What is inverse oflaplace (s+1)/(s^2(s+2)^3) ?

The answer to inverse oflaplace (s+1)/(s^2(s+2)^3) is -1/16 \H(t)+t/8+1/16 e^{-2t}-(e^{-2t}t^2)/8

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inverse oflaplace (s+1)/(s^2(s+2)^3)

inverse laplace

\frac{s + 1}{s ^{2} ( s + 2 ) ^{3}}

- \frac{1}{16} H (t) + \frac{t}{8} + \frac{1}{16} e^{- 2 t} - \frac{e ^{- 2 t} t ^{2}}{8}

Solution steps

L^{- 1} {\frac{s + 1}{s ^{2} ( s + 2 ) ^{3}}}

Take the partial fraction of

\frac{s + 1}{s ^{2} ( s + 2 ) ^{3}} : - \frac{1}{16 s} + \frac{1}{8 s ^{2}} + \frac{1}{16 ( s + 2 )} - \frac{1}{4 ( s + 2 ) ^{3}}

= L^{- 1} {- \frac{1}{16 s} + \frac{1}{8 s ^{2}} + \frac{1}{16 ( s + 2 )} - \frac{1}{4 ( s + 2 ) ^{3}}}

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}{16 s}} + L^{- 1} {\frac{1}{8 s ^{2}}} + L^{- 1} {\frac{1}{16 ( s + 2 )}} - L^{- 1} {\frac{1}{4 ( s + 2 ) ^{3}}}

L^{- 1} {\frac{1}{16 s}} : \frac{1}{16} H (t)

L^{- 1} {\frac{1}{8 s ^{2}}} : \frac{t}{8}

L^{- 1} {\frac{1}{16 ( s + 2 )}} : \frac{1}{16} e^{- 2 t}

L^{- 1} {\frac{1}{4 ( s + 2 ) ^{3}}} : \frac{e ^{- 2 t} t ^{2}}{8}

= - \frac{1}{16} H (t) + \frac{t}{8} + \frac{1}{16} e^{- 2 t} - \frac{e ^{- 2 t} t ^{2}}{8}

\int 5 t d t

7 sin (\frac{3}{2} t)

y^{^{''}} - y^{^{'}} + 5 y = 0

t e^{- 2 t}

\frac{\partial}{\partial x} (\frac{y}{x + 1})

What is inverse oflaplace (s+1)/(s^2(s+2)^3) ?
The answer to inverse oflaplace (s+1)/(s^2(s+2)^3) is -1/16 \H(t)+t/8+1/16 e^{-2t}-(e^{-2t}t^2)/8