In Problems\(21 - 30\), determine \({\mathcal{L}^{ - 1}}\{ F\} \).

\(F(s) = \frac{{7{s^3} - 2{s^2} - 3s + 6}}{{{s^3}(s - 2)}}\)

\(F(s) = \frac{{7{s^3} - 2{s^2} - 3s + 6}}{{{s^3}(s - 2)}} \Rightarrow \frac{A}{s} + \frac{B}{{{s^2}}} + \frac{C}{{{s^3}}} + \frac{D}{{s - 2}}\quad \)

Use the method of partial fractions to split up the fraction. Since there are repeated linear factors \(\left( {{s^3} = s*s*s} \right)\), we need to split up the denominator in a specific way with a different numerator over each one (as shown on the left). There is also an unrepeated linear factor, which we can handle the usual way.

\(A\left( {{s^2}} \right)(s - 2) + B(s)(s - 2) + C(s - 2) + d\left( {{s^3}} \right) = 7{s^3} - 2{s^2} - 3s + 6\)

Multiply each term by the original denominator, \(\left( {{s^3}} \right)(s - 2)\) Note: A, B and C will be multiplied by the term that is not in their denominator \((s - 2)\), and however many more \(s\) are necessary so to equal \({s^3}\). For instance, since \({\rm{A}}\) is over \(s,\;{\rm{A}}\) will be multiplied by \({s^2}\) because \(s\left( {{s^2}} \right) = {s^3}\)

\(\frac{{7{s^3} - 2{s^2} - 3s + 6}}{{\left( {{s^3}} \right)(s - 2)}} \Rightarrow \)\(7{s^3} - 2{s^2} - 3s + 6\) when multiplied by the original denominator.

\(A{s^3} - 2A{s^2} + B{s^2} - 2Bs + Cs - 2C + D{s^3} = 7{s^3} - 2{s^2} - 3s + 6\quad \)

Expand the equation by distributing A, B, C, and D.

Set any terms that contain an \({s^3},{s^2}\), or \(s\) on the left side of the equation equal to any terms that have \(s\) to the corresponding degree on the right side, and do the same thing for the constants.

\(\begin{array}{c}A{s^3} + D{s^3} = 7{s^3}\\B{s^2} - 2A{s^2} = - 2{s^2}\\ - 2Bs + Cs = - 3s\\ - 2C = 6\quad \end{array}\)

Divide each equation by the appropriate number of \(s\) 's so there are only coefficients left to solve.

\(\begin{array}{l}A + D = 7;\\B - 2A = - 2;\\ - 2B + C = - 3;\\ - 2C = 6\quad \end{array}\)

Solve the system of equations

\(\begin{array}{l}A = 1\\B = 0\\C = - 3\\D = 6\quad \end{array}\)

Plug the values of the variables into the original

\(F(s) = \frac{1}{s} + \frac{{ - 3}}{{{s^3}}} + \frac{6}{{s - 2}}\)fractions we found

Solve the inverse Laplace’s for \(f(t)\) using a table.

\({\mathcal{L}^{ - 1}}\left\{ {\frac{1}{s}} \right\} + {\mathcal{L}^{ - 1}}\left\{ {\frac{{ - 3}}{{{s^3}}}} \right\} + {\mathcal{L}^{ - 1}}\left\{ {\frac{6}{{s - 2}}} \right\} = 1 - \frac{3}{2}{t^2} + 6{e^{2t}}\)

Thus, \({\mathcal{L}^{ - 1}}\left\{ F \right\} = 1 - \frac{3}{2}{t^2} + 6{e^{2t}}\)