Question

Alright, another Dirac-Delta. Getting the hang of this. Posted below in a moment.

Answer 1

\[y''-y=\delta(t-2\pi),\ \ \ y(0)=0, \ y'(0)=1\]

Answer 2

\[[s^{2}Y(s)-sy(0)-y'(0)]-Y(s)=e^{-2\pi s}\]

Answer 3

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

Answer 4

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

Answer 5

Going to separate the numerator and evaluate them separately

Answer 6

yup :)

Answer 7

\[\frac{e^{-2\pi s}}{s^2-1}+\frac{1}{s^2-1}=\sinh(t-2\pi)\mathcal{U}(t-2\pi)+\sinh(t)\]

Answer 8

Looking over that, not sure about it

Answer 9

correct :) well done again

Answer 10

Eghhh, I think we made a mistake somewhere, heh, yeah, we (I) did

Answer 11

where?

Answer 12

Answer is close, but not identical\[y=\sin(t)+\sin(t)\mathcal{U}(t-2\pi)\]

Answer 13

I'm not sure

Answer 14

another typo in your book they missed that hyperbolic again

Answer 15

I made a mistake, too! It was my fault, it was y'' + y

Answer 16

Good thing is, I understand the process and am getting it, w/o typos I am doing things right, just need to make sure I cover all of my bases

Answer 17

oh ok now we are talking :P

Answer 18

Heh, alright, taking a look at it again before I leave it, shouldn't take too long to fix

Answer 19

What I still don't understand (I can understand the change from them to sine trig functions, is why the sine function attached to the unit step function term doesn't have its argument altered in the problem.

Answer 20

yes argument of the sine have to be altered by 2pi

Answer 21

Alright, col! Going to move on to another problem.

Answer 22

okaay :)