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

The answer to inverse oflaplace 1/(s*(s+5)^2) is 1/25 \H(t)-1/25 e^{-5t}-(e^{-5t}t)/5

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

inverse laplace

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

\frac{1}{25} H (t) - \frac{1}{25} e^{- 5 t} - \frac{e ^{- 5 t} t}{5}

Solution steps

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

Take the partial fraction of

\frac{1}{s ( s + 5 ) ^{2}} : \frac{1}{25 s} - \frac{1}{25 ( s + 5 )} - \frac{1}{5 ( s + 5 ) ^{2}}

= L^{- 1} {\frac{1}{25 s} - \frac{1}{25 ( s + 5 )} - \frac{1}{5 ( s + 5 ) ^{2}}}

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= L^{- 1} {\frac{1}{25 s}} - L^{- 1} {\frac{1}{25 ( s + 5 )}} - L^{- 1} {\frac{1}{5 ( s + 5 ) ^{2}}}

L^{- 1} {\frac{1}{25 s}} : \frac{1}{25} H (t)

L^{- 1} {\frac{1}{25 ( s + 5 )}} : \frac{1}{25} e^{- 5 t}

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

= \frac{1}{25} H (t) - \frac{1}{25} e^{- 5 t} - \frac{e ^{- 5 t} t}{5}

What is inverse oflaplace 1/(s*(s+5)^2) ?
The answer to inverse oflaplace 1/(s*(s+5)^2) is 1/25 \H(t)-1/25 e^{-5t}-(e^{-5t}t)/5