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

The answer to inverse oflaplace 1/(s^2(s+3)) is -1/9 \H(t)+t/3+1/9 e^{-3t}

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

inverse laplace

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

- \frac{1}{9} H (t) + \frac{t}{3} + \frac{1}{9} e^{- 3 t}

Solution steps

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

Take the partial fraction of

\frac{1}{s ^{2} ( s + 3 )} : - \frac{1}{9 s} + \frac{1}{3 s ^{2}} + \frac{1}{9 ( s + 3 )}

= L^{- 1} {- \frac{1}{9 s} + \frac{1}{3 s ^{2}} + \frac{1}{9 ( s + 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}{9 s}} + L^{- 1} {\frac{1}{3 s ^{2}}} + L^{- 1} {\frac{1}{9 ( s + 3 )}}

L^{- 1} {\frac{1}{9 s}} : \frac{1}{9} H (t)

L^{- 1} {\frac{1}{3 s ^{2}}} : \frac{t}{3}

L^{- 1} {\frac{1}{9 ( s + 3 )}} : \frac{1}{9} e^{- 3 t}

= - \frac{1}{9} H (t) + \frac{t}{3} + \frac{1}{9} e^{- 3 t}

\frac{1}{( s ^{2} + 4 s + 5 )}

n = 0 \sum \infty \frac{1}{4 ^{n}}

x \to 3 lim (f (x) (x - 1))

\frac{\partial}{\partial y} (x - y)

What is inverse oflaplace 1/(s^2(s+3)) ?
The answer to inverse oflaplace 1/(s^2(s+3)) is -1/9 \H(t)+t/3+1/9 e^{-3t}