so in the last class we have seen what are laplace transforms and how to find laplace
transform of some simple functions like one t or e k power a t so what is laplace
transform any function f t laplace transform any function f t is nothing but zero to infinity
f t e k power minus p t d t and we are calling it f p function of p so also we have seen
that laplace transform satisfy linearity property so now in this lecture we will find laplace
transform of some standard functions so first what is laplace transform of t k power n so
laplace transform of t k power n let us try to find it laplace or t k power
n so using the definition of laplace transform laplace t k power n nothing but zero to infinity
t k power n e k power minus p t d t now let p t is equals to some variable say z so it
will be nothing but zero to infinity z by p whole power n e k power minus z and d t
is nothing but d z upon p so it is nothing but one upon p t k power n plus one integral
zero to infinity z k power n e k power minus z into d z and this function is nothing but
gamma function so it is one upon p n plus one gamma of n plus one
now as gamma n plus one is factorial n when n is a positive integer therefore laplace
transform of t k power n will be factorial n upon p k power n plus one when n is a positive
integer now let us try to solve the two problems given below laplace transform of
three t k power three by two minus plus five into t minus four e k power two t now
again by the linearity property of laplace transform this is nothing but three times
laplace of three by two t k power three by two plus five laplace of t minus four laplace
of e k power two t so this is three into now laplace of t k power n as in the first expression
we are having it is gamma n plus one upon p k power n plus one
so here n is not a integer here n is positive integer so we have to use the first definition
of t k power n it is gamma plus one upon t k power n plus one so it is gamma here n is
three by two so it is five by two upon p k power five by two plus five into laplace of
t we have already seen that it is one by p square also we can obtain here from here also
when we put n equal to one in the sorry in the second expression so it is one by p square
minus four into laplace of e k power a t is one upon p minus a so it will be nothing
but one upon p minus two so now now now gamma n plus one we already
know it is n gamma n gamma n plus one nothing but n gamma n and gamma is half under root
pi this we already know so three into now gamma five by two so here it is three by two
plus one so it is three by two gamma three by two upon p k power five by two plus five
p square five by p square four upon p minus two is equal to three into three by two again
we will use a same expression instead of n we have one by two because one by two plus
is c by two so one by two plus one is c by two it is one by two gamma one by two so one
by two gamma one by two p k power five by two plus five upon p square minus four upon
p minus two so it is nothing but nine upon four and gamma half is under root pi so it
is under root pi upon p k power to five by two plus five upon p square minus four upon
p minus two so this is the laplace transform of this function
now now how to find laplace inverse of the this expression p q plus two upon p
k power five so laplace inverse of p cube plus two upon p k power five so it is laplace
inverse of one by p square plus two into laplace inverse of one by p k power five now recall
the definition laplace transform t k power n is given as gamma n plus one upon p k power
n plus one so what will be the laplace inverse of laplace inverse of one upon p k power n
plus one from here it is nothing but gamma n plus one is a constant quantity so it will
put it will go to the right hand side so it will be noting but one upon gamma n plus one
into t k power n by that definition so laplace inverse of one by p square so
one by p square means n is one when n is one so it is nothing but t k power one upon gamma
two plus two into now for this expression n is four when n is four so it will be nothing
but t k power four upon gamma five it is t now gamma n plus one is factorial n when n
is the positive integer so it is gamma two to this factorial one plus two t k power four
upon factorial four so it is t plus two t k power four upon four into three into two
into one so two two cancels out so it nothing but twelve t plus t k power four divided by
twelve so this will be the value of laplace inverse of this function
now how to find laplace of sin a t so now we will see how to find laplace of sin a t
so sin a t as we already know is nothing but e k power iota a t minus e k power minus iota
a t upon two iota this is sin a t now laplace of sin a t using linearity property of laplace
transform will be nothing but one upon two iota laplace of e k power iota a t minus one
upon two iota laplace of e k power minus iota a t so it is nothing but one upon two iota
as we already know let laplace of e k power a t is one upon p minus a ok here instead
of a we have iota a so we will replace a by iota a one upon p minus iota a minus one upon
two iota again here instead of a we are having minus
iota a so we will replace a by minus iota a so we will get one upon p plus iota a so
we will take one upon two iota a is common so and the numerator will get we will get
two a upon p square minus a square two iota a sorry so two iota will cancels out so it
is nothing but it is p square plus a square when we multiply this two so it is a upon
p square plus a square so this is laplace transform of sin a t it is a upon p square
plus a square now so therefore if you want to find out laplace
inverse of this function this f p this will be nothing but sin at so what will be laplace
inverse of one upon p square plus a square it will be nothing but one by a sin a t now
now now laplace transform of cos a t in the similar way we can find out laplace
transform of cos a t also so what is called cos a t in terms of e so cos a t will be nothing
but e k power iota a t plus e k power minus iota a t upon two this we already known so
to find laplace transform of cos a t again we will apply linearity property of laplace
transform laplace transforms of cos a t will be nothing but one upon two laplace transform
of e k power iota a t plus one by two laplace transform of e k power minus iota a t
so this is nothing but one upon two laplace transform of e k power a t is one upon p minus
a here instead of a we have iota a it is one upon p minus iota a plus one by two one upon
p plus iota a so when we simplify it so we will get one by two into two p upon p square
plus a square so two two cancels out so laplace transforms of cos a t is nothing but p upon
p square plus a square similarly laplace inverse of this f p p upon p square plus a square
is nothing but cos a t now now what is laplace transform of sin hyperbolic a t now we will
try to find out this expression to prove this expression in fact so what is cos hyperbolic
at sorry sin hyperbolic a t sin hyperbolic a t is nothing but e k power
a t minus e k power minus a t upon two this we already know so how to find laplace transform
of sin hyperbolic a t again we will use linearity property of laplace transforms so
laplace transform of sin hyperbolic a t will be nothing but one by two laplace transform
of e k power a t minus one by one by two laplace form of e k power minus a t so it is one by
two laplace transform of e k power a t is one upon p minus a minus one upon two one
upon p plus a so when we simplify this so we will get one by two it is p plus a minus
p plus a upon p square minus a square this p p cancels out so it is nothing but a upon
p square minus a square so this would be the laplace of sin hyperbolic
a t similarly if you want to find out laplace inverse of this expression this f p so it
will be nothing but sin hyperbolic a t now what is laplace transform of cos hyperbolic
a t in the similar way on the same lines we can obtain the second expression the this
expression for cos hyperbolic a t so what is cos hyperbolic a t cos hyperbolic a t is
nothing but e k power a t plus e k power minus a t upon two this we already know
now laplace transform of cos hyperbolic a t will be one by two laplace transform of
e k power a t plus one by two laplace transform of e k power minus a t so it is nothing but
one by two one upon p minus a laplace of e k power a t we already know it is one upon
p minus a plus one by two it is one upon p plus a so it is one by two numerator we
get two p upon p square minus a square two two cancels out so this will be the laplace
of cos hyperbolic a t now laplace inverse of p upon p square minus a square is similarly
cos hyperbolic a t because this laplace of f t is p f then laplace inverse of f p will
be nothing but f t that we already know so this we already seen laplace inverse of
these three expressions we already discussed now how to find laplace of this two function
this two simple function let us let us see how to find laplace of these functions laplace
of cos two t minus three so we know the laplace of laplace transformation of cos a t laplace
transform of cos a t is p upon p square plus a square but how to find laplace transform
of cos two t minus three that is a t plus b types how to find out the laplace of this
so it is cos a minus b so cos a minus b is nothing but cos a cos b plus sin a sin b
now cos three and sin three are the constants free from t so can be taken out because of
the linearity property of laplace transforms so it is nothing but cos three laplace transforms
of cos two t plus sin three laplace transform of sin three t sin two t so this is nothing
but cos three and laplace transform of cos two t is p upon p square plus four and plus
sin three laplace transform of sin two t is two upon p square plus four p square plus
a square a is two so this is nothing but p into cos three plus two into sin three upon
p square plus four so this will be the laplace transformation of cos two t minus three
now the second problem and now the next problem laplace transform of sin three t into
cos square t so laplace transform of sin three t into cos square t
now we know how to find laplace transform of sin three t but we dont know how to find
out laplace transform of product of two functions when both involves t so first we try to break
this function now sin three t what is cos square t so cos square t can be written as
one plus cos two t by two now this is nothing but laplace transform of sin three t upon
two plus sin three t into cos two t upon two this can be written as one by two laplace
transform of sin three t plus one by two laplace transform of sin three t into cos two t so
laplace of transform of sin three t we already know this is this is three upon p square
plus three square plus but again we have a problem here how to find laplace transform
of this so we multiply and divide by two in this expression
now two sin a cos b we know that it is equal to sin a plus b plus sin a minus b so it is
three by two p square plus nine plus one by four laplace of it is sin five t a plus b
two t plus three t five t plus this three t minus two t is t
so now we can find out the laplace of this expression so this is three by two p square
plus nine plus one by four laplace of sin five t is a upon p square plus five square
plus sin t is one upon p square plus one square so this will be the laplace transform of this
function sin three t into cos square t so now
now how to find laplace inverse of this simple f p the simple function so laplace inverse
of it is p minus three upon p square plus four so this is simple to find out you
just split this into two parts p into p square plus four minus three upon p square plus four
now by the linearity property of inverse laplace transforms it is this into p upon p square
plus four minus three times laplace inverse of one upon p square plus four so this laplace
inverse we already know this laplace inverse of this expression is cos two t because it
is four two square so it is cos two t because laplace of cos two t is p upon p square plus
two square that is p upon p square plus four now it is minus now now laplace inverse
of one upon p square plus a square is one upon sin a t so here a is two so it is minus
three by two sin two t so this will be the laplace inverse of this f p
now how to find laplace inverse of this problem now now here f p is what f p is
p plus two upon p minus one into p square plus four so this is a linear part and this
is a quadratic part so how to find out laplace inverse of this f p so this can be
written as now we will make use of partial fractions first we will try to reduce into
partial fractions and then we will try to find out its laplace inverse so it can
written as a upon p minus one plus now since it is a quadratic part so in the
numerator we will having a linear term linear in p so it will be something b p plus c upon
p square plus four so compare the coefficients from the both side we will this implies p
plus two will equals to this into this that is a into p square plus four and plus b p
plus c into p minus one now we have to compute the values of a b and c first so we can make
use of compare the coefficients of the both sides here the coefficient of p square is
a from here it is b and here there is no term involving p square so a plus b will be zero
and the coefficient of p here is zero and the coefficient of p here is minus b plus
c and here it is one so it will be one and the constant here is four a minus c that here
constant is two so we are having three equations with three
unknowns so we can find the values of a b and c so a here we have already obtained
actually so a is three by five b is minus three by five and c is two by five this can
easily compute by solving these three equations ok so once we obtained the values of a b and
c then it is nothing but sorry it is nothing but three by five times this it is minus
three plus two upon five now the laplace inverse of this f p will be nothing but three by five
times laplace inverse of one upon p minus one minus three by five times laplace inverse
of p upon p square plus four plus two by five times laplace inverse of one upon p square
plus four so this is nothing but three upon five now
laplace inverse of one upon t minus a e k power a t here a is one so laplace inverse
of one upon p minus a t is e k power t minus three by five laplace inverse of this is cos
two t ok so it will be cos two t a is two and here the for this it is plus two by five
laplace inverse of this expression a is two so it will be nothing but one by two times
sin two t so the final answer is e k power three t e into three e k power t minus three
cos two t this two cancels out plus sign two t upon five
so we have seen that how we can laplace transform of some standard functions like sin a t cos
a t sin hyperbolic a t or cos hyperbolic a t and using these laplace transforms how
we can find out laplace transform of some functions or inverse laplace transforms
thanks