# Thread: Help with Laplace Transform

1. ## [UNSOLVED]Help with Laplace Transform

So... Got my second diff eq test back today. 113/150.. Lost 31 points because it seems, contrary to my personal belief, I can't do Laplace transforms to save my life.

Therefore, I was hoping someone could help me. Here's the two functions I got very very wrong:

1. $\displaystyle L\{e^t*U(t-2)\}$
2. $\displaystyle L\{Y = \frac {1-e^{-2s}}{(s-3)(s^2)}\}$

Some quick notes. Y = Laplace of y,U(t-2) is that switch function that is zero when t < 2 and 1 when t>=2, and L = laplace transform

Here are my answers/ways I tried to solve them:

1. $\displaystyle L\{e^t*U(t-2)\}$
$\displaystyle L\{e^t\}*L\{U(t-2)\}$
$\displaystyle \frac {e^{-2s}}{s-1}$
It seems that is entirely wrong. I'm not sure why you can't do that and I would love some input.

2. $\displaystyle L\{Y = \frac {1-e^{-2s}}{(s-3)(s^2)}\}$
$\displaystyle y = L\{\frac {1}{(s-3)(s^2)}\} - L\{\frac {1}{(s-3)(s^2)}\}*L\{{e^{-2s}}\}$
$\displaystyle y = L\{\frac {1}{s-3}\} * L\{\frac {1}{s^2}\} - L\{\frac {1}{s-3}\} * L\{\frac {1}{s^2}\} * L\{e^{-2s}\}$
$\displaystyle y = e^{3t} * t - e^{3(t-2)} * (t-2) * U(t-2)$
It seems that is also entirely wrong. As I said earlier. I would LOVE some help!

Regards,
Harrison Jones

2. Originally Posted by harrisonhjones
So... Got my second diff eq test back today. 113/150.. Lost 31 points because it seems, contrary to my personal belief, I can't do Laplace transforms to save my life.

Therefore, I was hoping someone could help me. Here's the two functions I got very very wrong:

1. $\displaystyle L\{e^t*U(t-2)\}$
2. $\displaystyle L\{Y = \frac {1-e^{-2s}}{(s-3)(s^2)}\}$
Some quick notes. Y = Laplace of y,U(t-2) is that switch function that is zero when t < 2 and 1 when t>=2, and L = laplace transform

Here are my answers/ways I tried to solve them:

1. $\displaystyle L\{e^t*U(t-2)\}$
$\displaystyle L\{e^t\}*L\{U(t-2)\}$
$\displaystyle \frac {e^{-2s}}{s-1}$
It seems that is entirely wrong. I'm not sure why you can't do that and I would love some input.

You seem to be confusing the point-wise product of functions and their convolution here. If you intend "$\displaystyle *$" to denote a convolution then:

$\displaystyle \mathcal{L}\{e^t*U(t-2)\}$

is not the same as:

$\displaystyle \mathcal{L}\{e^t\}*\mathcal{L}\{U(t-2)\}$

Nor does it work if you mean it to be the point-wise product.

CB

3. Thanks for answering back. No, those were not convolutions. They are simple multiplications. If anybody would be so kind could they explain how to correctly solve those two problems? I would really really appreciate it. The rest of my diff eq class uses laplace so I need to understand this stuff!

4. Originally Posted by harrisonhjones
Thanks for answering back. No, those were not convolutions. They are simple multiplications. If anybody would be so kind could they explain how to correctly solve those two problems? I would really really appreciate it. The rest of my diff eq class uses laplace so I need to understand this stuff!
Then your notation is confused, when working with transforms never use * for the multiplication operator, it generally means something else.

Now you want:

$\displaystyle \mathcal{L} \{e^t U(t-2)\}(s)$

you can start by just writing out what this means:

$\displaystyle \mathcal{L} \{e^t U(t-2)\}(s)=\int_0^{\infty} e^t U(t-2)e^{-st}\ dt=\int_2^{\infty} e^{(-s+1)t}\ dt$

Which is elementary.

CB

5. Ok got it...