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

The answer to inverse oflaplace (s-1)/(s^2(s+3)) is 4/9 \H(t)-t/3-4/9 e^{-3t}

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

inverse laplace

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

\frac{4}{9} H (t) - \frac{t}{3} - \frac{4}{9} e^{- 3 t}

Solution steps

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

Take the partial fraction of

\frac{s - 1}{s ^{2} ( s + 3 )} : \frac{4}{9 s} - \frac{1}{3 s ^{2}} - \frac{4}{9 ( s + 3 )}

= L^{- 1} {\frac{4}{9 s} - \frac{1}{3 s ^{2}} - \frac{4}{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{4}{9 s}} - L^{- 1} {\frac{1}{3 s ^{2}}} - L^{- 1} {\frac{4}{9 ( s + 3 )}}

L^{- 1} {\frac{4}{9 s}} : \frac{4}{9} H (t)

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

L^{- 1} {\frac{4}{9 ( s + 3 )}} : \frac{4}{9} e^{- 3 t}

= \frac{4}{9} H (t) - \frac{t}{3} - \frac{4}{9} e^{- 3 t}

t \to a lim (t)

x^{2} - 2 x

\frac{\partial}{\partial y} (\frac{7 y}{x})

7 x^{2} - 8 x + 3

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

What is inverse oflaplace (s-1)/(s^2(s+3)) ?
The answer to inverse oflaplace (s-1)/(s^2(s+3)) is 4/9 \H(t)-t/3-4/9 e^{-3t}