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Question Number 143523 by lapache last updated on 15/Jun/21

$$Determiner{l}^{\prime }originedelaplace$$$$1-F\left(p\right)=\frac{p}{{(p+2)}^2}$$

Answered by Dwaipayan Shikari last updated on 15/Jun/21

$$F\left(p\right)=1-\frac{p}{{(p+2)}^2}$$$${\mathcal{L}}^{-1}\left(F\left(p\right)\right)={\mathcal{L}}^{-1}(1-\frac{p}{{(p+2)}^2})$$$$f\left(t\right)=p-{\mathcal{L}}^{-1}\left(\frac{1}{p+2}\right)+{\mathcal{L}}^{-1}\left(\frac{2}{{(p+2)}^2}\right)$$$$={pt}^0-e^{-2t}-2{te}^{-2t}$$

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