Question

Laplace Transforms Question

Answer 1

\[\mathcal L (1*t^3)\] evaluate

Answer 2

Formula for \[L({t^n}) = {n!}/s^{n+1}\]

Answer 3

that isn't multiplaction it's like a dot product

Answer 4

special instance

Answer 5

\[\mathcal L(1*t^3)=\mathcal L (1) \mathcal L (t^3)\]

Answer 6

Oh a convolution......hmmm. gimme a sec.

Answer 7

\[\mathcal L(1*t^3)=\mathcal L (1) \mathcal L (t^3)\]

This will be a normal multiplication @Outkast3r09

Right ??

Answer 8

No it wouldn't. Laplace transforms don't work like that last time I checked.

Answer 9

I mean to say that if we find L(1) and L(t^3)

Then we will multiply it simply ??

Answer 10

use this
\[\Large L(t^nf(t)=(-1)^n*\frac{d^n}{ds^n}(F(s))\]
where
f(t)=1
so
F(s)=1/s

Answer 11

\[\mathcal {L}(1) = \int\limits _{0}^{\infty}(1)(e^{-st}) dt \implies \left| -\frac{e^{-st}}{s} \right|_0^{\infty} \implies \frac{1}{s}\]

Answer 12

also Laplace is Linear operator so
\[\Large L(A.B)\neq L(a)*L(b)\]

Answer 13

\[\large \mathcal {L} (t^n) = \frac{n!}{s^{n+1}}\]

Answer 14

\[\mathcal{L} \{f(t)*g(t)\}=\mathcal{L}\{f(t)\}\mathcal{L}\{g(t)\}\]You silly geese! "*" represents convolution!!

Answer 15

which =

\[F(s)G(s)\]

Answer 16

so multiplication of the two transforms

Answer 17

\[\mathcal L (1*t^3)=\mathcal L(1) \mathcal L (t^3)=\frac{1}{s}*\frac{6}{s^4}=\frac{6}{s^5}\]