. a . Calculate L^-1 (1)/(s^3)-7s^2+10s)

To calculate the inverse Laplace transform of $\frac{1}{s^3 - 7s^2 + 10s}$, we can use partial fraction decomposition.<br /><br />First, let's factor the denominator:<br />$s^3 - 7s^2 + 10s = s(s^2 - 7s + 10) = s(s-2)(s-5)$<br /><br />Now, we can decompose the fraction into partial fractions:<br />$\frac{1}{s^3 - 7s^2 + 10s} = \frac{A}{s} + \frac{B}{s-2} + \frac{C}{s-5}$<br /><br />Multiplying both sides by the common denominator $s(s-2)(s-5)$, we get:<br />$1 = A(s-2)(s-5) + Bs(s-5) + Cs(s-2)$<br /><br />Expanding and simplifying, we have:<br />$1 = As^2 - 7As + 10A + Bs^2 - 5Bs + Cs^2 - 2Cs$<br /><br />Combining like terms, we get:<br />$1 = (A+B+C)s^2 - (7A-5B-2C)s + 10A$<br /><br />Comparing coefficients, we have the following system of equations:<br />$A+B+C = 0$<br />$-7A-5B-2C = 0$<br />$10A = 1$<br /><br />Solving this system, we find:<br />$A = \frac{1}{10}$<br />$B = \frac{7}{10}$<br />$C = -\frac{1}{10}$<br /><br />Now, we can rewrite the original expression as:<br />$\frac{1}{s^3 - 7s^2 + 10s} = \frac{1/10}{s} + \frac{7/10}{s-2} - \frac{1/10}{s-5}$<br /><br />Taking the inverse Laplace transform of each term separately, we get:<br />$L^{-1}\left\{\frac{1/10}{s}\right\} = \frac{1}{10}t$<br />$L^{-1}\left\{\frac{7/10}{s-2}\right\} = \frac{7}{10}e^{2t}$<br />$L^{-1}\left\{\frac{-1/10}{s-5}\right\} = -\frac{1}{10}e^{5t}$<br /><br />Therefore, the inverse Laplace transform of $\frac{1}{s^3 - 7s^2 + 10s}$ is:<br />$L^{-1}\left\{\frac{1}{s^3 - 7s^2 + 10s}\right\} = \frac{1}{10}t + \frac{7}{10}e^{2t} - \frac{1}{10}e^{5t}$