Question Number 143523 by lapache last updated on 15/Jun/21
$${Determiner}\:{l}'{origine}\:{de}\:{laplace} \\ $$$$\mathrm{1}−{F}\left({p}\right)=\frac{{p}}{\left({p}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$
Answered by Dwaipayan Shikari last updated on 15/Jun/21
$${F}\left({p}\right)=\mathrm{1}−\frac{{p}}{\left({p}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$\mathscr{L}^{−\mathrm{1}} \left({F}\left({p}\right)\right)=\mathscr{L}^{−\mathrm{1}} \left(\mathrm{1}−\frac{{p}}{\left({p}+\mathrm{2}\right)^{\mathrm{2}} }\right) \\ $$$${f}\left({t}\right)={p}−\mathscr{L}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{p}+\mathrm{2}}\right)+\mathscr{L}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\left({p}+\mathrm{2}\right)^{\mathrm{2}} }\right) \\ $$$$={pt}^{\mathrm{0}} −{e}^{−\mathrm{2}{t}} −\mathrm{2}{te}^{−\mathrm{2}{t}} \\ $$