# Find Laplace Transform: (1)/(s^2) (1-e^(-st_0))^2=(4)/(s^2) e^(-st_0) sinh^2 (1/2 st_0)

Laplace Transform: $\frac{1}{{s}^{2}}{\left(1-{e}^{-s{t}_{0}}\right)}^{2}=\frac{4}{{s}^{2}}{e}^{-s{t}_{0}}{\mathrm{sinh}}^{2}\left(\frac{1}{2}s{t}_{0}\right)$
Verify for the Laplace transform of
$F\left(t\right)=\left\{\begin{array}{ll}t& 0\le t<{t}_{0}\\ 2{t}_{0}-t& {t}_{0}\le t\le 2{t}_{0}\\ 0& t>2{t}_{0}\end{array}$

I was able to verify everything up to $\frac{1}{{s}^{2}}{\left(1-{e}^{-s{t}_{0}}\right)}^{2}$, but I don't see how $\frac{1}{{s}^{2}}{\left(1-{e}^{-s{t}_{0}}\right)}^{2}=\frac{4}{{s}^{2}}{e}^{-s{t}_{0}}{\mathrm{sinh}}^{2}\left(\frac{1}{2}s{t}_{0}\right)$?
You can still ask an expert for help

## Want to know more about Laplace transform?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

elulamami
$\left(1-{e}^{-x}{\right)}^{2}=4\left(\frac{1-{e}^{-x}}{2}{\right)}^{2}=4{e}^{-x}\left(\frac{{e}^{x/2}-{e}^{-x/2}}{2}{\right)}^{2}=4{e}^{-x}{\mathrm{sinh}}^{2}\left(\frac{x}{2}\right)$
Here $x=s{t}_{0}$