What is inverse oflaplace k/((s-4)^3) ?

The answer to inverse oflaplace k/((s-4)^3) is e^{4t}t(2t+1)

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inverse oflaplace k/((s-4)^3)

inverse laplace

\frac{k}{( s - 4 ) ^{3}}

e^{4 t} t (2 t + 1)

Solution steps

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

Apply inverse transform rule:

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

then

L^{- 1} {s F (s)} = \frac{d}{d t} f (t) + f (0)

For

\frac{k}{( s - 4 ) ^{3}} : F (s) = \frac{1}{( s - 4 ) ^{3}}

= \frac{d}{d t} (L^{- 1} {\frac{1}{( s - 4 ) ^{3}}}) + L^{- 1} {\frac{1}{( s - 4 ) ^{3}}} (0)

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

\frac{d}{d t} (\frac{e ^{4 t} t ^{2}}{2}) = e^{4 t} t (2 t + 1)

Evaluate

\frac{e ^{4 t} t ^{2}}{2}

at point

0 = 0

= e^{4 t} t (2 t + 1)

t^{3} e^{2 t}

\frac{d}{d x} (sin^{2} (3 x^{2} - 1))

\frac{3 ( x - 1 )}{4 x x}

t e^{- 2 t} cos (3 t)

\int \frac{1}{k s} d s

What is inverse oflaplace k/((s-4)^3) ?
The answer to inverse oflaplace k/((s-4)^3) is e^{4t}t(2t+1)