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

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

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

inverse laplace

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

\frac{1}{9} H (t) + \frac{2 t}{3} - \frac{1}{9} e^{- 3 t}

Solution steps

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

Take the partial fraction of

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

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

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

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

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

= \frac{1}{9} H (t) + \frac{2 t}{3} - \frac{1}{9} e^{- 3 t}

y^{^{'''}} - 4 y^{^{''}} + y^{^{'}} + 6 y = 0

x^{3} - 10 x, 6 x

\frac{0}{s ^{2} + 2 s + 3}

ln (x + x^{2} + a^{2})

x \to 4 lim (arcsin (\frac{x}{4}))

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