Question

Use the definition of laplace transform to find the transform of f (t)=4^t

Answer 1

\[\Large\rm \mathscr{L}\left(4^t\right)=\int\limits^0^{\infty} 4^t e^{-st}dt\]Hey there :)
So this is how we set up our definition, yes?

Answer 2

Woah my boundaries tweeked out :O sec...

Answer 3

\[\Large\rm \mathscr{L}\left(4^t\right)=\int\limits\limits_0^{\infty} 4^t e^{-st}dt\]

Answer 4

If you change the base on the exponential, it will be easier to integrate.
Recall that:\[\Large\rm \color{orangered}{a}=e^{\ln(\color{orangered}{a})}\]We want to apply this idea to the 4^t.

Answer 5

\[\Large\rm \color{orangered}{4^t}=e^{\ln(\color{orangered}{4^t})}\]

Answer 6

And then applying your exponent rule:\[\Large\rm 4^t=e^{t \ln(4)}\]

Answer 7

\[\Large\rm \mathscr{L}\left(4^t\right)=\int\limits\limits\limits_0^{\infty} e^{t \ln(4)}e^{-st}dt\]Again use rules of exponents to combine the terms :)

Answer 8

\[\Large\rm \mathscr{L}\left(4^t\right)=\int\limits\limits\limits\limits_0^{\infty} e^{(\ln4-s)t}dt\]

Answer 9

And it should be easier to work with from here I think.. hmm

Answer 10

this looks like a trick question as 4^t > e^t

Answer 11

I think it's still ok.
It still converges if we assume that \(\Large\rm s\gt \ln4\).
Just need to give a proper restriction to s, yah? :o

Answer 12

Exactly !

Answer 13

Remember how to integrate an exponential keynote? :)\[\Large\rm \int\limits e^{at}dt=\frac{1}{a}e^{at}\]We just divide by that coefficient on the t.

Answer 14

Keynooooote, where you at?
Play us some music! Jam those notes!

Answer 15

t.t

Answer 16

I am right here. I went out . Am just confused

Answer 17

About which part? :U

Answer 18

I mean, there is still a bit of work to do in order to wrap this up.
But I want to know what you're confused on up to this point.

Answer 19

Oh oh we also want to factor a negative out of each term in the exponential so it looks like this:\[\Large\rm \mathscr{L}\left(4^t\right)=\int\limits\limits_0^{\infty} e^{-(s-\ln4)t}dt\]

Answer 20

Gimme a minute. Lemme try to follow along one more time

Answer 21

How do you do the integration

Answer 22

You really want to be comfortable with this idea:\[\Large\rm \int\limits\limits e^{\color{#CC0033}{a}t}dt=\frac{1}{\color{#CC0033}{a}}e^{\color{#CC0033}{a}t}\]Yes, you can do a u-substitution if you prefer,
but that's really burdensome. You gotta get quick with these :)

Answer 23

Notice that we got our problem into this form:\[\Large\rm \mathscr{L}\left(e^{at}\right)=\int\limits\limits_0^{\infty} e^{\color{#CC0033}{-(s-a)}t}dt\]So using that rule for integrating exponential we get:\[\Large\rm =\frac{1}{\color{#CC0033}{-(s-a)}}e^{\color{#CC0033}{-(s-a)}t}\quad|_0^{\infty}\]

Answer 24

So for our problem:\[\Large\rm \int\limits\limits\limits_0^{\infty} e^{\color{#CC0033}{-(s-\ln4)}t}dt\quad=\quad \frac{1}{\color{#CC0033}{-(s-\ln4)}}e^{\color{#CC0033}{-(s-\ln4)}t}\quad|_0^{\infty}\]

Answer 25

From there we have to evaluate it at the limits.

Answer 26

Okay that's better coz I was getting something different