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

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

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

inverse laplace

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

\frac{1}{4} H (t) - \frac{1}{4} e^{- 2 t} - \frac{e ^{- 2 t} t}{2}

Solution steps

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

Take the partial fraction of

\frac{1}{s ( s + 2 ) ^{2}} : \frac{1}{4 s} - \frac{1}{4 ( s + 2 )} - \frac{1}{2 ( s + 2 ) ^{2}}

= L^{- 1} {\frac{1}{4 s} - \frac{1}{4 ( s + 2 )} - \frac{1}{2 ( s + 2 ) ^{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}{4 s}} - L^{- 1} {\frac{1}{4 ( s + 2 )}} - L^{- 1} {\frac{1}{2 ( s + 2 ) ^{2}}}

L^{- 1} {\frac{1}{4 s}} : \frac{1}{4} H (t)

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

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

= \frac{1}{4} H (t) - \frac{1}{4} e^{- 2 t} - \frac{e ^{- 2 t} t}{2}

y^{^{''}} = - 9 y

x ln (7 x), 1

\int \frac{4 ln ( x )}{x ^{3}} d x

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

\frac{d}{d y} (y^{- 2})

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