Laplace transform of the dirac delta function

[november1992]

Homework Statement

L[t^{2} - t^{2}δ(t-1)]

Homework Equations

L[ t^{n}f(t)] = (-1^{n}) \frac{d^{n}}{ds^{n}} L[f(t)]

L[δ-t] = e^-ts

The Attempt at a Solution

My teacher wrote \frac{2}{s^{3}} -e^{s} as the answer.

I got \frac{2}{s^{3}} + \frac{e^-s}{s} + 2 \frac{e^-s}{s^2} + \frac{2e^-s}{s^3}

 

[Mute]

You should show us how you derived your answer. That way we can best help you see where you went wrong. Right now we have to guess.

My guess is that you tried to use the identity under "Relevant equations" on the Dirac delta term, which likely made your problem more complicated than it needed to be.

Are you aware of the rule $f(t)\delta(t-t_0) = f(t_0)\delta(t-t_0)$? This will simplify your calculation.

 

[november1992]

I don't see that formula in my textbook.

When I plug in the numbers I get: \frac{2}{s^3} + \frac{e^-s}{s}

This is what I did to get my first answer.
L[t^2 - t^2 δ(t-1))]


\frac{2}{s^3} + \frac{d}{ds} (\frac{d}{ds} (\frac{e^(-s)}{s})




\frac{2}{s^3} +\frac{d}{ds} (\frac{e^-s * s + e^-s}{s^2} + \frac{e^-s*s^2 + 2s*e^-s}{s^2})

\frac{2}{s^{3}} + \frac{e^-s}{s} + 2 \frac{e^-s}{s^2} + \frac{2e^-s}{s^3}


Okay I just checked the answer on wolframalpha and what my teacher had was the correct answer. Why is it that L[t^n*d(t-a)] = e^-as

Edit:
Nevermind, I've found my mistake.