I got that the inverse Laplace of $\frac{1}{s-\lambda}$ is $e^{\lambda t}$. Does this look correct?
Inverse Laplace transform of $\frac1{s-\lambda}$ is $e^{\lambda t}$
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laplace-transform
1 Answers
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Your answer is correct. The inverse Laplace transform is $ \mathcal{L}\{F(s)\} = \frac{1}{2\pi i}\int_{\gamma - i\infty}^{\gamma + i\infty}F(s)e^{st}ds. $ In your problem, we have that $F(s) = \frac{1}{s-\lambda}$ so we have a simple pole in the $s$ plane at $s = \lambda$. \begin{align} \frac{1}{2\pi i}\int_{\gamma - i\infty}^{\gamma + i\infty}F(s)e^{st}ds &= \frac{1}{2\pi i}\int_{\gamma - i\infty}^{\gamma + i\infty}\frac{e^{st}}{s-\lambda}ds\\ &= \sum\text{Res}\\ &= \lim_{s\to\lambda}(s-\lambda)\frac{e^{st}}{s-\lambda}\\ &= e^{t\lambda} \end{align} as you have found.