Problem 예제.

Multiple Choice • Views 728 • Comments 0 • Last Updated at 1 month ago

$\mathcal{L}^{-1} \left[ \dfrac{s}{(s+1)^2} \right]$ 는?

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$e^{t}-te^{-t}$

$e^{-t}+te^{-t}$

$e^{-t}-te^{-t}$

$e^{-t}+2te^{-t}$

Solution

$\dfrac{s}{(s+1)^2} = \dfrac{s+1}{(s+1)^2}-\dfrac{1}{(s+1)^2}= \dfrac{1}{s+1}-\dfrac{1}{(s+1)^2}$

$\mathcal{L}^{-1} \left[ \dfrac{1}{s+1}-\dfrac{1}{(s+1)^2} \right]= e^{-t}-te^{-t}$

Correct answer: 3