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I was doing my laplace transform revision and I came across this problem. How to decompose this partial fraction ??

$$ \frac{145s}{(s^2+4)(s^2+4s+13)} +\frac{9s+55}{s^2+4s+13}$$

How to write the partial decomposition form and i much appreciated if you can show me the full step. I really forgot the partial decomposition... Thanks in advance =)

given answer is $$\frac{16}{s^2+4} + \frac{9s}{s^2+4} + \frac{3}{s^2+4s+13}$$

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1 Answer

up vote 2 down vote accepted

$s^2+4$ and $s^2+4s+13$ are irreducible quadratics, so you’re really just decomposing

$$\begin{align*} \frac{145s}{(s^2+4)(s^2+4s+13)}&=\frac{As+B}{s^2+4}+\frac{Cs+D}{s^2+4s+13}\\\\ &=\frac{(As+B)(s^2+4s+13)+(Cs+D)(s^2+4)}{(s^2+4)(s^2+4s+13)}\;. \end{align*}$$

Equate numerators:

$$\begin{align*} 145s&=(As+B)(s^2+4s+13)+(Cs+D)(s^2+4)\\ &=(A+C)s^3+(4A+B+D)s^2+(13A+4B+4C)s+(13B+4D)\;, \end{align*}$$

so

$$\left\{\begin{align*} &A+C=0\\ &4A+B+D=0\\ &13A+4B+4C=145\\ &13B+4D=0\;. \end{align*}\right.\tag{1}$$

Solving the system $(1)$ is mildly annoying but routine; the solution is $A=9,B=16,C=-9$, and $D=-52$. Thus,

$$\frac{145s}{(s^2+4)(s^2+4s+13)}=\frac{9s+16}{s^2+4}-\frac{9s+52}{s^2+4s+13}\;,$$

and

$$\begin{align*} \frac{145s}{(s^2+4)(s^2+4s+13)}+\frac{9s+55}{s^2+4s+13}&=\frac{9s+16}{s^2+4}+\frac3{s^2+4s+13}\\\\ &=\frac{9s}{s^2+4}+\frac{16}{s^2+4}+\frac3{s^2+4s+13}\;. \end{align*}$$

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Brian thanks for your great effort and help. You did very well. Thank you very much –  Garett Jan 20 '13 at 12:05
    
@Garett: You’re very welcome. –  Brian M. Scott Jan 20 '13 at 12:06
    
you have any facebook account or email. I like to add you as my friend. –  Garett Jan 20 '13 at 12:09
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