What is inverse oflaplace ((s+1))/(s(s^2+4s+4)) ?

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

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inverse oflaplace ((s+1))/(s(s^2+4s+4))

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

\frac{d}{d x} (\frac{x ^{2}}{x + y})

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

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

\frac{d y}{d x}, y^{2} = x

\frac{\partial}{\partial y} (4 x^{2} y^{2})

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