Can't finish a Laplace Initial Value Problem.

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Discussion Overview

The discussion revolves around a Laplace initial value problem, focusing on the application of Laplace transforms and the use of partial fractions in solving differential equations. Participants share their approaches, challenges, and suggestions related to the problem presented.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested
  • Homework-related

Main Points Raised

  • One participant expresses difficulty in transitioning to partial fractions due to complex denominators.
  • Another suggests using partial fractions on the first term only and proposes that the second term can be manipulated to resemble a sine function.
  • There is a discussion about the Laplace transforms of exponential functions multiplied by sine and cosine, with specific forms provided for rewriting Y(s).
  • Some participants question the correctness of completing the square for the denominator and challenge the assumptions made in earlier posts.
  • There is a request for clarification on how to equate different forms of Y(s) and compute coefficients C1, C2, and C3.
  • One participant mentions running the problem through Matlab, indicating a potential solution but does not confirm it as definitive.
  • Several participants express confusion regarding the manipulation of fractions and the need for a common denominator.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best approach to the problem, with multiple competing views on how to handle the Laplace transforms and partial fractions. Some participants agree on the need for a common denominator, while others express uncertainty about the steps involved.

Contextual Notes

There are unresolved mathematical steps regarding the completion of the square and the manipulation of fractions. Participants rely on specific forms of Laplace transforms that may not be universally accepted or verified.

Who May Find This Useful

Students or individuals studying differential equations, particularly those interested in Laplace transforms and their applications in solving initial value problems.

chief10
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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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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
 
chief10 said:
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
 
FOIWATER said:
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..
 
Last edited:
JJacquelin said:
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
 
I mean s²+2s+5 = (s+1)²+2² = (s+a)²+b², then a=1 and b=2
 
JJacquelin said:
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!

https://www.physicsforums.com/showthread.php?t=639980
 
chief10 said:
how does s²+2s+5 = (s+1)²+2²?
that's not right is it?
Can you develop (s+1)²+2² = ?
 
JJacquelin said:
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.jpgBTW, 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
 
Last edited:
  • #10
chief10 said:
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.
 
  • #11
JJacquelin said:
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.
 
  • #12
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
 
  • #13
any ideas on which step I should take next?
 
  • #15
wait so are you saying to equate both of the Y(s) equations?
 
  • #16
Of course, obviously, YES ! :zzz:
 
  • #17
lol alright, i'll do that and post back in a bit
 
  • #18
but then how do you solve that? I'm a little lost?

i tried it out but I've got brackets and fractions everywhere? sorry to seem a little out of touch here.
 
  • #19
Do you have the same denominator for ALL fractions ?
 
  • #20
JJacquelin said:
Do you have the same denominator for ALL fractions ?

nope

in this one: http://i46.tinypic.com/2z4aaup.png - The Y(s) function has two different denominators so you can't cancel them out on both sides if you equated the Y(s) equations.

i mean even if you did equate them and cancel them out, does one side just become 10+5? (15)
 
  • #21
<< The Y(s) function has two different denominators >> ! ! !
Oh my God ! Can't you make all fractions with the same denominator s((s+1)²+4) just like you did before for the three other fractions ?
 
  • #22
JJacquelin said:
<< The Y(s) function has two different denominators >> ! ! !
Oh my God ! Can't you make all fractions with the same denominator s((s+1)²+4) just like you did before for the three other fractions ?

yeah I thought that's what you meant, no issue

lol relax man, it's confusing over the net with just text and stuff, i appreciate the help, i'll post back in a bit.
 
  • #23
JJacquelin said:
<< The Y(s) function has two different denominators >> ! ! !
Oh my God ! Can't you make all fractions with the same denominator s((s+1)²+4) just like you did before for the three other fractions ?

thanks a lot mate I think I'm pretty sure I worked it out! with your help!

much appreciated dude, I've attached it below

http://i48.tinypic.com/2ep3t3c.jpg
 
  • #24
chief10 said:
thanks a lot mate I think I'm pretty sure I worked it out! with your help!

much appreciated dude, I've attached it below

http://i48.tinypic.com/2ep3t3c.jpg

You are right to be pretty sure ! :approve:
 
  • #25
haha :)
 

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