Question

Another IVP; am trying to come to the conclusion independently but can already see some issues, will post my workings as I go through the problem.

Answer 1

\[y''-y'=e^tcos(t), \ \ \ y(0)=0, \ y'(0)=4\]

Answer 2

doing good

Answer 3

\[s^2Y(s)-sy(0)-y'(0)-[sY(s)-y(0)]= \frac{1}{(s-1)^2+1}\]

Answer 4

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

Answer 5

Now here's where I can see things getting really bad in the denominator of the fractions that would result when I isolate Y(s); is this all correct so far, and is it really supposed to look this way, lol?

Answer 6

Seems fine to me so far.

Answer 7

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

Answer 8

Is it a good idea to expand the denominator and then try to simplify it in a different form here, or what should be the next step, using only t or x-axis shifting methods (exponential) to change this? No Heaviside, no convolution, etc.

Answer 9

\[\frac{1}{[(s^2-2s+1)+1](s^2-s)}+\frac{4}{s^2-s}=\frac{1}{(s^2-2s+2)(s^2-s)}+\frac{4}{s^2-s}\]

Answer 10

ok u want to use any other properties?

Answer 11

other than partial fractions

Answer 12

Only allowed to use those shifting ones xP

Answer 13

Oh, if you mean algebraic ones, then yeah

Answer 14

I don't see why you cant use partial fractions. Thats pretty much what laplace transform problems are, exercises in partial fractions, lol.

Answer 15

Can't use Heaviside, can't use Convolution, just reminding; purely algebraic properties to get it into a preferable form, if you see one.

Answer 16

ok \[F(s)=\frac{ 1 }{ s(s-1)[(s-1)^2+4] }+\frac{ 4 }{ s(s-1) }\\=[\frac{ 1 }{ s-1 }-\frac{ 1 }{ s }][\frac{ 1 }{ (s-1)^2+4 }]+4[\frac{ 1 }{ s-1 }-\frac{ 1 }{ s }]\]

Answer 17

\[=\frac{ 1 }{ s-1 }*\frac{ 1 }{ (s-1)^2+4 }-\frac{ 1 }{ s }*\frac{ 1 }{ (s-1)^2+4 }+\frac{ 4 }{ s-1 }-\frac{ 4 }{ s }\]
getting this?

Answer 18

Man, Siddhartha, you are like a math God, I can't even understand how you're able to figure this stuff out so quickly half of the time. I don't understand what you just did, but I'm going to keep looking at it for a minute. The first post confused me.

Answer 19

Alright, I'm seeing it, however

Answer 20

i just broke
\[\frac{ 1 }{ s(s-1) }=\frac{ 1 }{ s-1 }-\frac{ 1 }{ s }\]
in the first step

Answer 21

I'm not immediately seeing how the four is turning up in the denominator of the first fraction, yeah one sec:

Answer 22

because u gotta use some shifting property

Answer 23

Wait whoah

Answer 24

Are you using convolution in there later on?

Answer 25

now first only taking this term
\[\frac{ 1 }{ s-1 }*[\frac{ 1 }{ (s-1)^2+4 }]\]
dont worry i wont use conv

Answer 26

I just don't get where that four came from in the denominator, in the very beginning.

Answer 27

\[\frac{ 1 }{ [(s-1)^{2}+1](s^{2}-s) }+ \frac{ 4 }{ s^{2} - s } = \frac{ -s }{ (s-1)^{2} + 1 } + \frac{ 5 }{ s-1 }- \frac{ 9 }{ 2s }\]

This is another partial fraction result if this is more comfortable looking to ya.

Answer 28

so first use shifting on this one\[e^{t}*L^{-1}[\frac{ 1 }{ s }*\frac{ 1 }{ s^2+4 }]\]

Answer 29

\[[(s-1)^2+1](s^2-s) = s(s-1)[(s-1)^2+1]\]

Answer 30

That's what I see happening to the denominator in the beginning, not sure where that four came from

Answer 31

i know @concentrationalizing that partial fraction will work ,but i'm just trying some thing else to solve it :)

Answer 32

Yeah, Im curious how you're messing with it, but alright, lol.

Answer 33

u, sorry that 4 wont be there its a typo

Answer 34

yes so lets deal with the first term,\[\frac{ 1 }{ s-1 }*\frac{ 1 }{ (s-1)^2+1 }\]
thats it right?

Answer 35

I'm just confused that you are using the sign typically used for convolution for multiplication; that's what's going on, yes? And-oh, okay, the four is a typo, I think I can follow, let me reread everything with that in mind

Answer 36

ok sorry it not conv simple multiplication

Answer 37

When you did\[\frac{1}{s(s-1)}=\frac{1}{s-1}-\frac{1}{s}\], that was partial fractions in the first place, yes? Or was that something else?

Answer 38

yes a little one :P

Answer 39

That part is partial fractions, yes. The rest he seems to be making fancy, lol.

Answer 40

Alright, then one sec, rereading with that in knowledge. Just filling in the gaps.

Answer 41

yes now i have just multiplied this with terms in the right hand side

Answer 42

Just writing out the math myself, give me a moment.

Answer 43

ok take the first term in RHs\[[\frac{ 1 }{ s-1 }-\frac{ 1 }{ s }].[\frac{ 1 }{ (s-1)^2+1 }]\\=\frac{ 1 }{ s-1 }.\frac{ 1 }{ (s-1)^2 +1}-\frac{ 1 }{ s }.\frac{ 1 }{ (s-1)^2+1 }\]

Answer 44

\[\frac{1}{[(s-1)^2+1](s^2-s)}-\frac{4}{(s^2-s)}\]\[\frac{1}{s(s-1)[(s-1)^2+1]}-\frac{4}{s(s-1)}=\left\{\vphantom{}\right\}\]

It's 1 AM, I think I'm going to sleep for a little and then wake back up and work on this-sorry for suddenly leaving.

Answer 45

ok sleep well but beware laplace will haunt you in your dreams
bwahahaha

Answer 46

lol :)

Answer 47

Nevermind, can't sleep like this anyways, finishing this off.

Answer 48

\[\frac{1}{s(s-1)[(s-1)^2+1]}-\frac{4}{s(s-1)}=\]\[\Bigg[\frac{1}{s-1}-\frac{1}{s}\Bigg]\Bigg[\frac{1}{(s-1)^2+1}\Bigg]-4\Bigg[\frac{1}{s-1}-\frac{1}{s}\Bigg]\]

Answer 49

I'll just finish it out the way I was setting it up :) So given that I can split it into partial fractions and get:
\[\frac{ -s }{ (s-1)^{2} + 1 }+ \frac{ 5 }{ s-1 } - \frac{ 9 }{ 2s }\]

So the last two fractions don't require any fancy manipulation at all. we just have
\(\frac{5}{s-1}\) matching the form \(F(s) = \frac{1}{s-a} = e^{at}\) and we have \(\frac{-9}{2s}\) matching the form \(F(s) = \frac{1}{s} = 1\). Of course the 1's in the F(s) formulas will be replace with our actual coefficients.

The first fraction is close to a form that has a nice inverse transform, we just have to add something to it. So I'll do this:
\[\frac{ -s }{ (s-1)^{2} + 1 } = \frac{ -s+1-1 }{ (s-1)^{2} + 1 } = \frac{ -(s-1) - 1 }{ (s-1)^{2} + 1 } = -\frac{ (s-1) }{ (s-1)^{2} + 1 } - \frac{ 1 }{ (s-1)^{2} + 1 }\]


So splitting this up gives me the forms:
\(F(s) = \frac{s-a}{(s-a)^{2} + k^{2}} = e^{at}cos kt\) and \(F(s) = \frac{k}{(s-a)^{2} + k^{2}} = e^{at}sin kt\)
So with that, I can do the inverse laplace transform now to all of our fractions and get:
\[-L^{-1}[\frac{ (s-1) }{ (s-1)^{2} + 1 }] - L^{-1}[\frac{ 1 }{ (s-1)^{2} + 1 }] + L^{-1}[\frac{ 5 }{ s-1 }]-L^{-1}[\frac{ 9 }{ 2s }]\]
\[= -e^{t}cost - e^{t}sint + 5e^{t} - \frac{ 9 }{ 2 }\]

Just my way of doing it, though.

Answer 50

yes that seems ok
from here u can multiply then use some shifting properties ,

Answer 51

but i think this will be more difficult because at some point u have to use some integral because of that (1/s) term
\[1/s[f(s)]=\int\limits_{0}^{t}f(t)dt\]
thats why partial fraction is more suitable i think :)

Answer 52

sorry, i'll have to head out for college now

Answer 53

i'll post the solution when i'll get back :)

Answer 54

\[\Bigg[\frac{1}{(s-1)[(s-1)^2+1]}\Bigg]-\Bigg[\frac{1}{s[(s-1)^2+1]}\Bigg]-4\Bigg[\frac{1}{s-1}-\frac{1}{s}\Bigg]\]

Answer 55

yes that fine :)

Answer 56

Typo, plus sign supposed to be there, my bad

Answer 57

Alright, now next

Answer 58

next use shifting property, and sorry i gotta go now :(

Answer 59

Sure thing! Ttyl, thank you, both you guys.

Answer 60

Im still randomly around if you wanna mess with it more, lol. Seems like you're more along the lines of sid's approach, but I can still watch on and keep track of your steps and such.

Answer 61

I'm along the line of any approach, I'd like to try both, just sticking with that one and finishing it before the other, heh.

Answer 62

(Just thinking about how to move forward, gimme a minute)

Answer 63

I dunno, shifting aside, I've never really dealt with inverse Laplace Transforms of this general form; how should I move forward?

Answer 64

Well, when sid starting splitting it up as he did, I thought he was going along the lines of a convolution as well. That's why it's hard for me to see this being done without further partial fractions.