Question

Now dealing with Dirac-Delta functions in ODE and their laplaces/inverses. Posted below momentarily.

Answer 1

\[y'-3y=\delta(t-2), \ \ \ y(0)=0\]

Answer 2

Taking the Laplace of both sides, \[sY(s)-y(0)-3Y(s)=e^{-2s}\]

Answer 3

u got it

Answer 4

\[Y(s)=\frac{e^{-2s}}{s-3}\]

Answer 5

\[f(t)=\mathcal{L^{-1}}\left\{\vphantom{}\frac{e^{-2s}}{s-3}\right\}= \ ?\]

(Haven't dealt with inverse laplaces with e^(something)'s in their argument for a while, lookuing up how to do so)

Answer 6

It's some sort of shifting, right? Shifting on the t-axis?

Answer 7

yes same thing

Answer 8

Alright,\[e^{3t}\mathcal{U}(t-2)\]I think

Answer 9

Whoops, 3t-2

Answer 10

\[L\frac{ 1 }{ s-3 }=e^{3t}\]

Answer 11

3(t-2)

Answer 12

\[e^{3(t-2)}\mathcal{U}(t-2)\]

Answer 13

now just make it lag by -2

Answer 14

yeah, think I got it (not sure if my posts are lagging)

Answer 15

yeah that looks fine

Answer 16

Alright! Continual progress!!!!!

Answer 17

yes same here lagging

Answer 18

ur replies are dancing here

Answer 19

Gonna keep going, don't let me keep you @sidsiddhartha , but thank you, again.

Heh, I saw mine doing the same earlier, yeah

Answer 20

lol

Answer 21

i can stay a little longer :)