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Can't finish a Laplace Initial Value Problem.

 
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Sep30-12, 02:33 AM   #1
 

Can't finish a Laplace Initial Value Problem.


I've had to take diff eqtns now and I'm trying to get my head around Laplace again.. it's been a while. I can't seem to transition to the simplest step of partial fractions, my denominators are tough to figure out.

If someone could point me to the next step that'd be great!

Thanks a lot guys and girls.

What I have is in the link below, I'd embed it but it's too big and will mess with the formatting of the page.

http://i46.tinypic.com/2z4aaup.png
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Sep30-12, 10:47 AM   #2
 
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my suggestion would be to use partial fractions on the first term only. The second term can be forced to resemble sin. 5 divided by (s^2+2s+5) completing the square on the denominator gives (S+2)^2+1 which when inversed gives you 5e^-2tsint I believe? then it's just a matter of the first term..

I believe I haven't checked it more than once may have made a mistake but it definitely seems that it would be easiest to handle each term on its own. Maybe the first one can also be forced to resemble either sin or cosine without using partial fractions
Sep30-12, 11:08 AM   #3
 
Quote by chief10 View Post
I've had to take diff eqtns now and I'm trying to get my head around Laplace again.. it's been a while. I can't seem to transition to the simplest step of partial fractions, my denominators are tough to figure out.
If someone could point me to the next step that'd be great!
Thanks a lot guys and girls.
What I have is in the link below, I'd embed it but it's too big and will mess with the formatting of the page.
http://i46.tinypic.com/2z4aaup.png
Hi !

The Laplace transform of exp(-at)sin(bt) is b/((s+a)²+b²)
The Laplace transform of exp(-at)cos(bt) is (s+a)/((s+a)²+b²)
So, you have to rewrite Y(s) on the form :
Y(s) = C1/s + C2* b/((s+a)²+b²)+ C3* (s+a)/((s+a)²+b²)
First, compute a and b, then C1, C2 and C3
Sep30-12, 11:18 AM   #4
 

Can't finish a Laplace Initial Value Problem.


Quote by FOIWATER View Post
my suggestion would be to use partial fractions on the first term only. The second term can be forced to resemble sin. 5 divided by (s^2+2s+5) completing the square on the denominator gives (S+2)^2+1 which when inversed gives you 5e^-2tsint I believe? then it's just a matter of the first term..

I believe I haven't checked it more than once may have made a mistake but it definitely seems that it would be easiest to handle each term on its own. Maybe the first one can also be forced to resemble either sin or cosine without using partial fractions
Do you mean [5(e^-2) x tsint]? you're not raising the sine function are you, just multiplying it by the exp?
The second component appears to be solvable in that sense, however your completion of the square I believe is incorrect. You can't simplify the denominator to my knowledge.

That specific Laplace transform wasn't on my given formula sheet, that's a bummer.

Hmmm the first term is a little trickier! any ideas on that one?

10/s(s^2+2s+5)

transforming that is quite a task..
Sep30-12, 11:22 AM   #5
 
Quote by JJacquelin View Post
Hi !

The Laplace transform of exp(-at)sin(bt) is b/((s+a)²+b²)
The Laplace transform of exp(-at)cos(bt) is (s+a)/((s+a)²+b²)
So, you have to rewrite Y(s) on the form :
Y(s) = C1/s + C2* b/((s+a)²+b²)+ C3* (s+a)/((s+a)²+b²)
First, compute a and b, then C1, C2 and C3
in the C3 function, could I ask why you put (s+a) in the numerator? thanks mate
Sep30-12, 11:53 AM   #6
 
I mean s²+2s+5 = (s+1)²+2² = (s+a)²+b², then a=1 and b=2
Sep30-12, 12:02 PM   #7
 
Quote by JJacquelin View Post
I mean s²+2s+5 = (s+1)²+2² = (s+a)²+b², then a=1 and b=2
how does s²+2s+5 = (s+1)²+2²?

that's not right is it?


while you're here though, if you could take a look at my other thread involving Laplace that'd be great!!!

http://www.physicsforums.com/showthread.php?t=639980
Sep30-12, 12:11 PM   #8
 
Quote by chief10 View Post
how does s²+2s+5 = (s+1)²+2²?
that's not right is it?
Can you develop (s+1)²+2² = ?
Sep30-12, 12:18 PM   #9
 
Quote by JJacquelin View Post
Can you develop (s+1)²+2² = ?
excuse me it's 3am here in Aus, lol.. my bad.. got it

do you mind elaborating on the partial fraction a little if you don't mind?

i've drawn up the initial equation you gave me below, how does the completed square factor in here?

http://i49.tinypic.com/w39cj.jpg


BTW, I just ran it through Matlab and this is the answer it gave me to the ODE, seems like we're on the right track.

(3*sin(2*t))/(2*exp(t)) - (2*cos(2*t))/exp(t) + 2
Sep30-12, 01:06 PM   #10
 
Quote by chief10 View Post
i've drawn up the initial equation you gave me below, how does the completed square factor in here?
http://i49.tinypic.com/w39cj.jpg
That's OK. Then you have to compute C1, C2, C3
Rewrite all with only one denominator s((s+a)²+b²) and compare to Y(s) also rewriten with the same denominator.
Sep30-12, 01:25 PM   #11
 
Quote by JJacquelin View Post
That's OK. Then you have to compute C1, C2, C3
Rewrite all with only one denominator s((s+a)²+b²) and compare to Y(s) also rewriten with the same denominator.
Alright I think I get what you're saying, i posted a link below.

http://i48.tinypic.com/2zits9l.jpg

I won't post back for a few hours now though because it's 4:15am here so i'll head off to bed now. I really appreciate all of your help. Hopefully you'll be here tomorrow to help me too mate :)

I await your reply! thanks bud.
Sep30-12, 07:06 PM   #12
 
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oh wow sorry - yeah i completed that square wrong - should have been as you said, sorry. so, sin2t instead of t right?

And yeah as you said it's not raised to the power it's a multiplication
Oct1-12, 06:24 AM   #13
 
any ideas on which step I should take next?
Oct1-12, 06:45 AM   #14
 
In http://i48.tinypic.com/2zits9l.jpg you have writen Y(s) on a form in which C1, C2, C3 appear.
Do you forget that you already have writen Y(s) on another form whitout C1, C2, C3 : http://i46.tinypic.com/2z4aaup.png
Compare both in order to compute C1,C2, C3
Oct1-12, 06:49 AM   #15
 
wait so are you saying to equate both of the Y(s) equations?
Oct1-12, 07:46 AM   #16
 
Of course, obviously, YES !
Oct1-12, 07:50 AM   #17
 
lol alright, i'll do that and post back in a bit
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