Question

Doing another integrodifferential equation now:

Answer 1

\[f(t)+\int\limits_{0}^{t}f(\tau)d \tau = 1\]

Answer 2

Making the identification that this is the same sort of situation as \[\int\limits_{0}^{t}f(\tau)d \tau = \mathcal{L^{-1}}\left\{\vphantom{}\frac{F(s)}{s}\right\}\]

Answer 3

Alright, uhh, Oh! This is really odd, can I do this? Let's see...

Answer 4

yes this is easier than the first one

Answer 5

LAtEx Isn't playing nice (whoops lol), but I think we're on the same page, inverse and forward operation cancel each other out

Answer 6

you can think of \(\int\limits_0^t f(\tau)d\tau \) as \(f(t)*1\) right ? why to identiy it as some other situation ?

Answer 7

ughh gonna rewrite stuff, one moment
\[F(s)+\mathcal{L}\left\{\mathcal{L^{-1}}\left\{\vphantom{}\frac{F(s)}{s}\right\} \right\}=1\]

Answer 8

\[F(s)+\frac{F(s)}{s}=1\]

Answer 9

you need to transform right side also

Answer 10

yes

Answer 11

forgot that 1

Answer 12

Whoops! Thank you, heh\[F(s)\bigg[1+\frac{1}{s}\bigg]=\frac{1}{s}\]

Answer 13

yeah that fine now just inverse

Answer 14

\[F(s)=\frac{1+s}{s}\](Trying to go quickly, might've jsut made an algebra error)

Answer 15

yes lol

Answer 16

\[F(s)\frac{ 1+s }{ s }=1/s\]

Answer 17

\[F(s)=\frac{ 1 }{ s+1 }\]

Answer 18

i would multiply \(s\) through out so that fractions disappear as i hate fractions

Answer 19

Ah, I see what you did, yeah, makes it way easier

Answer 20

\[\mathcal{L^{-1}}\left\{\vphantom{}\frac{1}{s+1}\right\}=e^{-t}\]

Answer 21

woo-hoo

Answer 22

Alright, ~4 hours left

Answer 23

https://www.youtube.com/watch?v=oOM-91dydPU

Answer 24

Time to start a new problem.

Answer 25

I can see you're ready !

Answer 26

*for the exam

Answer 27

Heh, trying my best. We'll see, this professor is pretty tough.

Answer 28

ohh i like that one :)