Question

Next Laplace Question

Answer 1

EValuate the Convulution

\[\mathcal L \{e^{-t}*e^tcost\}=\mathcal L\{e^{-t}\}\mathcal L\{e^tcost\}\]

Answer 2

@waterineyes

Answer 3

@jim_thompson5910

Answer 4

\[\mathcal{L}(e^{-t}) = \int\limits_0^{\infty}(e^{-t} \cdot e^{-st}) dt \implies \int\limits_0^{\infty}(e^{-t(s + 1)}dt \implies |\frac{e^{-t(s+1)}}{-(s+1)}|_0^{\infty} \implies \frac{1}{s+1}\]

Answer 5

Generally:

\[\mathcal{L} (e^{-at}) = \frac{1}{s + a}\]

Answer 6

In general:

\[\mathcal{L}(\cos(at)) = \frac{s}{s^2 + a^2}\]

Answer 7

\[\mathcal L \{e^{-(-t)}cost\}=\frac{s+1}{(s+1)^2+1}\]

Answer 8

Wow...

Answer 9

so the answer in the back book should be

\[\frac{1}{s+1}\frac{s+1}{(s+1)^2+1}\]

Answer 10

unless they simplified it hold on let me check

Answer 11

ahhh gahd i messed up somewhere lol

Answer 12

But I think that is not right..

Answer 13

there is a sign error somewhere

Answer 14

For positive a, you will do s - 1..

Answer 15

ahh i put a -

Answer 16

*s - a

Answer 17

yeah

Answer 18

Yeah can you do now ??

Answer 19

yeah the harder part is the inverses of convulutions some of the identities thy use are really old or completely wrong in the book

Answer 20

\[\mathcal{L} (e^{at} \cdot \cos(wt)) = \frac{s - a}{(s-a)^2 + w^2}\]

Answer 21

For positive a we will use s - a

For negative a that -a we will use s+a..