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Performing fractional decomposition on the fraction

$\frac{a(x-c)+b}{(x-y_1)(x-y_2)}=\frac{A}{x-y_1} + \frac{B}{x-y_2}$ $\space$(1)

where $x$ is a variable and the rest are constants, you find

$A = \frac{a(y_1-c)+b}{y_1-y_2}$

$B = \frac{a(y_2-c)+b}{y_2-y_1}$

If you try to check these values by inserting $A$ and $B$ into equation (1), you arrive at this equation

$\frac{a(x-c)+b}{(x-y_1)(x-y_2)}= \frac{(y_1-y_2)^2[a(x-c)+b]}{(x-y_1)(x-y_2)}$

which implies that $(y_1-y_2)^2 = 1$, despite these being just arbitrary constants. Is this true? Where does this come from? I came upon this from checking the values obtained from fractional decomposition for a dynamic model on a first order system.

Andy
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1 Answers1

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Your $A$ and $B$ are correct. Then you had a mistake somewhere. Inserting $A$ and $B$ on the RHS gives the following steps:

$$ \frac{(a(y_1-c)+b)}{(y_1-y_2) (x-y_1)} - \frac{(a(y_2-c)+b)}{(y_1-y_2) (x-y_2)} \\ = \frac{(a(y_1-c)+b) (x-y_2)}{(y_1-y_2) (x-y_1)(x-y_2)} - \frac{(a(y_2-c)+b)(x-y_1)}{(y_1-y_2) (x-y_1)(x-y_2)} \\ = \frac{a(x-c)(y_1-y_2) +b(y_1-y_2) }{(y_1-y_2) (x-y_1)(x-y_2)} \\ = \frac{a(x-c) +b }{ (x-y_1)(x-y_2)} $$

Andreas
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