What is inverse oflaplace 1/((s-5)^4) ?

The answer to inverse oflaplace 1/((s-5)^4) is (e^{5t}t^3)/6

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inverse oflaplace 1/((s-5)^4)

inverse laplace

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

\frac{e ^{5 t} t ^{3}}{6}

Solution steps

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

Apply inverse transform rule:

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

then

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

For

\frac{1}{( s - 5 ) ^{4}} : a = 5, F (s) = \frac{1}{s ^{4}}

= e^{5 t} L^{- 1} {\frac{1}{s ^{4}}}

L^{- 1} {\frac{1}{s ^{4}}} : \frac{t ^{3}}{6}

= \frac{e ^{5 t} t ^{3}}{6}

y^{^{'}} + tan (2 \cdot x) y = 0

\int 2 x (x^{2} - 2) d x

f (x) = (x + 2) 2 x - 3, at x = 2

n = 1 \sum \infty (\frac{π}{5})^{n}

y = \frac{1}{x}, y = x, x = 3

What is inverse oflaplace 1/((s-5)^4) ?
The answer to inverse oflaplace 1/((s-5)^4) is (e^{5t}t^3)/6