we then
we will have minus sin here two two becomes four i c and we get e to the power minus i
c x here we get e to the power plus i c x so e to the power i c x plus e to the power
minus i c x and this is equal to one over two i c x times now e to the power i theta
minus e to the power minus i theta by two is i sin theta so we get i times sin c x and
then here we have minus one over four i c e to the power i theta plus e to the power
minus i theta is two cos theta so two cos c x and thus we have x sin c x by two c
and we have i this two will cancel with that two here so here cos c x divided by four c
square now now our differential equation is of the type our differential equation is
d square plus a square y equal to cos c x d square plus c square y equal to cos c x
so auxiliary equation will be m square plus c square equal to zero so complementary function
y h x will be equal to a cos c x plus b sin c x
and then when we will operate now we will operate one over d c square on cos c x to
find the particular integral we get x sin c x over two c plus cos c x over four c square
so this term cos c x over four c square can be observed in the term a cos c x occurring
in the complementary function and therefore when we write the result for one over d
square plus c square cos c x we just retain x sin c x by two c so we say that since
of cos c x over four c square can be observed in y h x we have one over d square plus c
square cos c x equal to x sin c x by two c and similarly we can prove similarly we
can show one over d square plus c square sin c x equal to minus x by two a cos c x
now with these formulas we can then find the particular integral easily in the case where
phi minus a square becomes zero for example one o d let us look at the equation d square
plus one y equal to cos square one by two x so particular integral y p x is equal to
one over d square plus one operating on cos square x one by two which we can write as
one plus cos x by two which is equal to one by two one over d square plus one operating
on one and then we have one over d square plus one operating on cos x one can be regarded
as e to the power zero x so this is one over two d over d is replaced by a zero so we have
one and then one over d square plus c square cos c x gives x sin c x over two c so c
is equal to one here so we have x sin x divided by two by using the formula which we have
just proved and similarly we can solve example two
in the case of example two when we find the particular integrals sin square x by two
is written as one minus cos x so over two and then we apply the formula same formula
one over d square plus c square cos c x equal to x sin x by two c to obtain the particular
integral now let us look at another formula which is
given by one over f d e to the power a x into v where v real function of x and a is a
is some constant so this when one over f d operates on e to the power a x into v
what we get is e to the power a x into one upon f d plus a operating on v so this
formula is very useful in the case earlier we have considered one over f d operating
on e to the power alpha x this formula gives us one over f alpha e to the power alpha x
provided f alpha is non zero but if if alpha if f alpha becomes zero then we cannot apply
this formula so the case where this f alpha become zero can be handled by the formula
which we are now going to prove so let us see how we prove this result
when you find the derivative of the function e to the power a x into v you get e to the
power a x into derivative of v plus a times e to the power a x derive[ative] into v and
this can be expressed as e to the power a x d plus a operating on v and then you again
differentiate that is you find the second derivative of e to the power a x into v what
you get is if you find d square e to the power a x into v then this is d of and derivate
of e to the power a x into v we have found as e to the power a x d plus a operating
on v let us operate by d so derivate of e to the power a x is a times e to the power
a x then d plus a
operating on v then e to the power a x d operating on d plus a v now this can be expressed as
e to the power a x then we have a d v a square d so we can write it as d plus a whole square
v d d is d square v and then a d v and then a d v becomes two a d v and then we have a
square v so a to the power a x d plus a whole square we get and thus we if we apply f d
operator if we apply suppose f d is equal to d square plus alpha d plus beta ok
let f d v equal to d square plus alpha d plus beta then f d operating on
e to the power a x into b will give you d square plus alpha d plus beta operating on
this and so what you will get actually d square plus alpha d plus beta operating on e to the
power a x into v so this will give you e to the power a x e to the power a x d plus
a whole square operating on v then alpha times d plus a e to the power a x e to the power
a x d plus a v and then e to the power a x beta b which is equal to e to the power a
x f d plus a operating on v so this is what we get
and so thus ultimately we get f d operating on e to the power a x into v it is equal to
e to the power a x f d plus a into v so this suggests us this formula one over f d operating
on e to the power a x into v equal to e to the power a x one over f d plus a into v now
let us see how we get this so [vocalised-noise] we we have to prove one over f d e to the
power a x into v equal to e to the power a x one over f d plus a operating on v now
what we do is let us operate by f d on both sides so operating in by f d
f d in one over f d are inverse operators so we get e to the power a x into v equal
to f d operating on e to the power a x one over f d plus a v now let us look at the
right hand side
so let us assume that one over f d plus a operating on v is equal to v one ok let us
assume that one over f d plus a operating on v is equal to v one so we shall get f d
right hand side right hand side here will be equal to f d operating on e to the power
a x into v one and when f d operates on e to the power a x into v what we had got
earlier that if this is e to the power a x f d plus a v one we have proved this earlier
now this implies that f d plus a when you operate on both sides you get f d plus a v
one equal to v so this will be equal to e to the power a x into v and this is what we
had to prove this is the left hand side so left right hand side left hand side are same
and therefor the formula one over f d e to the power a x into v equal to e to the
power a x one over f d plus a operating on v is true
now let us apply this formula to this eq[uation] to solve this equation d square plus four
d plus four which we can write as d plus two whole square y equal to e to the power two
x minus e to the power minus two x i will just find particular integral here complementary
function is easy to find so this is one over d plus two whole square e to the power two
x minus e to the power minus two x now this is equal to when one over d plus two operator
a square operates on e to the power two x we get two plus two four square so we get
one by sixteen into e to the power two x minus Refer Time: 20:00) now when one over d plus
two whole square operates on e to the power minus two x on replacing d y minus to d
plus two become zero so we will apply this formula so we will let us see how we get this
one over d plus two whole square v operates separately on e to the power let us operate
on e to the power minus two x so this is one over d plus two one over d plus two operating
on e to the power minus two x we can we can assume here one so one over d plus two
operating on e to the power minus two x into one so v we are taking as one so this is equal
to one over d plus two e to the power minus two x d will be replaced by d minus two and
this cancels we get one over d plus two e to the power minus two x one over d is d means
derivative one over d means inverse of the derivative that is integral integral of one
is x now again we will apply the the formula which
we have just proved two two e to the power minus two x into x let us take v equal to
two x so then we will have e to the power minus two x d will be replaced by d minus
two now one over d is again integral of we have to make integral of x so that is x
square by two o x square by two e to the power minus two x and thus the particular integral
is so this is how we can make use of this formula in the case where under placing
d by a in the expression one by f d e to the power a x f a becomes zero now let us
find another result so now we discuss how to find the particular
integral in case r x is of we form x to the power m where m is a positive integer
so when when we have to find but particular integral of this type where m is positive
integer what we do is in order to evaluate one over f d x to the power m we extend f
d in ascending powers of d and then operate on x to the power m since the derivative
m plus one of the derivative of x to the power m is zero an[d] all derivatives more than
one plus are also zero while writing the expression in the ascending powers of d we
need to write the derivate of the powers of v only up to m rest of the terms will
be zero so we can evaluate particular integral by writing in the ascending powers
of d let us see we how apply this formula suppose
we have the equation d square minus d minus six y equal to one plus x square so this is
let say lets find find particular integral here so one over d square minus d minus six
operating on one plus x square one can be taken it again by writing e to the power zero
x so one over d square d square minus d minus six e to the power zero x and then one over
d square minus d minus six x square so this is give by replacing d by zero we get minus
one over six and this is ok and one over d square minus d minus six when operates on
x to the power two we have to expand it in ascending powers of d so that we can do like
this one over minus one over six one plus d minus d square divided by six raise to the
power minus one now we will expand it like the expand one
plus x to the power minus one so this one will be
one plus x to the power minus one is one minus x plus x square
minus x cube now x cube i am not writing because x cube in the x cube when you do d minus d
square to the by six to the power cube will minimum power of d will be three but we have
x square here third derivative x square is zero so we need not write those terms we can
stop here and then ok so this operates on this operates
on x square and so what we will get is minus one by six minus one by six and then x square
and then d minus d square when operates on x square d or in square gives two x d square
on x square gives two so two x minus two divided by six and then we have d minus d square square
one we find we get d square then we get d cube then we get d to the power four
so d square one operates on x square gives you two so two by thirty six all other terms
will contribute zero because when they operate on x square we what we get is zero so this
is the particular integral in this case so when one over f d operates on x to the power
m where m is an integer we have to extend one over f d in ascending powers of d
now i will tell you how this expression is justified suppose we have one over one plus
d operate[n] on x to the power m and we are writing it as one minus d plus d square minus
d cube and so on we are expanding one over one plus d as one minus d plus d square minus
d cube and so on like we expand one plus x to the power m minus one so how this expansion
is valid now what you do is to prove this you operate by one plus d on both sides
so operating by one plus d we will have this ok now one plus d and one
over one plus d are inverse of operators so we will have x to the power m and in the right
side let us operate one plus d on one minus d plus d square minus d cube what you get
is when you operate by one you get one minus d plus d square minus d cube and so on and
when you operate by d you get plus d minus d square plus d cube minus and so on x to
the power m so these terms they go on cancelling and what we get is x to the power m so both
sides are equal and therefore one over one plus d if we write as one minus d plus
d square minus d cube like we expand bv binomial expansion this just extension is valid
justified so then we have to find one over f d operating on x to the power m we
have to expand one over f d in ascending powers of d
thank you