So, let us start the next class. In the earlier
class what we did certain properties we have
done of Laplace transform. Initially, let
us go through certain examples quickly, and
after that we will go through the other things.
The first one is applying the change of scale
property, find the Laplace transform of sins
hyperbolic 3 t. So, as you know a Laplace
transform of sin hyperbolic t is 1 by s square
minus 1, so f of s. Therefore, applying the
change of scale of property, I can write downs
Laplace transform of sin hyperbolic 3 t is
1 by 3 into f of s by 3, f of a sin hyperbolic
3 t is nothing but sin hyperbolic a t.
So, you see the advantage. Once I know the
sin hyperbolic t, then automatically if I
wish, I can simply a will be changed 1 by
a f of s by a using the formula, and we get
the result. So, this is one example. And I
can find out what is the value of f of s by
3 1 by 3 1 by s minus 3 whole square minus
1. So, immediately I can just calculate and
I can obtain the result as 3 by s square minus
9. So, the result of Laplace transform of
sin hyperbolic 3 t is here 3 by s square minus
9.
The next example is on the similar line, find
the Laplace transform of cos of 5 t, you know
the Laplace transform of cos t is s by s square
plus 1, so that again changing the scale property.
We can write down that Laplace transform of
cos 5 t is 1 by 5 f of s by 5, and this equals
1 by 5 into you just a replace s by s by 5,
so that you will obtain 1 by 5 into this and
the result will become that is by s square
plus 25.
Let us see the next example. Laplace transform
of G t, where G t is e power t minus a, whenever
t is greater than a, and t is less than a,
the value of the function G t is equals to
0. So, we are assuming initially that Laplace
Laplace transform of G t is g s. So, Laplace
transform of G t if the e power minus a s
into g s. So, please note this one by second
shifting property Laplace transform of G t
is e power minus a s into g s.
So, now we are assuming that F t is equals
to e power t say. Once F t is e power t, so
Laplace transform of F t that is Laplace transform
of e power t, which is known to us which is
equals 1 by s minus 1, which is f s where
s is greater than 1.
Now, we are assuming G t equals F of t minus
a. Your F t is e power t, therefore your G
t will be F of t minus a over here, which
is equals to e power t minus a for t greater
than a, and 0 for t less than a. So, if you
take Laplace transform of G t, in the earlier
slides we have shown that it is nothing but
e power minus a s into f s, where Laplace
transform of F t that is e power t is 1 by
s minus 1, so that the value becomes e power
minus a s by s minus 1.
Next states the next example that is F t equals
cos of t minus 2 by 3 pi, whenever t is greater
than 2 by 3, and t is less than 2 by 3. So,
just like earlier example, your F t is this,
so I have to take your one function the function
here is cos of t minus 2 by 3 into pi.
So, I will take a function phi t equals cos
t, whose Laplace transform is known to us.
And we will take F t equals this phi of t
minus 2 by 3 of pi that is nothing but your
cos of this thing will come t minus 2 by 3
of pi.
Now, Laplace transform of phi t that is Laplace
transform of cos t, which is known to us equals
s by s square plus 1, which is equals to f
s. And Laplace transform of F t will be here,
e power minus 2 by pi 2 pi by 3 into s, here
your a is 2 pi by 3.
Basically, we have made t greater than a and
t greater than t less than a, and the value
of a here is 2 pi by 3, so that it becomes
e power this particular thing this one becomes
your this particular case here your a is 2
pi by 3, so that you will obtain Laplace transform
of F t equals e power minus twice pi by 3
into f s using the second shifting property.
So, whenever I am using the property, it becomes
very easy for us to find out the Laplace transform
of unknown functions, when I know the Laplace
transform of certain known functions.
The next one is 
your this thing Laplace transform of phi t
equals this.
And some alternative method you can use quickly
let us see. Laplace transform of F t if I
do not want to use the property, then I can
break it into 2 parts 0 to 2 pi by 3 plus
2 pi by 3 to infinity into e power minus s
t for the 1st part the value is 0, for the
2nd part the function F t is cos of t minus
2 pi by 3. But, here you see you have to evaluate
this particular integral this one e power
minus s t cos of t minus 2 by 3 into pi, and
we are substituting this t minus 2 by 3 equals
pi simply evaluation of an integral in this
case.
And after evaluating the integral, you are
coming to Laplace transform of cos t. So,
if see if my if I know the property earlier
in that case directly by knowing Laplace transform
of cos t, I can tell what will be the Laplace
transform of cos this function, and I do not
have to evaluate the integral.
The next one is a function is given like this
sin of t minus pi by 3 and 0, which is a similar
to your last example. So, I am just giving
the examples, so that afterwards you can just
use it, you can see go through these things
and quickly you can find the solution. So,
without going in a details Laplace transform
of sin t is your 1 by s square plus 1. And
from here I can find out Laplace transform
of F t is equals to e power minus pi by s
into 3 pi sorry pi s by 3 into f s using the
second shifting property. Now, from this f
s equals to 1 by s square plus 1, so that
your result is value of your Laplace transform
of the function is this quantity e power minus
pi s by 3 into 1 by s square plus 1.
Now, come to the another property that is
Laplace transform of derivatives of F t. Let
capital F t be a continuous function for all
t greater than equals 0. And be of exponential
order a as t approaches infinity. And if F
dash t is of plus a, if please note F dash
t is also piecewise of class a means, piecewise
continuous and is of exponential order.
Because, if you recall class a we define like
that where class a will be those type of functions,
which are piecewise continuous and is of exponential
order. So, note that F t is continuous and
is of exponential order a as t approaches
infinity, and the first derivative of the
function F t that is F dash t is of class
a that is is piecewise continuous, and is
of exponential order. Then Laplace transform
of the derivative of f that is f dash t exist,
when s greater than a.
And we can say that Laplace transform of F
dash t, this is equals s into Laplace transform
of F t minus F 0. So, please note Laplace
transform of derivative of a function is equals
to s into Laplace transform of F t minus F
0. Let us see the proof of this one.
Now, so at first the two cases can arise your
case 1 is your case 1. In this case, you are
assuming that F dash t is continuous, we can
think of two cases, it may be continuous for
all t greater than equals 0 or F dash t is
the piecewise continuous. So, we are considering
the first function F dash t is continuous
for all t greater than 0.
Then Laplace transform of F dash t, this is
equals you can write down from the definition
of Laplace transform 0 to infinity e power
minus s t into F dash t d t. And this equals
you can break it into by parts e power minus
s t into F t will come here, F dash t will
be replaced, where value of t will vary from
0 to infinity plus s into 0 to infinity e
power minus s t into F t d t.
Now, this value I can write it here that one
part will be 0, the other part will be limit
s t approaches infinity e power minus s t
e power minus s t into F t. And if I integrate
this one, in this case I will get for 0 it
will be f 0, so for the first part limit s
t approaches infinity e power minus s t into
F t at t equals 0, this e power s t will vanish,
so that it will become minus s 0.
And this part is s into Laplace transform
of this F t this note this thing that whenever
I am writing this, I know the F 0 has a finite
value s Laplace transform of F t, this Laplace
transform of F t is known to us since it is
continuous So, let us see we have to only
check this part limit t approaches infinity
e power minus s t F t F t, whether this is
finite or not. If I the limiting value exists,
then we can say that Laplace transform of
F dash t exists, and it will have some value.
So, for this one let us we know this thing
absolute modulus of F t, this is less than
equals M into e power a t for all t greater
than equals 0. This we are getting since effect
F t is of exponential order a in the theorem
itself, we have told that F t is continuous,
and is of exponential order a.
So, from there we are writing absolute value
modulus of F t less than equals M into e power
a t for all t greater than 0, and for some
constants a and capital M for some constants
will come a and capital M, so that you can
write down e power minus s t into F t modulus
of this one is less than equals e power minus
s t into modulus of F t this you are getting
directly this is equals this is less than
equals again if using the earlier inequality,
this we can write down less than equals e
power minus s t into M into e power a t this
M into e power a t we are getting from here
So, this equals you can write down M into
e power minus s minus a into t. Now, you see
this quantity M into e power minus s minus
a into t, this approaches 0 as t approaches
infinity. Of course, if s is greater than
a, whenever s is greater than they this term
e power minus s by s minus a by into t approaches
0 as t approaches 0.
Therefore, you can say that limit t approaches
infinity e power minus s t into F t, this
is equals to 0 from earlier one e power you
can get t approaches 0, this is 0 where for
s greater than a. So, from this term from
these two, we can conclude that from this
equation we can conclude that Laplace transform
of F dash t this one exists.
Laplace transform of F dash t and exists and
what will be the value of this Laplace transform
of F dash t the earlier thing which we got
that is s into Laplace transform sorry this
will be s into Laplace transform of F t minus
this value that is F 0, and the limiting value
which approaches 0. So, this completes the
proof whenever you are having the function
F t is a continuous function, then we are
showing that Laplace transform of F dash t
exists, and Laplace transform of F dash t,
which is equals to s into Laplace transform
of F t minus F 0.
Now, come to the case-2 that is the second
case in case-2 what we are showing that is
in the first case we have assumed that F dash
t is continuous here we are saying that F
dash t is piecewise continuous. So, what you
can conclude from this in this case the given
integral whatever was there that is 0 to infinity
e power minus s t into F dash t d t this is
your Laplace transform of F dash t, which
I wrote earlier also.
So, this Laplace transform of F dash equals
0 to infinity e power minus s t into F dash
t. Since, F dash t is piecewise continuous
that means, in each particular sub domain
the function F dash t will be continuous.
So, this integral I can break it into n number
of finite sub integrals, say 0 to a 1 plus
a 1 to a 2 plus a 2 to a 3 like this way plus
a n to infinity.
And in each of them the function F dash t
is each of those F dash t is continuous. So,
since F dash t is continuous in each of the
sub intervals. Therefore, just like case-1
we can prove in the similar way. In each of
the sub intervals, the integral exists and
it has a value and if we calculate, we will
get the value as same as we got it for the
case-1 that is Laplace transform of F dash
t, this is equals s into Laplace transform
of F t minus f 0.
So, please note that once I know the Laplace
transform of F t, then very easily using this
theorem, I can say what is the Laplace transform
of derivative of that function also, which
is nothing but S into Laplace transform of
F t, which is again known to me minus F 0.
So, this is the Laplace transform of derivative
of the function.
Now, come to the this one, this we have just
written like this whatever I described you
can go through these slides.
And you can see it over here, the you are
getting it and from here you are saying e
power modulus of e power minus s t F t approaches
0. So, from 2 here concluding this and is
Laplace transform of F dash t exists, and
the value is it is Laplace transform of F
t minus F 0.
For case-2 again F dash t is piece-wise continuous,
so integral 1 may be broken as the sum of
integrals in different ranges such that F
dash t is continuous in each of the subdomains.
And proceeding as in case-1, we can prove
the same thing.
Now, let us take the next one that is Laplace
transform of nth derivative of F t means,
once I know the derivative of Laplace transform,
what can be the Laplace transform of nth derivative
of a function. This is equals s to the power
n Laplace transform of F t F t minus s to
the power n minus 1 F 0 minus s to the power
n minus 2 F dash 0 like this way minus F n
minus 1 0.
So, let us see how we can prove this thing.
One thing I know, already we have proved Laplace
transform of F dash t, this is equals s into
Laplace transform of F t minus F 0. So, this
we already know.
From here if you apply again what will be
Laplace transform of F double dash t, it should
be s into Laplace transform of F t minus F
dash 0, because F 0 is replaced by F dash
here. In this we have replaced F by F dash
your F dash is the function again Laplace
transform of F t is known to us. So, we can
write down s into sorry this will be also
F dash, because F t is replaced by F dash
t.
So, now I know the Laplace transform of F
dash t from here that is s into Laplace transform
of F t minus F 0 and minus this F dash 0 will
come over here. Once F dash 0 is coming over
here, so this is s square into Laplace transform
of F t minus s F 0 minus F dash 0. So, whatever
nth derivative, we talked about that is this
one.
So, Laplace transform of F double dash t,
which we are showing as s square Laplace transform
of F t minus s F 0 minus F dash 0. Now, whenever
we try to find out Laplace transform of F
triple t that is 3rd derivative of the function
F Laplace transform of 3rd derivative of the
function F.
We can go on the similar process. And we can
write down Laplace transform of third derivative
of F of t using the property s into Laplace
transform of F double dash t minus f double
dash 0 simply we can write down, because F
is replaced by F triple dash t.
Now, this F double dash t you can simply write
it as again I know it in the last one we have
done s square into Laplace transform of F
t minus s into F 0 minus F dash 0 using the
2nd derivative, we are getting minus F double
dash 0. Now, everything is known since Laplace
transform of F t is known to us. So, you can
write down s cube into Laplace transform of
F t minus s square into F 0 from here minus
s F dash 0 minus F double dash 0.
So, like this way if we proceed in that case
we can tell that Laplace transform of F n
t, this will be equals to s to the power for
3, it was s to the power 3, so it will be
s to the power n Laplace transform of F t
minus this will be s to the power 3 was there,
then 2 was there, so it will be n minus 1
into F 0 minus s n minus 2 into F dash 0 like
this way, it will proceed.
And ultimately the last term, which you will
get F n minus 1 0. Here when it was 3 e was
getting a double dash 0, so that here you
will obtain F n minus 1 0. So, Laplace transform
of F n t also we can find out by this way
if the derivatives are known to us.
So, the by the derivative whenever you are
telling when the Laplace transform of a function
is known to us, I can find out what can be
the Laplace transform of the first derivative
of a function. And 2nd derivative of a function,
and nth derivative of a function also. Similarly,
we can find out from here.
Now, just we have shown this thing F dash
t, I am showing it in some other way. So,
F double dash t, we have shown here.
Then we are showing F triple t, and then Laplace
transform of this one. So, this is the derivative
of a function. The next one is the Laplace
sorry you just see this thing we have written
in terms of summations. s n Laplace transform
of F t minus summation r equals 0 to s n minus
1, s to the power n minus 1 minus r F to the
power r 0. So, r if varying from 0 to n minus
1, so that last term this n will vanish, and
only this term F n minus 1 0 will be there.
The next one is Laplace transform of integrals
of a function. If F t is piece-wise continuous
and satisfies absolute value modulus of F
t less than equals M into e power a t or in
other sense M F t is the exponential order
or a, for t approaches infinity, for some
constants a and M.
Then Laplace transform of 0 to t F x d x,
this is equals 1 by s Laplace transform of
F t. So, Laplace transform of integral of
the function is equals to 0 integration of
that function is equals to 1 by s Laplace
transform of F t or in other sense Laplace
transform of 0 to t F x is equals to this
quantity.
So, let us see how we can give the proof.
Your G t is equals to 0 to t F x d x we are
assuming, so that whenever you are writing
this G 0 0 to 0, so that the value will be
0. Now, what is your G dash t your G dash
t is nothing but d d t of G t and d d t of
G t is this one 0 to t F x d x, so that it
will be some function of t ultimately, because
it is from 0 to t.
So, what is your Laplace transform of F t,
Laplace transform of F t you know it Laplace
transform of F t is Laplace transform of G
dash t, F t is nothing but G dash t, we are
assuming. So, this is equals again using the
earlier theorem that is derivative
Laplace transform of derivative of a function,
you can write down s into Laplace transform
of G of t minus G of 0 your G of 0 is 0 from
here, so that you are getting is into Laplace
transform of G of t or in other sense I can
from here I can tell 1 by s into f s Laplace
transform of F t 1 by s into f s 1 by s is
coming here, and Laplace transform of F t
I am assuming as f s. This is nothing but
Laplace transform of G t. And what is your
G t, G t is nothing but G of this thing Laplace
transform of Laplace transform of of 0 to
t F x into d x.
So, from here if you see Laplace transform
of F t, we are getting s into Laplace transform
of G t, and then we are proving that Laplace
transform of of 0 to t F x d x is nothing
but 1 by s into f s, where f s is the Laplace
transform of this one. So, therefore if I
know that Laplace transform of F t is f s,
then Laplace transform of d d t of 0 to t
F x d x this is equals 1 by s into f s. So,
this is the theorem on the or in other sense,
if I know the Laplace transform of a function,
then I can find out what is the Laplace transform
of derivative of that function also.
Thank you.
