- #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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- Thread starter cabellos
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- #1

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

how do you then find A and B?

- #2

Hootenanny

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- #3

arildno

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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.

You've made a mistake somewhere.

- #6

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- #7

OlderDan

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

Not sure about the second part????

- #9

OlderDan

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If you can use tables, go here

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

Not sure about the second part????

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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- #11

OlderDan

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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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Sure you can.how does that change anything........still can't solve A and B??

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

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

s + 1 = As - 2A + B

A = 1

This will give you two equations for your two unknows A and B.

1 = -2A + B

B = 1 + 2A = 3

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