prove theorem 1 Lim t rightarrow infinity est yt 0 Perhaps

Lim t rightarrow infinity e^-st y(t) = 0 Perhaps the the most useful property of the Laplace Transform is that it turns differentiation of f(t) into multiplication by the Laplace variable s of the transformed function F(s), in a sense that is made clear by the following theorem. Theorem 1 Let y(t) be continuously differentiable on (0, infinity), and denote the Laplace Transform of y(t) by Y(s). Then {y\'(t)}(s) = sY(s)-y( 0)

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