# Solving partial fraction expansion with all variables

Okay so I have an equation in my book which is as follows.. $$\frac {a}{s(s+a)}$$ it says "using partial fractions this can be expanded to $$\frac {1}{s} + \frac {-1}{s+a}$$

My usual method would be to cross multiply and do something like this $$\frac {a}{s(s+a)} = \frac {A(s+a)}{s(s+a)} + \frac {B(s)}{s(s+a)}$$

Then cancel off the denominators and solve..

$$a = A(s+a) + B(s)$$

usually though the a would be some constant but here I have no values to play around with.. how has he done it in the book?

• Sometimes you have $s+a$ in the denominator; sometimes $s+1$. Which is correct? – Empy2 Nov 3 '13 at 14:33
• Sorry I copied it down wrongly.. I have redone it now – Shasam Nov 3 '13 at 14:38

We have:

$$\dfrac{a}{s(s+a)} = \dfrac{A}{s}+\dfrac{B}{s+a}$$

So,

$$a = A(s+a) + Bs = (A+B)s + A a$$

we have $A = 1, B = -1$

Final result:

$$\dfrac{a}{s(s+1)} = \dfrac{1}{s}-\dfrac{1}{s+a}$$

• Ahh right I get it.. Making A+B = 0 and making A = 1.. Leaves us with a = 0 + 1*a, which means a=a.. Thanks! – Shasam Nov 3 '13 at 14:49
• You got it, but I look at it slightly different. We see the LHS is $a$ by itself, right off that tells us what $Aa$ is, so $A = 1$. Now I need for the $(A+B)s$ to cancel, and already know $A$, so $B = -1$. Clear? Regards – Amzoti Nov 3 '13 at 14:50

Let $s=0$, so $a=Aa$. Let $s=1$, so $a=A(a+1)+B$
Solve for $A$ and $B$.

The equality you have is $$a=(A-B)s+aA.$$ This suggests taking $A=B=1$.

• Sorry i totally copied it down wrong.. I have changed it now – Shasam Nov 3 '13 at 14:41