What is inverse oflaplace (e^{-7s})/((s-3)^3) ?

The answer to inverse oflaplace (e^{-7s})/((s-3)^3) is (e^{3(t-7)}\H(t-7)(t-7)^2)/2

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inverse oflaplace (e^{-7s})/((s-3)^3)

inverse laplace

\frac{e ^{- 7 s}}{( s - 3 ) ^{3}}

\frac{e ^{3 (t - 7)} H ( t - 7 ) ( t - 7 ) ^{2}}{2}

Solution steps

L^{- 1} {\frac{e ^{- 7 s}}{( s - 3 ) ^{3}}}

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 - 7) L^{- 1} {\frac{1}{( s - 3 ) ^{3}}} (t - 7)

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

= H (t - 7) \frac{e ^{3 (t - 7)} ( t - 7 ) ^{2}}{2}

= \frac{e ^{3 (t - 7)} H ( t - 7 ) ( t - 7 ) ^{2}}{2}

\int \frac{1}{x ^{4} ( x ^{3} + 1 ) ^{2}} d x

\int x \cdot x^{2} + 3 d x

\int x^{6} (x^{7} - 3)^{\frac{3}{4}} d x

x \to 0 + lim ((ln (\frac{1}{x}))^{x})

\frac{d}{d x} (10 cos (\frac{3}{10} x))

What is inverse oflaplace (e^{-7s})/((s-3)^3) ?
The answer to inverse oflaplace (e^{-7s})/((s-3)^3) is (e^{3(t-7)}\H(t-7)(t-7)^2)/2