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$\frac{ 2(s^2 + 9(s-1)}{ s (s^2 -9) } = \frac{A}{S} + \frac{ Bs + C}{s^2 -9} $

$ 2(s^2 + 9(s-1) ) = A (s^2 -9) + Bs+C $

Let $S = 0$

$ 2 (-18) = A(-9)$ , so $A = 2$

I sub $A = 2$ into the equation to get

$2s^2 + 18s - 18 = s^2 (2 + B) + Cs - 18$

By comparing coefficients,

$ B = 0$ and $ C= 18$

Why am I wrong here ?

Clayton
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user185692
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    $s(s^2 - 9) = s(s + 3)(s - 3)$, so you can use three linear factors. – N. F. Taussig Aug 17 '18 at 14:49
  • @N.F.Taussig does this mean my partial fractions is wrong ? Or I can’t do it like that ? – user185692 Aug 17 '18 at 14:52
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    When you put $s=0$, you end up with $-18=-9A+C$. You ignored the $C$. Also, as N.F. points out, you can't use partial fractions this way since $s^2-9$ is reducible (specifically, not irreducible). You always have to factor the denominator completely. – Clayton Aug 17 '18 at 14:55
  • @user185692 Try $$\frac{2(s^2+9(s-1))}{s(s^2-9)}=\frac{A}{S}+\frac{B}{s-3}+\frac{C}{s+3}$$ – Nosrati Aug 17 '18 at 16:54

1 Answers1

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you should factorise the denominator as far as you can while doing partial fraction

$\hspace{10pt}$ $$\frac{2(s^2+9(s-1))}{s(s-3)(s+3)}=\frac{A}{s}+\frac{B}{s-3}+\frac{C}{s+3}$$ $\hspace{10pt}$

multiply both the sides by $s(s-3)(s+3)$ your equations becomes $\hspace{12pt}$

$$2(s^2+9(s-1))=A(s-3)(s+3)+Bs(s+3)+Cs(s-3)$$ $$2s^2+18s-18=A(s^2-9)+B(s^2+3s)+C(s^2-3s)$$ $$2s^2+18s-18=s^2(A+B+C)+s(3B-3C)-9A$$

thus $$A+B+C=2$$ $$3B-3C=18$$ $$-9A=-18 \implies A=2$$

solve for $B, C$ you will get $B=3$,$C=-3$

your partial fraction is $$\frac{2(s^2+9(s-1))}{s(s-3)(s+3)}=\frac{2}{s}+\frac{3}{s-3}-\frac{3}{s+3}$$

Deepesh Meena
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