What is inverse oflaplace 1/((s+5)^2) ?

The answer to inverse oflaplace 1/((s+5)^2) is e^{-5t}t

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inverse oflaplace 1/((s+5)^2)

inverse laplace

\frac{1}{( s + 5 ) ^{2}}

e^{- 5 t} t

Solution steps

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

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 ) ^{2}} : a = - 5, F (s) = \frac{1}{s ^{2}}

= e^{- 5 t} L^{- 1} {\frac{1}{s ^{2}}}

L^{- 1} {\frac{1}{s ^{2}}} : t

= e^{- 5 t} t

\frac{d}{d x} (\frac{2 x ^{5} + 3 x ^{4} - x ^{3} + 2}{x ^{2}})

\int_{0}^{3} 2 x - 1 d x

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

\int \frac{sin ( ln ( 4 x ^{2} ) )}{x} d x

x \to 1 lim (- 2 1 - x + 2)

What is inverse oflaplace 1/((s+5)^2) ?
The answer to inverse oflaplace 1/((s+5)^2) is e^{-5t}t