Math Help - Laplace Transform and Heaviside Function

1. Laplace Transform and Heaviside Function

Hello, This is my first post here, after seeing that a lot of great help is on hand, so i thought i would give it a try as i am having difficulty.

question: Find the Laplace Transform of the following from first principles;

$

f(x)=(x+2)^{2}H(x-3)

$

After looking at some other examples on the forum, I have tried letting $g(t)=(t+5)$ in order to make the two expressions inside the brackets the same, ie $(x-3)$ so that standard rules can be applied, though I am not quite sure if this is right or where to go from here. ANy help would be much appreciated!

2. Progress so far

In order to give you a batter idea of where my train of thought is, this is the answer I have got, through the method below;

let $g(t)=(t^{2}+4t+1)$

therefore $f(x)=(x-3)H(x-3)$

so, from first principles, although, using rule: 

\begin{aligned}L\left\{f(x)\right\}&=e^{-3s}\bigg[L\left\{x-3\right\}\bigg]\\ &=e^{-3s}\left(1/s^2-3/s\right) \\ &=\frac{e^{-3s}}{s^2}-\frac{3e^{-3s}}{s}\end{aligned}

The question also asks to "Indicate for what values of 's' is the transform valid?" I am not sure what this means, is it something involving the solution of the $g(t)$ function?

Again, thanks a lot for any help or pointers or letting me know I am barking completely up the wrong tree!

3. Originally Posted by arickard1988
Hello, This is my first post here, after seeing that a lot of great help is on hand, so i thought i would give it a try as i am having difficulty.

question: Find the Laplace Transform of the following from first principles;

$

f(x)=(x+2)^{2}H(x-3)

$

After looking at some other examples on the forum, I have tried letting $g(t)=(t+5)$ in order to make the two expressions inside the brackets the same, ie $(x-3)$ so that standard rules can be applied, though I am not quite sure if this is right or where to go from here. ANy help would be much appreciated!

Form first principles:

$\mathcal{L}f(s)=\int_{x=0}^{\infty}(x+2)^2H(x-3)e^{-sx}\;dx=\int_{x=3}^{\infty}(x+2)^2e^{-sx}\;dx$

which you will now need to evaluate.

CB

4. Originally Posted by CaptainBlack
Form first principles:

$\mathcal{L}f(s)=\int_{x=0}^{\infty}(x+2)^2H(x-3)e^{-sx}\;dx=\int_{x=3}^{\infty}(x+2)^2e^{-sx}\;dx$

which you will now need to evaluate.

CB
And integration of the final integral found by CB can be avoided by making the substitution $u = x - 3$, take out the 'constant' of $e^{-3s}$, expand $(u + 5)^2$ and then express the integral as three seperate integrals (each of which is recognised as the Laplace transform of a simple function).

Edit: Addition in italics (added to clarify my remark in light of CB's quite correct subsequent post).

5. Originally Posted by mr fantastic
And integration can be avoided by making the substitution $u = x - 3$, take out the 'constant' of $e^{-3s}$, expand $(u + 5)^2$ and then express the integral as three seperate integrals (each of which is recognised as the Laplace transform of a simple function).
To take the LT from first principles requires that he construct the integrals given in my post, I don't care how he then evaluates them (but he cannot look them up in a table of LTs as that would not be doing this from first principles).

CB

6. Originally Posted by CaptainBlack
Form first principles:

$\mathcal{L}f(s)=\int_{x=0}^{\infty}(x+2)^2H(x-3)e^{-sx}\;dx=\int_{x=3}^{\infty}(x+2)^2e^{-sx}\;dx$

which you will now need to evaluate.

CB
CB,

Many thanks! Its now crystal, and has helped with the rest of the paper.

Alex

7. Originally Posted by mr fantastic
And integration of the final integral found by CB can be avoided by making the substitution $u = x - 3$, take out the 'constant' of $e^{-3s}$, expand $(u + 5)^2$ and then express the integral as three seperate integrals (each of which is recognised as the Laplace transform of a simple function).

Edit: Addition in italics (added to clarify my remark in light of CB's quite correct subsequent post).
Fantastic,

Thankyou for the add-on advice, has wrapped it up nicely, seems trivial now, as always!

I look forward to using this forum again, a GREAT tool in helping me learn.

Alex