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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

\frac{d}{d x} ((lo g_{10} (x))^{2})

\frac{d}{d x} (arctan (3 x^{2} - 1))

y^{^{''}} + 3 y = 5

y^{^{''}} + y = \frac{1}{2 k ^{2}} (1 + cos (2 t))

\frac{d}{d x} ((x^{5} + 1)^{4})

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