Question

Case closed wew

Answer 1

Oo this one is a little tricky :)
Integration by parts business ya?

Answer 2

yas sure hahaha

Answer 3

I need to review everything

Answer 4

\[\large\rm \mathscr L\{t^2\}\quad=\quad \int\limits_0^{\infty}t^2 e^{-st}dt\]

Answer 5

\(\large\rm u=t^2,\qquad\qquad dv=e^{-st}dt\)

Answer 6

du and v? :)

Answer 7

du= 2t

Answer 8

i remember spending a ton of time practicing those things while in the class..., convolution , step function, shift, exe...
now i can't remember a dam thing, well the integral definition i did

Answer 9

good good good.

Answer 10

prolly returns quickly if you practice a little

Answer 11

@DanJS Haha i even forgot everything 'bout integ

Answer 12

\[\large\rm \mathscr L\{t^2\}\quad=\quad \int\limits\limits_0^{\infty}t^2 e^{-st}dt\]\[\large\rm \mathscr L\{t^2\}\quad=\quad uv-\int\limits v~du\]\[\large\rm \mathscr L\{t^2\}\quad=\quad -\frac{1}{s}t^2e^{-st}-\int\limits\left(-\frac{1}{s}\right)2t~e^{-st}~dt\]Something like this, yes?

Answer 13

We'll have to evaluate everything at 0 and infinity,
maybe we can do that now...
or later...
whatever you prefer.

Answer 14

Cause it looks like we have to apply Integration by Parts again :)

Answer 15

Oh imma don't like integ by parts. Imma sure ill be getting lost

Answer 16

what was the point of changing from the ' t '-space to that 's'-space anyway, besides solving a DE easier in that land

Answer 17

Let's bring any constant garbage out front,
and combine the negatives,\[\large\rm \mathscr L\{t^2\}\quad=\quad -\frac{1}{s}t^2e^{-st}+\frac{2}{s}\color{orangered}{\int\limits\limits t e^{-st}~dt}\]So we're just applying parts to this orange portion.

Answer 18

Oh now that looks easier i think

Answer 19

Similar assignment for u and dv would be appropriate, ya?
\(\large\rm u=t,\qquad\qquad\qquad dv=e^{-st}dt\)

Answer 20

du= dt, v= e^-st/s

Answer 21

\[\large\rm \mathscr L\{t^2\}\quad=\quad -\frac{1}{s}t^2e^{-st}+\frac{2}{s}\color{orangered}{\int\limits\limits\limits t e^{-st}~dt}\]\[\large\rm \mathscr L\{t^2\}\quad=\quad -\frac{1}{s}t^2e^{-st}+\frac{2}{s}\color{orangered}{\left[uv-\int\limits v~du\right]}\]\[\large\rm \mathscr L\{t^2\}\quad=\quad -\frac{1}{s}t^2e^{-st}+\frac{2}{s}\color{orangered}{\left[-\frac{1}{s}t e^{-st}-\int\limits \left(-\frac{1}{s}\right)e^{-st}dt\right]}\]

Answer 22

Sorry if I'm stealing all the fun here :)
I can slow down on the integrals lol.

Answer 23

It really okay haha

Answer 24

That last integral is pretty straight forward, just another -1/s, in front of the exponential, right?

Answer 25

\[\large\rm \mathscr L\{t^2\}\quad=\quad -\frac{1}{s}t^2e^{-st}+\frac{2}{s}\color{orangered}{\left[-\frac{1}{s}t e^{-st}-\left(-\frac{1}{s}\right)\left(-\frac{1}{s}\right)e^{-st}\right]}\]

Answer 26

Yes yes im following you

Answer 27

How could i get 2/s^3?

Answer 28

Let's distribute to the orange stuff,\[\large\rm =-\frac{t^2e^{-st}}{s}-\frac{2te^{-st}}{s^2}-\frac{2e^{-st}}{s^3}\qquad |_0^{\infty}\]

Answer 29

So that's all the stuff we have.
You might see where we'll EXPECT to get the 2/s^3 from, right? :)

Answer 30

Quiet confused with this -2e^-st /s^3

Answer 31

Ah yes I got it now swear haha

Answer 32

You have to evaluate all this mess at 0 and infinity.
Maybe start with the FIRST TERM, cause you'll see something interesting happen.

Answer 33

Not sure how can i get the transform 2/s^3 haha sorry

Answer 34

Let's switch the limits and get rid of the negative in front\[\large\rm -\frac{t^2e^{-st}}{s}|_0^{\infty}\quad=\quad \frac{t^2e^{-st}}{s}|_{\infty}^0\quad=\quad \frac{0^2e^{-0}}{s}-\lim_{b\to\infty}\frac{b^2e^{-sb}}{s}\]

Answer 35

0+0+ 2/s^3 hahaha finally

Answer 36

Do you understand how to show that the first two terms are 0 though? :)
Requires L'Hopital's rule.

As long as you understand that, then yes we can say 0+0+(the last part).

Answer 37

Oh sorry internet sucks. Thank you so much @zepdrix :)