Answer to Question #153827 in Differential Equations for Ra

Question

Calculate the inverse Laplace transform of the function :

F(s) = (s+1)/ (s³+s²-6s)

Answer 1

$F(s)=\frac{s+1}{s^3+s^2-6s}\\ \therefore F(s)\\ =\frac{s+1}{s^3+s^2-6s}\\ =\frac{(s+1)}{s(s-2)(s+3)}\\ =\frac{3}{10}\times \frac{1}{s-2}-\frac{1}{6}\times \frac{1}{s}-\frac{2}{15}\times \frac{1}{s+3}$

$So,$

$\mathcal{L}^{-1}\{F(s)\} \\=\mathcal{L}^{-1}\{\frac{3}{10}\times \frac{1}{s-2}-\frac{1}{6}\times \frac{1}{s}-\frac{2}{15}\times \frac{1}{s+3}\} \\=\frac{3}{10}\mathcal{L}^{-1}\left( \frac{1}{s-2}\right)-\frac{1}{6}\mathcal{L}^{-1}\left( \frac{1}{s}\right)-\frac{2}{15}\mathcal{L}^{-1}\left( \frac{1}{s+3}\right)\\ =\frac{3}{10}e^{2t}-\frac{1}{6}\times1-\frac{2}{15}e^{-3t}\\ =\frac{3}{10}e^{2t}-\frac{2}{15}e^{-3t}-\frac{1}{6}$

$\pmb\therefore$ Inverse laplace of of $\pmb{F(s) :=\frac{3}{10}e^{2t}-\frac{2}{15}e^{-3t}-\frac{1}{6}}$