Ronnie Baur

2021-11-17

Find the partial fraction expansion of $\frac{5{s}^{3}}{\left({s}^{2}+4\right)\left({s}^{2}+2s+2\right)}$

### Answer & Explanation

Parminquale

Step 1
The given fraction is,
$\frac{5{s}^{3}}{\left({s}^{2}+4\right)\left({s}^{2}+2s+2\right)}$
The denominator of the fraction has two factors .
Since the quadratic equations are not reducible, the partial fractions must have the form
$\frac{5{s}^{3}}{\left({s}^{2}+4\right)\left({s}^{2}+2s+2\right)}=\frac{As+B}{{s}^{2}+4}+\frac{Cs+D}{{s}^{2}+2s+2}$
Step 2
Simplify further as follows.
$\frac{5{s}^{3}}{\left({s}^{2}+4\right)\left({s}^{2}+2s+2\right)}=\frac{As+B}{{s}^{2}+4}+\frac{Cs+D}{{s}^{2}+2s+2}$
$5{s}^{3}=\left(As+B\right)\left({s}^{2}+2s+2\right)+\left(Cs+D\right)\left({s}^{2}+4\right)$
$5{s}^{3}={s}^{3}\left(B+D\right)+{s}^{2}\left(A+2B+C\right)+s\left(2A+2B+4D\right)+\left(2A+4C\right)$
Equate the coefficients on both sides as follows.
$B+D=5$
$A+2B+C=0$
$2A+2B+4D=0$
$2A+4C=0$
Step 3
On solving the above equations, .
Substitute in (1) as follows.
$\frac{5{s}^{3}}{\left({s}^{2}+4\right)\left({s}^{2}+2s+2\right)}=\frac{2s+\left(-8\right)}{{s}^{2}+4}+\frac{3s+4}{{s}^{2}+2s+2}$

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