Question

Taking a fairly ugly final IVP dealing with Dirac-Delta before moving onto my last subject to cover, Systems of Diff. Eqn's. Posted below in a moment.

Answer 1

Oh! Now I know why the argument was just sin(t)! It's because it makes a full round and it doesn't actually affect the outcome of the sine function on that last problem! sin(t-2pi)=sin(t)!

Answer 2

(Lol now onto this problem)

Answer 3

yeah lol god!!

Answer 4

\[y''+4y'+13y=\delta(t-\pi)+\delta(t-3\pi), \ \ \ y(0)=1, \ y'(0)=0\]

Answer 5

(That doesn't look as bad as I thought but w/e, might get bad soon)

Answer 6

\[[s^2Y(s)-sy(0)-y'(0)]+4[sY(s)-y(0)]+13Y(s)=e^{-\pi s}+e^{-3\pi s}\]

Answer 7

hmmmmmm............345678900987653673559

Answer 8

\[s^2Y(s)-s+4sY(s)-1+13Y(s)=e^{-\pi s}+e^{-3\pi s}\]
\[Y(s)[s^2+4s+13]=e^{-\pi s}+e^{-3\pi s}+1+s\]

Answer 9

\[Y(s)=\frac{e^{-\pi s}+e^{-3\pi s}+1+s}{s^{2}+4s+13}\]

Answer 10

(So much for "doesn't look too bad")

Answer 11

hey wait

Answer 12

Alright, now, splitting up the terms sounds sensible, but I have no idea what to do with that denominator, doesn't appear to be doable with partial fractions. Even if I don't solve this to the end, I just need to figure out a general idea.

(And if you gotta go, I understand, getting late there, heh. Don't mean to keep you)

Answer 13

Mistake?

Answer 14

no it fine go ahead

Answer 15

*its

Answer 16

Alright, now not sure what to do

Answer 17

brb (>2 mins, short time)

Answer 18

man this one is going to be (*dont know what to say*)
lol

Answer 19

\[s^2+4s+13=(s+2)^2+9\]

Answer 20

that'll help

Answer 21

Alright, back

Answer 22

\[\frac{ e^{-\pi s} }{ (s+2)^2+3^2 }+\frac{ e^{-3 \pi s} }{ (s+2)^2+3^2 }+\frac{ 1 }{ (s+2)^2+3^3 }+\frac{ s }{ (s+2)^2+3^3 }\]

Answer 23

Oh, okay, now evaluating {L^{-1}}

Answer 24

\[\mathcal{L^{-1}}\left\{\vphantom{}\frac{e^{-\pi s}}{(s+2)^2+3^2}\right\}=\](Using Heaviside and shifting)

Well, actually, no, I'm not so sure how to evaluate this, I'm not sure if it applies to both inverse laplace transforms, but don't all instances of s have to be paired with the same subtraction/addition?

Answer 25

yes thats the idea

Answer 26

So do I need a\[e^{-\pi(s+2)}\]in the denominator?

Answer 27

no u dont because it does'nt affect exponentials

Answer 28

Alright, great! So then I can just apply the shift as a multiple of e^{-2t} or something like that on the t-axis, or some kind of shifting again? (Gonna have to look it up, one moment)

Answer 29

Got it, First Translation Theorem, \[\mathcal{L}\left\{e^{at}f(t)\right\}=F(s-a)\]

Answer 30

yes
for the first term\[e^{-2(t-\pi)}\sin3(t-\pi).u(t-\pi)\]
that looks ok?

Answer 31

heh, lol, does that work out by complete chance? One sec, working it myself.

Answer 32

\[e^{-2(t-\pi)}\sin(3(t-2\pi))\mathcal{U}(t-2\pi)\]Somewhere we're missing a fraction, that whole thing is supposed to be multiplied by 1/3rd

Answer 33

Whoops, pi, not 2pi

Answer 34

I'm just wondering where we're missing that 1/3rd that the whole thing is multiplied by.

Answer 35

oooooohhhhhhhhhh yes

Answer 36

i forgot

Answer 37

\[L[\sin3t]=\frac{ 3 }{ s^2+9 }\]

Answer 38

look we did'nt multiplied that 1/3 to make it sib(3t)

Answer 39

Oh! I see!

Answer 40

Alright, well, cool, I'll have time to take time on my test/check for algebra mistakes, for now I'm just covering concepts; I'm going to move onto my final subject, systems of Diff Eqns. never done any, know nothing about it, just a quick read and working through practice problems.

Answer 41

yes those are not that hard

Answer 42

Alright, cool. Getting close to the end. Just let me know if you have to go, must be getting tired over there.

Answer 43

yes sorry i've to go :( its now 2 am and gotta sleep :)
bye :)