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

L(y+∫ydt)=L(1−e^(−t) )

$${L}\left({y}+\int{ydt}\right)={L}\left(\mathrm{1}−{e}^{−{t}} \right) \\ $$

Answered by Olaf last updated on 24/Nov/20

L(y+∫ydt) = L(1−e^(−t) )  L(p)+((L(p))/p) = (1/p)−(1/(p+1))  L(p)(((p+1)/p)) = (1/(p(p+1)))  L(p) = (1/((p+1)^2 ))  With the Laplace tables :  y(t) = te^(−t)

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

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