Laplace transform, fraction expansion

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Homework Help Overview

The discussion revolves around the application of Laplace transforms in solving a differential equation, specifically focusing on the partial fraction expansion of the transformed function. The original poster presents a confusion regarding the values of coefficients obtained during the expansion process.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the process of partial fraction expansion and the need to separate terms in the numerator. There are questions about the correct formulation of the partial fractions, particularly when dealing with quadratic terms in the denominator.

Discussion Status

Some participants have offered guidance on how to approach the partial fraction expansion, while others have pointed out potential errors in the original poster's setup. Multiple interpretations of the problem are being explored, particularly regarding the coefficients in the expansion.

Contextual Notes

There are indications of confusion regarding the application of Laplace transforms and the appropriate methods for partial fraction decomposition, especially in the context of educational background and terminology differences.

sibiryk
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I have y"+y=t , y(0)=1, y'=0

After Laplace transformation a got:

(S^3+1)/(S^2(S^2+1))
After I made a partial fraction expansion

(S^3+1)/(S^2(S^2+1))=a1/S^2+a2/S+a3/(S^2+1) (1)

It comes to a system where

a2=1
a1+a3=0 (2)
a2=0
a1=1


Here I am getting confused because according to (1) and (2)
a2 is equal to 0 and 1. What I should use for inverse transformation: zero, one, or both?
 
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First of all check out the LaTex tutorial thread https://www.physicsforums.com/showthread.php?t=8997".

Your problem is to exapand

[tex]\frac {1 + s^3} {s^2 (s^2+1)}[/tex]

Try seperating the polynomial in the numerator to get this

[tex]\frac 1 {s^2 (s^2+1)} + \frac s {(s^2+1)}[/tex]

IIRC the second fraction has a well known trasform, now use partial fractions on the first piece, it is pretty straight forward.
 
Last edited by a moderator:
I tried that and it looks like it works for me.

Thank you!
 
Your basic formula for the "partial fractions" was wrong. If you have a quadratic, that cannot be factored into real factors, in the denominator, such as [tex]\frac{1}{s^2+1}[/tex], you will need a linear term in the numerator, not a constant: [tex]\frac{As+ B}{s^2+1}[/tex].
 
May I be so bold as to ask someone to be so kind to explain to me why/where/when/ Laplace Transforms are teached in K-12 grade? Really, that's so impressive.:smile: I expect great things from someone, well just some of them I suppose, alright, maybe one in a . . . several . . . thousand I guess, who study LT at such an early age.:smile:
 
saltydog said:
May I be so bold as to ask someone to be so kind to explain to me why/where/when/ Laplace Transforms are teached in K-12 grade? Really, that's so impressive.:smile: I expect great things from someone, well just some of them I suppose, alright, maybe one in a . . . several . . . thousand I guess, who study LT at such an early age.:smile:

may be it is time for me to move up:smile:

I'm not really familiar with this forum. Where should I post with stuff like that?
 
HallsofIvy said:
Your basic formula for the "partial fractions" was wrong. If you have a quadratic, that cannot be factored into real factors, in the denominator, such as [tex]\frac{1}{s^2+1}[/tex], you will need a linear term in the numerator, not a constant: [tex]\frac{As+ B}{s^2+1}[/tex].

I tried that. I must did something wrong because I was getting too many unknowns to solve equetions.
 
sibiryk said:
I'm not really familiar with this forum. Where should I post with stuff like that?

In the College Level section.

I found out not too long ago that to people outside the US, college=high school. But here college=university, which is why the other Homework Forum is named the way it is.
 
HallsofIvy said:
Your basic formula for the "partial fractions" was wrong. If you have a quadratic, that cannot be factored into real factors, in the denominator, such as [tex]\frac{1}{s^2+1}[/tex], you will need a linear term in the numerator, not a constant: [tex]\frac{As+ B}{s^2+1}[/tex].
That's it! I was posting form work last night and did not have my book shelf to draw on. Looks like my solution gets the job done. There is always more then one way to approach problems.

I was lucky, in general you need to learn the correct (Hall's ) approach. It will, and does in this case, always get you to the correct solution. What I did worked and in someways is actually a simpler method, but it is not general and may not be simpler or even work in every case.
 
  • #10
Your method worked out, but I can't figure a way HallsofIvy suggested.

I'm getting

(S-1)/(S^2+1) + 1/S^2.

The second part of it is known combination. The first is not.
Am I doing something wrong?
 

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