Inverse Laplace Transform of $e^{-as}(\frac{1}{s^2}- \frac{1}{s^2+1})$

Question

I'm having troubles finding the inverse laplace transform of $$e^{-as}\left({1\over s^2} - {1\over {s^2+1}}\right)$$ where $a$ is a constant of-course.

Help with explanations would be appreciated!

Answer 1

You need this:

$\mathcal{L}^{-1}\left\{F(s)e^{-as}\right\} =\left[\mathcal{L}^{-1}\{F(s)\}\right]_{t\to t-a}U(t-a)$

For example

$$ \mathcal{L}^{-1}\left\{\frac{se^{-\pi s}}{s^2+1}\right\}=\cos(t-\pi)U(t-\pi)$$

Answer 2

Second Shifting Property: If $~\mathcal{L}^{-1}\left\{F(p)\right\} =f(t)~$, then $~\mathcal{L}^{-1}\left\{F(p)~e^{-ap}\right\} =g(t)~$ where $$~g(t)=\begin{cases} f(t-a) \quad \text{if} ~~~t~\gt~a\\ 0 \quad ~~~~~~~~~~~~\text{if} ~~~t~\lt ~a\end{cases}~$$

Here we have to find $$\mathcal{L}^{-1}\left\{e^{-as}\left({1\over s^2} - {1\over {s^2+1}}\right)\right\} $$

Here $$F(s)={1\over s^2} - {1\over {s^2+1}}$$

So $$f(t)=\mathcal{L}^{-1}\left\{F(s)\right\} =t-\sin t$$as $~\mathcal{L}(t)=\frac{1}{s^2}~$and $~\mathcal{L}(\sin at)=\frac{a}{a^2+s^2}~$

Hence $$\mathcal{L}^{-1}\left\{e^{-as}\left({1\over s^2} - {1\over {s^2+1}}\right)\right\}=\begin{cases} (t-a)~-~\sin(t-a) \quad \text{if} ~~~t~\gt~a\\ 0 \quad ~~~~~~~~~~~~\text{if} ~~~t~\lt ~a\end{cases}~ $$