Question

@ganeshie8

Answer 1

\[y''+y=u_{\pi/2}(t)+3\delta(t-3\pi/2)-u_{2\pi}(t); ~~~~~y(0)=0,y'(0)=0\]

Answer 2

\(\mathcal{L}(y'')+\mathcal{L}(y)=\mathcal{L}(u_{\pi/2}(t))+3\mathcal{L}(\delta(t-3\pi/2))-\mathcal{L}(u_{2\pi}(t))\)

\(\mathcal{L}(y'')=s^2Y(s)-sy(0)-y'(0)\)
\(\mathcal{L}(y)=Y(s)\)
\(\mathcal{L}(u_{\pi/2}(t)) = \frac{ e^{-\pi/2s} }{ s }\)
\(3\mathcal{L}(\delta(t-3 \pi/2)) = e^{-3\pi/2s}\)
\(\mathcal{L}(u_{2\pi}(t))=\frac{ e^{-2\pi s} }{ s }\)

That should be all the laplace transformations, ok so putting it together..

Answer 3

\[\large s^2 Y(s)-sy(0)-y'(0)+Y(s) = \frac{ e^{-\pi/2s} }{ s }+e^{-3\pi/2s}+\frac{ e^{-2\pi s} }{ s }\]

Answer 4

\[\large Y(s) = \frac{ e^{-\pi/2s} }{ s(s^2+1) }+\frac{ e^{-3\pi/2s} }{ s^2+1 }+\frac{ e^{-2 \pi s} }{ s(s^2+1) }\] haha me too

Answer 5

Well I see no reason for partial fractions decomposition, so just take the inverse?

Answer 6

\[\large y(t) = Y^{-1}(s)(\frac{ e^{-\pi/2s} }{ s(s^2+1) }+\frac{ e^{-3\pi/2s} }{ s^2+1 }+\frac{ e^{-2 \pi s} }{ s(s^2+1) })\]

Answer 7

Ok here I'm not so sure, cause in my notes I have stuff like \[\mathcal{L}^{-1}(e^{-as}F(s))=u_a(x)f(x-a)\] but where did the \(u_a(x)\) come from mhm.

Answer 8

Oh I'm an idiot

Answer 9

It's on the table

Answer 10

@AlexandervonHumboldt2 making me think you're ganeshie

Answer 11

i did haha

ganeshie is here lol

Answer 12

shouldn't that be->
\(\large s^2 Y(s)-sy(0)-y'(0)+Y(s) = \frac{ e^{-\pi/2s} }{ s }+e^{-3\pi/2s}\color{red}{-}\frac{ e^{-2\pi s} }{ s }\)

Answer 13

Yes!

Answer 14

then ->
\(\large Y(s) = \frac{ e^{-\pi/2s} }{ s(s^2+1) }+\frac{ e^{-3\pi/2s} }{ s^2+1 }\color{red}{-}\frac{ e^{-2 \pi s} }{ s(s^2+1) }\)

then u use partial fractions ig..

Answer 15

No need for partial fractions

Answer 16

wait a minute now qwerty is ganeshie

Answer 17

why
i saw the guy in vedio solving such ques using partials

Answer 18

Because I can just use inverse laplace

Answer 19

\[3\mathcal{L}(\delta(t-3 \pi/2)) = \color{red}{3}e^{-3\pi/2s}\]

Answer 20

Ah yeah I have this all right on paper haha, thanks

Answer 21

\[\mathcal{L}^{-1}(e^{-\pi/2 s} \times \frac{ 1 }{ s(s^2+1) })\]

Answer 22

Would that be just \[u_{\pi/2}(t)f(t-\pi/2)\]

Answer 23

Mhm no

Answer 24

astro

don't you have

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

?

Answer 25

yes thats one of them

Answer 26

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

Answer 27

I think I have to find the inverse of f(t-pi/2)

Answer 28

I mean if I know how to do this part should be good for rest

Answer 29

i can't see anything obvious in the tables in Boas, so i would do a convolution

but i have a funny feeling that the second time shift theorem thingy might apply [i am v rusty on this]