show that
L{f(t)}=f(s) then L{e^(kt)}.f{(t)}=f(s-k) where
k=costant
L=laplace transformation
well, just use the definition.
F(s) = ∫ e^(-st) f(t) dt
multiply that by e^(kt) and you now have
∫ e^(-(s-k)t) f(t) dt
L{f(t)}=f(s) then L{e^(kt)}.f{(t)}=f(s-k) where
k=costant
L=laplace transformation
F(s) = ∫ e^(-st) f(t) dt
multiply that by e^(kt) and you now have
∫ e^(-(s-k)t) f(t) dt