What is inverse oflaplace (e^{-5s})/((s-2)^4) ?

The answer to inverse oflaplace (e^{-5s})/((s-2)^4) is (e^{2(t-5)}\H(t-5)(t-5)^3)/6

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inverse oflaplace (e^{-5s})/((s-2)^4)

inverse laplace

\frac{e ^{- 5 s}}{( s - 2 ) ^{4}}

\frac{e ^{2 (t - 5)} H ( t - 5 ) ( t - 5 ) ^{3}}{6}

Solution steps

L^{- 1} {\frac{e ^{- 5 s}}{( s - 2 ) ^{4}}}

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 - 5) L^{- 1} {\frac{1}{( s - 2 ) ^{4}}} (t - 5)

L^{- 1} {\frac{1}{( s - 2 ) ^{4}}} : \frac{e ^{2 t} t ^{3}}{6}

= H (t - 5) \frac{e ^{2 (t - 5)} ( t - 5 ) ^{3}}{6}

= \frac{e ^{2 (t - 5)} H ( t - 5 ) ( t - 5 ) ^{3}}{6}

\int x^{2} + 3 x + 1 d x

\int 2 y d y

\int x sin^{2} (7 x) d x

f (x) = \frac{2}{x}, (3, \frac{2}{3})

f (x) = \frac{x + 6}{x - 4}

What is inverse oflaplace (e^{-5s})/((s-2)^4) ?
The answer to inverse oflaplace (e^{-5s})/((s-2)^4) is (e^{2(t-5)}\H(t-5)(t-5)^3)/6