How to find the inverse Laplace transform of the following fraction?

Question

Please help me to find the inverse Laplace transform of the following: $$\frac{s^2+6s+9}{(s-1)(s-2)(s-3)}$$ Do I have to divide the fractions and use partial fractions decomposition? Like this: $$\frac{s^2}{(s-1)(s-2)(s-3)}=\frac{A}{s-1}+\frac{B}{s-2}+\frac{C}{s-3}$$ $$\frac{6s}{(s-1)(s-2)(s-3)}=\frac{A}{s-1}+\frac{B}{s-2}+\frac{C}{s-3}$$ $$\frac{9}{(s-1)(s-2)(s-3)}=\frac{A}{s-1}+\frac{B}{s-2}+\frac{C}{s-3}$$

Answer 1

We perform partial fraction decomposition on the expression first: \begin{align*} \frac{s^2 + 6s+9}{(s-1)(s-2)(s-3)} &= \frac{A}{s-1} + \frac{B}{s-2} + \frac{C}{s-3} \\ &= \frac{A(s-2)(s-3) + B(s-1)(s-3) + C(s-1)(s-2)}{(s-1)(s-2)(s-3)} \\ s^2 + 6s+9 &= A(s^2-5s+6) + B(s^2-4s+3) + C(s^2-3s+2) \\ &= (A+B+C)s^2 + (-5A-4B-3C)s + (6A+3B+2C) \end{align*} So we have the following system of equations, which can be solved by row reduction (the simpler way) or matrix inversion (the harder way): \begin{align*} \begin{pmatrix} 1 & 1 & 1 \\ -5 & -4 & -3 \\ 6 & 3 & 2 \end{pmatrix} \begin{pmatrix} A \\ B \\ C\end{pmatrix} &= \begin{pmatrix} 1 \\ 6 \\ 9\end{pmatrix} \\ \begin{pmatrix} A \\ B \\ C\end{pmatrix} &= \begin{pmatrix} 8 \\ -25 \\ 18\end{pmatrix} \end{align*} So we have: \begin{align*} \frac{s^2 + 6s+9}{(s-1)(s-2)(s-3)} &= \frac{8}{s-1} - \frac{25}{s-2} + \frac{18}{s-3} \\ \mathcal{L}^{-1} \left\{ \frac{s^2 + 6s+9}{(s-1)(s-2)(s-3)} \right\} &= \mathcal{L}^{-1} \left\{ \frac{8}{s-1} - \frac{25}{s-2} + \frac{18}{s-3} \right\} \\ &=8 \cdot \mathcal{L}^{-1} \left\{ \frac{1}{s-1} \right\} - 25 \cdot \mathcal{L}^{-1} \left\{ \frac{1}{s-2} \right\} + 18 \cdot\mathcal{L}^{-1}\left\{ \frac{1}{s-3} \right\} \\ &= 8e^t - 25e^{2t}+18e^{3t} \end{align*}

Answer 2

The partial expansion is:

$$\frac{s^2 + 6 s + 9}{(s - 3) (s - 2) (s - 1)}=\frac{18}{s-3}-\frac{25}{s-2}+\frac{8}{s-1}$$

Inverting term-by-term we get:

$$\mathcal{L}_s^{-1}\left(\frac{18}{s-3}\right)=18 e^{3 n}$$

$$-\mathcal{L}_s^{-1}\left(\frac{25}{s-2}\right)=-25 e^{2 n}$$

$$\mathcal{L}_s^{-1}\left(\frac{8}{s-1}\right)=8 e^n$$

And, putting it all together...

$$8 e^n-25 e^{2 n}+18 e^{3 n}$$

EDIT:

To get the quantities, $A$, $B$, $C$, we need to crate the partial fraction template using the denominator $(s-3)(s-2)(s-1)$

$$\frac{s^2 + 6 s + 9}{(s-3)(s-2)(s-1)}=\frac{A}{s-3}-\frac{B}{s-2}+\frac{C}{s-1}$$

Next, multiply the equation by the denominator:

$$\frac{(s^2 + 6 s + 9)(s-3)(s-2)(s-1)}{(s - 3) (s - 2) (s - 1)}=\frac{A(s-3)(s-2)(s-1)}{s-3}-\frac{B(s-3)(s-2)(s-1)}{s-2}+\frac{C(s-3)(s-2)(s-1)}{s-1}$$

And, after we simplify, it turns out that:

$$s^2 + 6 s + 9=A (s - 2) (s - 1)-B(s - 3) (s - 1)+C(s - 3) (s - 2)$$

Solve for the unknown parameters by plugging in the roots of the denominators

For example, for root 3, plug in $s=3$ into the equation

$$3^2+3\dot6+9 = A(3-2))(3-1)+B(3-3)(3-1)+C(3-3)(3-2)$$

$$36=2A$$

$$A=18$$

Can you do the rest?