What is inverse oflaplace (e^{-4s})/((s-3)^2) ?

The answer to inverse oflaplace (e^{-4s})/((s-3)^2) is \H(t-4)e^{3(t-4)}(t-4)

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inverse oflaplace (e^{-4s})/((s-3)^2)

inverse laplace

\frac{e ^{- 4 s}}{( s - 3 ) ^{2}}

H (t - 4) e^{3 (t - 4)} (t - 4)

Solution steps

L^{- 1} {\frac{e ^{- 4 s}}{( s - 3 ) ^{2}}}

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {e^{- a s} F (s)} = H (t - a) f (t - a)

Where

H (t)

is Heaviside step function

= H (t - 4) L^{- 1} {\frac{1}{( s - 3 ) ^{2}}} (t - 4)

L^{- 1} {\frac{1}{( s - 3 ) ^{2}}} : e^{3 t} t

= H (t - 4) e^{3 (t - 4)} (t - 4)

What is inverse oflaplace (e^{-4s})/((s-3)^2) ?
The answer to inverse oflaplace (e^{-4s})/((s-3)^2) is \H(t-4)e^{3(t-4)}(t-4)