What is the inverse laplace transform of an inverted function?

Question

I have

$\mathcal{L}\{y'\}+4\mathcal{L}\{y\circledast e^{-4t}\}=\mathcal{L}\{1\}$ with $y(0)=0$

So we get:

$sY+4Y\frac{1}{1+4}=\frac{1}{s}$

$\iff Y = \frac{1}{s^2+\frac{4s}{s+4}}$

Now I could rewrite this as $Y=\frac{1}{F(S)}$ but I am not aware of any identity for $\mathcal{L}^{-1}\{\frac{1}{F(s)}\}$.

How can I find the Laplace inverse of this expression?

Answer 1

You don't need an identity for $\mathcal{L}^{-1}\{\frac{1}{F(s)}\}$.

Rewrite it:

$$Y = \frac{1}{s^2+\frac{4s}{s+4}} = \dfrac{s + 4}{s^2 (s + 4) + 4s}= \dfrac{s + 4}{s (s + 2)^2}$$

Do a partial fraction decomposition

$$ Y = \dfrac{s + 4}{s (s + 2)^2} = \dfrac{A}{s} + \dfrac{B}{s + 2} + \dfrac{C}{(s + 2)^2} $$

and you should be able to do the inverse transformation from a basic transform table.

Answer 2

It is worth noting that, as of the date of posting, that there are some potential errors in the analysis by the proposer of the problem. This error comes to light when taking the transform of the convolution term, ie \begin{align} \int_{0}^{\infty} e^{-s t} \, e^{-4 t} \, y(t) \, dt &= \int_{0}^{\infty} e^{- (s + 4) t} \, y(t) \, dt = \overline{y}(s+4) \end{align} which makes the equation $$s \, \overline{y}(s) + 4 \overline{y}(s + 4) = \frac{1}{s}.$$

It is of note to mention that the solution to the first order differential equation is $$y(t) = e^{e^{-4 t}} \, \int_{0}^{t} e^{- e^{-4 u}} \, du.$$