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Question Number 123294 by roniee last updated on 24/Nov/20

$$L(y+\int ydt)=L(1-e^{-t})$$

Answered by Olaf last updated on 24/Nov/20

$$\mathcal{L}(y+\int ydt)=\mathcal{L}(1-e^{-t})$$$$\mathcal{L}\left(p\right)+\frac{\mathcal{L}\left(p\right)}{p}=\frac{1}{p}-\frac{1}{p+1}$$$$\mathcal{L}\left(p\right)\left(\frac{p+1}{p}\right)=\frac{1}{p(p+1)}$$$$\mathcal{L}\left(p\right)=\frac{1}{{(p+1)}^2}$$$$\mathrm{With}\mathrm{the}\mathrm{Laplace}\mathrm{tables}:$$$$y\left(t\right)={te}^{-t}$$

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