91Ó°ÊÓ

We want to show that \(A_{k} = \frac{P(r_{k})}{Q'(r_{k})}\) for k = 1, ... , n. Using the hint given in the problem, let's multiply the partial fraction expansion equation by \((s-r_{k})\): $$ (s-r_{k}) \frac{P(s)}{Q(s)} = A_1 (s-r_{k})\frac{1}{s-r_1} + ... + A_k (s-r_{k})\frac{1}{s-r_k} + ... + A_n (s-r_{k})\frac{1}{s-r_n} $$ Now take the limit as s approaches \(r_k\): $$ \lim_{s \to r_k}(s-r_k) \frac{P(s)}{Q(s)} = \lim_{s \to r_k} \left[A_1 (s-r_k)\frac{1}{s-r_1} + ... + A_k (s-r_{k})\frac{1}{s-r_k} + ... + A_n (s-r_{k})\frac{1}{s-r_n}\right] $$ Outside of the k-th term, all other terms will go to 0 as s approaches \(r_k\). The k-th term becomes: $$ \lim_{s \to r_k} A_k (s-r_{k})\frac{1}{s-r_k} = A_k $$ By also taking the limit of the left side of the equation, we determine: $$ A_k = \lim_{s \to r_k} (s-r_k)\frac{P(s)}{Q(s)} $$ L'Hôpital's Rule can be applied, since both numerator and denominator approach 0: $$ A_k = \frac{P(r_k)}{Q'(r_k)} $$ Hence, we have proven that \(A_{k}=\frac{P(r_{k})}{Q'(r_{k})}\) for k = 1, ... , n.

Now that we found the coefficients \(A_k\), let's compute the inverse Laplace transform of \(F(s)\): $$ \mathcal{L}^{-1}\{F(s)\} = \mathcal{L}^{-1} \left\{ \sum_{k=1}^{n} \frac{A_{k}}{s-r_{k}} \right\} $$ Since the inverse Laplace transform is a linear operator, we can write it as a sum of the inverse Laplace transforms of the individual terms: $$ \mathcal{L}^{-1}\{F(s)\} = \sum_{k=1}^{n} \mathcal{L}^{-1}\left\{ \frac{A_{k}}{s-r_{k}} \right\} $$ Using the inverse Laplace transform formula for exponential functions, \(\mathcal{L}^{-1}\left\{ \frac{A_{k}}{s-r_{k}} \right\} = A_{k}e^{r_{k}t}\), we get: $$ \mathcal{L}^{-1}\{F(s)\} = \sum_{k=1}^{n} A_{k}e^{r_{k}t} $$ Substitute the expression for \(A_k\) we found in Part 1: $$ \mathcal{L}^{-1}\{F(s)\} = \sum_{k=1}^{n} \frac{P(r_{k})}{Q'(r_{k})} e^{r_{k}t} $$ Thus, we have shown that the inverse Laplace transform of F(s) can be expressed as \(\mathcal{L}^{-1}\{F(s)\}=\sum_{k=1}^{n} \frac{P(r_{k})}{Q'(r_{k})} e^{r_{k} t}\).