Laplace Transform of t^1 x e^(3t): Solving the Dilemma

[Ry122]

Im trying to find the laplace transform of t^1 x e^(3t)
but looking it at the table, it looks like there's two different possible solutions for it.
one is for t^(n) x f(t)
and the other is for e^(at) x f(t)
which one do i choose?

 

[Defennder]

Do you get different answers for both? In this case, the table I'm using would point to the one for e^(at)f(t).

 

[Ry122]

i do get two different answers.
doesnt ur table have t^(n) x f(t)? wouldn't that also satisfy it?

 

[Defennder]

My table doesn't have that one. What are your answers for each?

 

[Ry122]

this is what the table has.
http://users.on.net/~rohanlal/2222.jpg

 

[HallsofIvy]

Okay, in your case n= 1 so it is just -F(s). What is F(s), the Laplace transform of e^3t?

Of course, you could use the basic formula for Laplace transform:
$$L(s)= \int_0^\infty t e^{3t}e^{-st}dt$$
using integration by parts.

 

[Ry122]

so i choose the other one for t^1 x e^(3t), the one that defender mentioned, because n = 1?
so if i used the one in my previous post, that would that be incorrect?

 

[Metaleer]

You could also use the other property, namely that

$$\displaystyle \mathcal{L}[e^{at}f(t)] = F(s-a),$$​

where $$F(s) = \mathcal{L}[f(t)].$$ Using this, you only need to get the Laplace transform of $$t$$, and evaluate it at $$s-3$$. You should get the same result with both properties.

Good luck.

 

[HallsofIvy]

The point is that all three methods:
a)$$\displaystyle \mathcal{L}[e^{at}f(t)] = F(s-a),$$
where F(s) is the Laplace transform of t.

b)$$\displaystyle \mathcal{L}[tf(t)]= -F'(s)$$
where F(s) is the Laplace transform of $e^{3t}$.

c)$$\displaystyle \mathcal{L}[te^{at}]= \int_0^\infty te^{(-s+3)t}dt$$

will give the same result.

It would be a good exercise to try each method and see.

 

[Ry122]

for the laplace transform of t^2 x e^(3t) (n is greater than 1)
would http://users.on.net/~rohanlal/2222.jpg be the correct one to use?

 

[Metaleer]

All three ways are correct, but I personally think the exponential property is the quickest, if you already know the Laplace transforms of polynomials.

 

[Ry122]

can you reread my previous post, i put in the wrong url for the image.