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leopard
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Separating a fraction
I don't remember what this method is named in English, but I want to write the fraction
[tex]\frac{1}{(s^2 + 1)(s-3)(s+2)}[/tex]
in the form
[tex]\frac{A}{s^2 + 1} + \frac{B}{s-3} + \frac{C}{s+2}[/tex]
I multiply A with (s-3)(s+2), B with (s^2 + 1)(s+2) and C with (s^2 + 1)(s-3), to get
[tex]As^2 - As - 6A + Bs^3 + 2Bs^2 + Bs + 2B + Cs^3 - 3Cs^2 + Cs - 3C = 1[/tex]
and the equations
(1) B + C = 0
(2) A + 2B - 3C = 0
(3) -A + B + C = 0
(4) -6A + 2B - 3C = 1
(1) and (3) gives A = 0
(2) then gives B = C = 0
This is obviously wrong. The correct answer is
A = 2s - 4
B = 2
C = -4
according to my book. There must be something I don't understand here...
I don't remember what this method is named in English, but I want to write the fraction
[tex]\frac{1}{(s^2 + 1)(s-3)(s+2)}[/tex]
in the form
[tex]\frac{A}{s^2 + 1} + \frac{B}{s-3} + \frac{C}{s+2}[/tex]
I multiply A with (s-3)(s+2), B with (s^2 + 1)(s+2) and C with (s^2 + 1)(s-3), to get
[tex]As^2 - As - 6A + Bs^3 + 2Bs^2 + Bs + 2B + Cs^3 - 3Cs^2 + Cs - 3C = 1[/tex]
and the equations
(1) B + C = 0
(2) A + 2B - 3C = 0
(3) -A + B + C = 0
(4) -6A + 2B - 3C = 1
(1) and (3) gives A = 0
(2) then gives B = C = 0
This is obviously wrong. The correct answer is
A = 2s - 4
B = 2
C = -4
according to my book. There must be something I don't understand here...
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