Question: Prove according to convolution property F(s) G(s) = F(s) L (g(t)) = F($) [e-stg(t) dt

Prove according to convolution property

F(s) G(s) = F(s) L (g(t)) = F($) [e-stg(t) dt

F(s) G(s) = F(s) L (g(t)) = F($) [e-stg(t) dt

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