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Find the inverse Laplace transform of $$\mathcal{L}^{-1}\left(\frac{2s+12}{ (s^2 + 5s + 6)(s+1)}\right)$$

I recognise I need to use partial fractions to solve it and that is where I got stuck. Here’s my working,

After factoring the denominator, I got a case of non repeating linear factors

$\dfrac{2s+12}{ (s+2)(s+3)(s+1)} = \dfrac{A}{s+2} + \dfrac{B}{s+3} + \dfrac{C}{s+1} $

$2s + 12 = A (s+3)(s+1) + B (s+2)(s+1) + C (s+2)(s+3) $

$2s + 12 = (As^2 + Bs^2 + Cs^2) + (4As + 3Bs + 5Cs) + (3A + 2B+6C) $

So...

$A + B + C = 0$

$ 4A + 3B + 5C = 2$

$3A + 2B + 6C = 12$

how do I solve this complicated equations ? This is where i got stuck and cannot continue.

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You can use the methods associated with the "Partial fraction decomposition" (see https://en.wikipedia.org/wiki/Partial_fraction_decomposition). After writing :

$$\frac{2s+12}{ (s+2)(s+3)(s+1)} = \frac{A}{s+2} + \frac{B}{s+3} + \frac{C}{s+1} $$

You multiply and the right and left hand side by $(s+2)$, which gives :

$$\frac{2s+12}{(s+3)(s+1)} =A + \frac{B(s+2)}{s+3} + \frac{C(s+2)}{s+1} $$

Taking $s = -2$, you find :

$$ \frac{2(-2)+12}{(-2+3)(-2+1)} = \frac{8}{-1} = -8 = A $$

You can find $B$ and $C$ using the same method (which is usually easier than solving the system).

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There is a shortcut way here, in this simple case. Let to mutiply the relation to $\color{red}{s+1}$, then $$\frac{2s+12}{ (s+2)(s+3)(s+1)} \color{red}{(s+1)}= \frac{A\color{red}{(s+1)}}{s+2} + \frac{B\color{red}{(s+1)}}{s+3} + \frac{C\color{red}{(s+1)}}{s+1}$$ or $$\frac{2s+12}{ (s+2)(s+3)} = \frac{A\color{red}{(s+1)}}{s+2} + \frac{B\color{red}{(s+1)}}{s+3} + C$$ now set $s=-1$ then $C=\dfrac{10}{2}$.

Can you proceed?

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$A + B + C = 0\quad[1]$

$ 4A + 3B + 5C = 2\quad[2]$

$3A + 2B + 6C = 12\quad[3]$

$ $

$[2]-3[1]$

$A+2C=2\quad[4]$

$ $

$[3]-2[1]$

$A+4C=12\quad[5]$

$ $

$[5]-[4]$

$2C=10$

$ $

$\therefore C=5$

$\therefore A=-8$

$\therefore B=3$

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