Homework Help: Laplace D.E Problem

1. May 26, 2012

Hiche

1. The problem statement, all variables and given/known data

2. Relevant equations

Laplace's transforms.

3. The attempt at a solution

Okay, so applying the Laplace on both sides yields:

$s^2Y(s) - sy(0) - y'(0) + 2sY(s) - 2y(0) + 10Y(s) = ? + 13 / s - 1$

Is $e^{-2s} / s$ the Laplace $\delta(t - 2)$? Our instructor gave us the result of the Lapace but he did not prove it. The only thing he gave us was that the Laplace of $f(t - a)\delta(t - a) = e^{-as}F(s)$. Can someone point me on how to prove this? It seems our instructor told us that we need to know the proof without him giving it to us. I know that this unit step function is defined to be 0 when 0 <= t < 2 and 1 when t >= 2.

2. May 26, 2012

vela

Staff Emeritus
Are you sure $\delta(t)$ is the unit step function? It typically denotes the Dirac delta function.

3. May 27, 2012

Hiche

Yes yes. I mixed up the function. Sorry about that.

So is the Laplace of $\delta(t - 2)$ typically $e^{-2s} / s$?

4. May 27, 2012

vela

Staff Emeritus
Which function are you talking about? If it's the delta function, then no, that's not correct. If it's the unit step, then that's right.

5. May 27, 2012

Hiche

The delta function. Then is it simply $e^{-2s}$?

If so, how to prove it starting with $\int^\infty_0 e^{-st}\delta(t - a)dt = e^{as}$? The proof is apparently required for our exam yet our instructor failed to provide the solution.

6. May 27, 2012

vela

Staff Emeritus
What's the defining property of the Dirac delta function?

7. May 27, 2012

Hiche

..that the $\int^{a+e}_{a-e} f(t)\delta(t - a)dt = f(a)$ for $e > 0$?

I appreciate your patience but I'm relatively 'new' to this concept.

8. May 27, 2012

vela

Staff Emeritus
Yup. Just apply that to the Laplace transform integral you have.