Partial fractions (for laplace)

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Homework Help Overview

The discussion revolves around the application of partial fractions in the context of Laplace transforms. The original poster expresses difficulty in breaking down a given expression using partial fractions.

Discussion Character

  • Exploratory, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the correctness of the initial setup for partial fraction decomposition and explore various substitutions for the variables involved. There is an attempt to clarify the next steps after the initial decomposition.

Discussion Status

Some participants have confirmed the original poster's setup as correct and have shared additional tools that may assist in the process. Multiple interpretations of the next steps are being explored, but there is no explicit consensus on the approach to take.

Contextual Notes

Participants are considering the implications of substituting specific values into the expression, and there is an acknowledgment of the complexity involved in the problem. The discussion is framed within the constraints of homework guidelines.

bakin
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Homework Statement


This problem is killing me.

I need to bust this thing up using partial fractions.

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Homework Equations

The Attempt at a Solution



I'm leaning towards it being separated like this. Is this correct?
fnfne1.jpg


If it is, I'm not exactly sure what I'm supposed to do next.
 
Last edited:
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145s/ [(s+2j)(s-2j)(s-2+3j)(s-2-3j)]
Make sure above is correct .. now

[145s / (s-2j)(s-2+3j)(s-2-3j) ]*1/ [(s+2j)]
+
[145s / (s+2j)(s-2+3j)(s-2-3j) ]*1/ [(s-2j)]
+
[145s / (s-2j)(s+2j)(s-2-3j) ]*1/ [(s-2+3j)]
+
[145s / (s-2j)(s+2j)(s-2+3j) ]*1/ [(s-2-3j)]

I believe what I did above should be self explanatory. Now next step is to substitute s=-2j to [145s / (s-2j)(s-2+3j)(s-2-3j) ]
s=2j to [145s / (s+2j)(s-2+3j)(s-2-3j) ]
s = 2-3j in the third, and s=2+3j .. so on, You leave one of the simplest possible factor in the denominator taking all others to the numerator and then substituting the root.

See if that works :biggrin:
 
I'll try that, thanks :approve:
 
bakin said:
I'll try that, thanks :approve:

Answering your question in the OP, you had that correct. But, I just wanted to share a new tool I discovered in Calc III class which I think is very helpful for partial fractions.
 

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