Question

y''+9y=g(t) y(0)=2 y'(0)=-3
H(s) for the system and the impulse response function h(t) and give a formula for the solution to the IVP. I have no idea what to do no examples like this worked in class :/

Answer 1

these are solutions to the problem...
http://prntscr.com/1kkisc

Answer 2

@Loser66 you know how to work these?

Answer 3

nope, I don't. My prof skip this part, just wrote out the formula as you did. He talked about impulsion in 1 minute

Answer 4

lol I don't think my professor talked about them at all. He was asking us in class about solving some problems the day after we learned them and was no one was answering questions and he was like "c'mon guys we did these yesterday....like two problems..." lol as if we are supposed to know how to work them after 10 min of exposure to them.

Answer 5

that's fine im meeting up with a class mate tomorrow and hopefully we can figure it out. thanks for taking a look though

Answer 6

ha, yours is new to me, hehe..

Answer 7

I can help you find h(s) and h(t). Other parts I don't know

Answer 8

@rperez36

Answer 9

just post @ash2326 we need it.

Answer 10

My apologies. This is my husband's post. He just stepped in the shower right quick, but he's getting out. Please help with what you can, I know he would appreciate the help.

Answer 11

We have
\[y'' (t)+9y=g(t)\]
Let's take laplace transform of this
we know Laplace transform of a derivative of function is s \times the Laplace of a function
\[L(f(t))=F(s)\]
\[L(f'(t))=s\times F(s)-f(0)\]
\[L(f''(t))=s^2\times F(s)-sf(0)-f'(0)\]

do you get this part?

Answer 12

im back @ash2326 and yes I get that part

Answer 13

just a min

Answer 14

the answer you have given is that correct? I mean is that given in textbook

Answer 15

yeah that is what is given

Answer 16

It seems that it's wrong, it'll be valid on if the initial conditions are zero, y(0)=0 and y'(0)=0
but here we don't have this

Answer 17

*only

Answer 18

I'll be back in 10 minutes, could you confirm the question and solution meanwhile?

Answer 19

no problem

Answer 20

http://prntscr.com/1kksex

Answer 21

ok, so the question seem correct.

Answer 22

if y(0)=0 and y'(0)=0
then
\[L(y''+9y)=L(g(t))\]
\[s^2 Y(s)+9Y(s)=G(s)\]
Transfer function is defined as
\[\frac{G(s)}{Y(s)}=H(s)\]
can you find this from here?

Answer 23

so h(s)=s2+9?

Answer 24

oops it's
\[\frac{(Y(s))}{(G(s))}=H(s)\]

Answer 25

I see it says to the ^-1 where is that from?

Answer 26

Now you try, you'll get

Answer 27

@rperez36 do you get this? sorry I made a mistake

Answer 28

it all makes sense except that its I don't see how its
\[(s^2+9)^{-1}\]

Answer 29

as the solution manual states

Answer 30

s^2+9 ok but to the ^-1 ????

Answer 31

oh cause your mistake?

Answer 32

yes, try with the changed formula

Answer 33

ok so I just take inverse laplace

Answer 34

yes, now take inverse laplace transform of H(s)

Answer 35

yes I got what soln says what about for the other solutions?

Answer 36

those two other parts, I'm sorry but I don' know :(

Answer 37

ah ok well thank you for helping with what you did. hopefully these wont be on Mondays test

Answer 38

ha ha :) I hope so