Dumb partial fractions question

  • Thread starter cabellos
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  • #1
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dumb partial fractions question....

suppose i get x+1=A(x-2)+B(x-2)

how do you then find A and B?
 

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  • #2
Hootenanny
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Which values of x could you chose so that the equation simplified to leave you with a single unknown?
 
  • #3
arildno
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By demanding you've got an IDENTITY there, and remembering that the functions f(x)=1 and g(x)=x are linearly INDEPENDENT functions.
This will give you two equations for your two unknows A and B.
 
  • #4
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But say I use 2 as my x value, it removes both A and B in this equation....
 
  • #5
arildno
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Oops, I didn't see that closely!
You've made a mistake somewhere.
 
  • #6
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i haven't made a mistake i have to find the inverse laplace transform of s+1/(s^2 - 4s + 4) and im using partial fractions to do this.........but im stuck now......:cry:
 
  • #7
OlderDan
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i haven't made a mistake i have to find the inverse laplace transform of s+1/(s^2 - 4s + 4) and im using partial fractions to do this.........but im stuck now......:cry:
The denominator is a perfect square. You need to use a different form for the sum of fractions. Have you seen this before?

Is that s+1/(s^2 - 4s + 4) or (s+1)/(s^2 - 4s + 4)??

I assume the second based on your original equation.
 
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  • #8
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ok thanks - does this mean i find the inverse laplace transform of 1/(s-2)^2 and add this to the inverse LT of s/(s-2)^2 ??????????

I can do the first that would be te^2t wouldnt it?

Not sure about the second part???? :cry:
 
  • #9
OlderDan
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ok thanks - does this mean i find the inverse laplace transform of 1/(s-2)^2 and add this to the inverse LT of s/(s-2)^2 ??????????

I can do the first that would be te^2t wouldnt it?

Not sure about the second part???? :cry:
If you can use tables, go here

http://www.vibrationdata.com/Laplace.htm

2.10 confirms your first term, and 2.15 cannot be used for a perfect square denominator. Partial fractions will get rid of the s in the numerator for you, and you will get a 2.9 form and a 2.10 form. Yes??
 
  • #10
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i can see the first relationship with 2.10 but what do i need to do to s/(s-2)^2 ? use partial fractions? Thats what i tried at the start but how do i find A and B if s = A(s-2) + B(s-2) ?
 
  • #11
OlderDan
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i can see the first relationship with 2.10 but what do i need to do to s/(s-2)^2 ? use partial fractions? Thats what i tried at the start but how do i find A and B if s = A(s-2) + B(s-2) ?
You cannot use that form when the denominator is a perfect square. Use
(s + 1)/(s - 2)² = A/(s - 2) + B/(s - 2)²
 
  • #12
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how does that change anything........still can't solve A and B??
 
  • #13
OlderDan
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how does that change anything........still can't solve A and B??
Sure you can.

(s + 1)/(s - 2)² = A/(s - 2) + B/(s - 2)²

s + 1 = A*(s - 2) + B

s + 1 = As - 2A + B
By demanding you've got an IDENTITY there, and remembering that the functions f(x)=1 and g(x)=x are linearly INDEPENDENT functions.
This will give you two equations for your two unknows A and B.
A = 1

1 = -2A + B

B = 1 + 2A = 3
 

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