So I have $$\mathcal{L}\{f(t)\}=\frac{1-e^{-3s}}{s(1+e^{-3s})}$$
$f(t)$ is the function whose laplace transform is above. Suppose Y(s) is the Laplace transform of the solution to the Initial Value Problem $$y'(t) = 10f(t) \quad \text{with} \quad y(0)=0$$
Find an expression for Y(s):
I got $$sY(s) - 0 = \frac{10(1-e^{-3s})}{s(1+e^{-3s})}$$ so $$Y(s) = \frac{10(1-e^{-3s})}{s^2(1+e^{-3s})}$$
Now, by writing $\frac{1-e^{-3s}}{1+e^{-3s}}$ as the infinite series $$\frac{1-e^{-3s}}{1+e^{-3s}} = \sum_{j=0}^\infty (-1)^j(e^{-3js} -e^{-3(j+1)s})$$
Find the solution to the Initial Value Problem above.
I know this is probably really simple, and I know how to integrate series etc, but I am struggling to see how I approach this.
Do I multiply the $\frac{10}{s^2}$ through the series and then integrate term by term using the Heaviside function result?
Hints preferred, although if suggesting one approach over another, a short explanation of why (in case it isn't obvious) please.
Notation: $\overline{f} = \overline{f}(s)$ is the transformed function and $f(t) \doteqdot f(s)$ shows the function and its transform.
Consider $y^{'}(t) = a \, f(t)$ the then Laplace transform is $$ s \, \overline{y} - y_{0} = a \, \overline{f} $$ with $$ f \doteqdot \frac{1 - e^{-b s}}{s \, (1 + e^{- b s})}. $$ Now, by using $$ \begin{cases} 0 & 0 < t < a \\ t - a & a < t < b \\ b-a & t > b \end{cases} \, = (t-a) \, H(t-a) + (b-t) \, H(t-b) \doteqdot \frac{e^{-a s} - e^{-b s}}{s^2}, $$ where $H(t)$ is the Heaviside step function, and \begin{align} f(t) &\doteqdot \frac{1 - e^{-b s}}{s \, (1 + e^{- b s})} \\ &\doteqdot \sum_{k=0}^{\infty} (-1)^k \, \frac{e^{- b k s} - e^{- b (k+1) s}}{s} \end{align} then \begin{align} \overline{y} &\doteqdot \frac{y_{0}}{s} + \frac{a \, \overline{f}}{s} \\ &\doteqdot \frac{y_{0}}{s} + a \, \sum_{k=0}^{\infty} (-1)^k \, \frac{e^{- b k s} - e^{- b (k+1) s}}{s^2} \\ y(t) &= y_{0} + a \, \sum_{k=0}^{\infty} (-1)^k \, \left[ (t-b k) \, H(t-b k) + (b(k+1)-t) \, H(t-b(k+1)) \right]. \end{align}
It may be possible to give a solution in terms of error functions, but the one presented here could be efficient enough to obtain properties and compute the series further.