1. ## step Laplace

Hi,
I am bit confused by the step function. I was under the impression that it is basically a plug and chug but I think I am doing something wrong and need some clarification and some guidance.

Find the L[usub2(t)e^(-t)cos(3t)
So I get e^(-2s)L[e^(-2t-4)cos(3t+6)]

I am almost sure that this is wrong but really don't have any examples that are similar. Can someone please help me with this?

2. Originally Posted by pberardi
Hi,
I am bit confused by the step function. I was under the impression that it is basically a plug and chug but I think I am doing something wrong and need some clarification and some guidance.

Find the L[usub2(t)e^(-t)cos(3t)
So I get e^(-2s)L[e^(-2t-4)cos(3t+6)]

I am almost sure that this is wrong but really don't have any examples that are similar. Can someone please help me with this?
$\mathcal{L} u_2(t) e^{-t}\cos(3t) =\int_2^\infty e^{-t}\cos(3t) e^{-st} dt$

Now change the variable to $\tau=t-2$ then we have:

$\mathcal{L} u_2(t) e^{-t}\cos(3t) =\int_0^\infty e^{-\tau-2}\cos(3\tau + 6) e^{-s\tau-2s} d\tau$

Simplify:

$\mathcal{L} u_2(t) e^{-t}\cos(3t) =e^{-2s-2}\int_0^\infty e^{-\tau}\cos(3\tau + 6) e^{-s\tau} d\tau=e^{-2s-2}\left[\mathcal{L}e^{-t}\cos(3t + 6)\right]$

Or you may prefereint in the form:

$\mathcal{L} u_2(t) e^{-t}\cos(3t)=e^{-2s}\left[\mathcal{L}e^{-t-2}\cos(3t + 6)\right]$

Alternativly fiddle with the translation theorem to get the same result.

CB

3. Well the step after that is actually the confusing part. How does the exponential with the t mesh with the exponential with the s? Also are you taking the Laplace of the exp with the t? How does that come out? Thank you for your assistance but it is from this part on that I get confused about.