Question

Alright, next problem in the long list that I couldn't figure out how to do using convolution (posted below in a moment).

Answer 1

\[\mathcal{L^{-1}}\left\{\vphantom{}\frac{1}{s^3(s-1)}\right\}\]

Answer 2

(Or, excuse me, I'm not supposed to use convolution in the regular sense, but I am supposed to use the F(s)/s convolution property):

Answer 3

\[\int\limits_{0}^{t}f(\tau)d \tau=\mathcal{L}\left\{\vphantom{}\frac{F(s)}{s}\right\}\](where g(t) or g(t-tau)=1)

Answer 4

Putting it in this general form:\[\mathcal{L}\left\{\vphantom{}\frac{1/s^2(s-1)}{s}\right\}\]

Answer 5

\[\mathcal{L^{-1}}\left\{\vphantom{}\frac{1}{s^2(s-1)}\right\}=?\]I can approach this using partial fractions, but I was wondering if somebody had a preferred or better way.

Answer 6

(Just dealing with F(s) by itself, above)

Answer 7

@sidsiddhartha , do you think there is a better way to do this inverse laplace at the bottom other than partial fractions, if the 1/s F(s) property has already been applied?

Answer 8

\[take\\F_1(s)=\frac{ 1}{ s^3 }\\and\\F_2(s)=\frac{ 1 }{ s-1 }\]
then find inverse and then conv.integral

Answer 9

Oh, so you can do it again? This was one of my earlier questions but I didn't outright ask it; this separating inverse laplace functions and using convolution is recursive, like,

Answer 10

(Oh well, enough words for now, just going to do it):\[\mathcal{L^{-1}}\left\{F_{1}(s) \right\}=t, \ \ \ \mathcal{L^{-1}}\left\{F_{2}(s)\right\}=e^t\]

Answer 11

\[F_{1}(s)F_{2}(s)=te^{t}\]

Answer 12

hey check again
L91/s^3]=?

Answer 13

oh typo

Answer 14

\[l[1/s^3] \neq t\]

Answer 15

Oh, read that as a 3, my bad, one sec\[\mathcal{L^{-1}}\left\{\vphantom{}\frac{1}{s^{3}}\right\}=\frac{t^2}{2}\]

Answer 16

yeah :)

Answer 17

Alright, now I'm getting confused because I'm trying to go quick because of time constraint and am prone to make some mistakes, lol, one sec.

Answer 18

\[\frac{t^{2}}{2}e^{t}\](Literally just entirely forgot what to do from here, one sec)

Answer 19

now
\[\int\limits_{0}^{t}F_1(\tau).F_2(t-\tau)d \tau\]

Answer 20

Alright, so just so we're on the same page, we're using two different instances of convolution? The first one to isolate F(s)/s, and the second one to find the laplace of F(s) itself in the form of F_19s) and F_2(s)?

Answer 21

\[\int\limits_{0}^{t}\frac{\tau^2}{2}e^{t-\tau}d \tau=e^t \int\limits_{0}^{t}\frac{\tau^2}{2}e^{\tau}d \tau\]

Answer 22

yes just a typo that e^{- tau}

Answer 23

Evaluating the integral here using integration by parts,\[\int\limits_{0}^{t}\frac{\tau^2}{2}e^{-\tau}d \tau=-\frac{\tau^{2}e^{-\tau}}{2} -\int\limits_{0}^{t}\ -\tau e^{-\tau}\]

Answer 24

Integration by parts again,\[-\frac{\tau e^{-\tau}}{2}+\int\limits_{0}^{t}\tau e^{-\tau}d \tau=-\frac{\tau e^{-\tau}}{2}+\Bigg(-\tau e^{-\tau}-\int\limits_{0}^{t}-e^{-\tau}d \tau \Bigg)\]

Answer 25

yes that looks fine :)

Answer 26

\[-\frac{\tau e^{- \tau}}{2}-\tau e^{-\tau}+e^{\tau}\]

Answer 27

Whoops, last exponent is still -tau

Answer 28

Evaluated (lol forgot to evaluate this whole time oh no)

Answer 29

\[e^{-t}\bigg[1-\frac{t}{2}-t\bigg]-\bigg[1-\frac{1}{2}\bigg]\]

Answer 30

Time to see how off this answer is from the real thing.

Answer 31

God, man, sometimes I feel like redoing this from scratch, real answer is \[e^{t}-\frac{1}{2}t^2-t-1\]Going to start over and try to finish this, then moving on to an integrodifferential equation problem, then dirac delta, then systems of linear diff eq, time is running out.

Answer 32

man u are something :P