Quick laplace transform question

[orangeincup]

Homework Statement

So I know 1/(s-a)=e^(a1), but why is say, 2/((s+4)^2) equal to 2xe^-4x? Do I just simply add an X if the numeration is a constant other than 1?

 

[Chestermiller]

Why don't you evaluate the Laplace transform of the expression and see what you get?

 

[orangeincup]

2*e^(-4*x) is what I get, I don't know where the other x came from.

Well I mean I know it's because it's squared, I just don't see how my laplace transform formula is giving me that

 

[Chestermiller]

2*e^(-4*x) is what I get, I don't know where the other x came from.

Well I mean I know it's because it's squared, I just don't see how my laplace transform formula is giving me that

Like I said, evaluate the LT of the expression with the x's and report back with what you got.

 

[orangeincup]

L(2x*e^(-4x)) = 2x* 1/(s-4)

 

[orangeincup]

Am I suppose to treat it as two different Laplace transforms combined? Not sure how that would work but I know the two individual formulas below
2/((s+4)^2)

1/s^n = t^n-1/(n-1)! = x*2
and
1/(s-a) = e^-4x

 

[Chestermiller]

L(2x*e^(-4x)) = 2x* 1/(s-4)

Let's see you do the integration. How come you didn't include the leading x in the integration?

Chet

 

[HallsofIvy]

By the way, your original statements, that "1/(s-a)=e^(at)" and that "2/((s+4)^2) is equal to 2xe^-4x" are non-sense. You knew that when you wrote them, didn't you? What you meant was that "1/(s- a)" is the Laplace Transform" of "e^(at)" and that "2/((s+ 4)^2)" is the Laplace Transform of "2xe^(-x)". It is always better to say what you mean!