How to find the inverse laplace transform of $\frac{48s+36}{s(6s^2+11s+6)}$?

Question

How to find the inverse laplace transform of this function? $$F(s) = \displaystyle \frac{48s+36}{s(6s^2+11s+6)}.$$ It has been hard to break down the denominator part. Any help from you guys would be appreciated!

Answer 1

$$\begin{aligned} F(s) &= \displaystyle \frac{48s+36}{s(6s^2+11s+6)}\\ &= \frac{8s+6}{s(s-\alpha-i\beta)(s-\alpha+i\beta)}, \quad \textrm{where } \alpha=-\frac{11}{12}, \textrm{ and } \beta = \frac{\sqrt{23}}{12}\\ &= \frac{8s+6}{s[(s-\alpha)^2+\beta^2]} \end{aligned}$$

Expressing this as a partial fraction expansion:

$$\begin{aligned} F(s) &= \frac{6}{s} - \frac{6(s-\alpha)}{(s-\alpha)^2+\beta^2}+\frac{5/2}{(s-\alpha)^2+\beta^2}\\&= \frac{6}{s} - 6\frac{s-\alpha}{(s-\alpha)^2+\beta^2}+\frac{30}{\sqrt{23}}\frac{\beta}{(s-\alpha)^2+\beta^2}\end{aligned} $$

So $$\begin{aligned}f(x)&=\mathcal{L}^{-1}\{F(s)\}\\&=6 - e^{-\alpha x}\left[6\cos \beta x - \frac{30}{\sqrt{23}}\sin \beta x \right]\\&= 6 - 6e^{-\frac{11}{12}x}\left[\cos \left(\frac{\sqrt{23}}{12} x\right)- \frac{5}{\sqrt{23}}\sin\left( \frac{\sqrt{23}}{12}x\right)\right].\end{aligned}$$

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$\textbf{UPDATE:}$ The second step, where we explicitly show the complex roots, is not necessary, but, by use of the quadratic formula, it is helpful to know that the roots are complex conjugates. We could have gone straight to the now third step, by completing-the-square. The main point is that the form we now have leads to a partial fraction expansion that is well known, and can be found in almost any table of Laplace transforms.

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