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I am currently stuck on a problem involving partial fraction decomposition. I think this one should be pretty easy, just that I can't wrap my head around it for some reason.

let $a, b$ and $k$ be constants. We suppose that $a < b$ and that $k$ is positive.

We have the expression

$\frac{1}{(a-y)(b-y)} = \frac{A}{y-a} + \frac{B}{y-b}$

This gives $A = \frac{1}{a-b}$ & $B = \frac{1}{a-b}$

Can someone please show me how to solve this using an equation system?

Thank you

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We want to write $$\dfrac1{(y-a)(y-b)} = \dfrac{A}{y-a} + \dfrac{B}{y-b}$$ The right hand side can be simplified as $$\dfrac{A}{y-a} + \dfrac{B}{y-b} = \dfrac{A(y-b) + B(y-a)}{(y-a)(y-b)} \,\,\,\,\,\,\,\,\,\,\,\, (\star)$$ We want $(\star)$ to equal $\dfrac1{(y-a)(y-b)}$. For this to happen, we need $$A(y-b) + B(y-a) = 1 \,\,\,\,\,\,\,\, \forall y\,\,\,\,\,\,\,\,\,\,\,\, (\dagger)$$ Now there are couple of ways to go about solving this. A simple way is as follows. Since $(\dagger)$ is true for all $y$, the equality holds for $y=a$ and $y=b$ as well.

Setting $y=a$, we get that $A(a-b) = 1 \implies A =\dfrac1{a-b}$.

Setting $y=b$, we get that $B(b-a) = 1 \implies B =\dfrac1{b-a}$.

A slightly more general way (which is very much related to the above procedure is as follows).

Note that $(\dagger)$ can be written as $$(A+B)y - (bA+aB) = 1 \,\,\,\,\,\,\,\, \forall y$$ Since this is true for all $y$, the coefficient of $y$ should be $0$ and $-(bA+aB) = 1$. Hence, we get that \begin{align} A+B & = 0\\ -(bA+aB) & = 1 \end{align} Solving the above equations, give us $$A = \dfrac1{a-b}, \,\,\,\,\,\,\,\,\,\, B = \dfrac1{b-a}$$

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  • $\begingroup$ Thank you for a great answer! $\endgroup$ – Lukas Arvidsson Nov 25 '12 at 20:59
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Assuming you have a typo in your formula ($y-b$ should be dividing $B$), you expand $$ \frac{A}{y-a}+\frac{B}{y-b}=\frac{A(y-b)+B(y-a)}{(y-a)(y-b)}=\frac{(A+B)y-Ab-Ba}{(y-a)(y-b)}. $$ You want this to be $1/(y-a)(y-b)$, so you need $(A+B)y-Ab+Ba=1$ for all $y$. This forces $A+B=0$, and $-Ab-Ba=1$. From the first equation, $B=-A$; substituting into the second one, we get $$ -Ab-(-A)a=1, $$ so $A(a-b)=1$, i.e. $A=1/(a-b)$. So $$ A=\frac1{a-b},\ \ B=-A=\frac{-1}{(a-b)}. $$

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  • $\begingroup$ Thank you for noticing the typo, and thank you for the answer! $\endgroup$ – Lukas Arvidsson Nov 25 '12 at 20:57

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