Can't finish a Laplace Initial Value Problem.

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Main Question or Discussion Point

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
 

Answers and Replies

  • #2
FOIWATER
Gold Member
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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
 
  • #3
798
34
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
 
  • #4
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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
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:
  • #5
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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
 
  • #6
798
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I mean s²+2s+5 = (s+1)²+2² = (s+a)²+b², then a=1 and b=2
 
  • #7
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  • #8
798
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how does s²+2s+5 = (s+1)²+2²?
that's not right is it?
Can you develop (s+1)²+2² = ?
 
  • #9
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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
 
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  • #10
798
34
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
78
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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
FOIWATER
Gold Member
434
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
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any ideas on which step I should take next?
 
  • #14
798
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  • #15
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wait so are you saying to equate both of the Y(s) equations?
 
  • #16
798
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Of course, obviously, YES ! :zzz:
 
  • #17
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lol alright, i'll do that and post back in a bit
 
  • #18
78
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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
798
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Do you have the same denominator for ALL fractions ?
 
  • #20
78
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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
798
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<< 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
78
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<< 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
78
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<< 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
798
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  • #25
78
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haha :)
 

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