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

The answer to inverse oflaplace s/((2s+1)((s+2)^2-1)) is 1/2 e^{-t}-1/5 e^{-t/2}-3/10 e^{-3t}

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

inverse laplace

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

\frac{1}{2} e^{- t} - \frac{1}{5} e^{- \frac{t}{2}} - \frac{3}{10} e^{- 3 t}

Solution steps

L^{- 1} ⎩ ⎨ ⎧ \frac{s}{( 2 s + 1 ) ( ( s + 2 ) ^{2} - 1 )} ⎭ ⎬ ⎫

Take the partial fraction of

\frac{s}{( 2 s + 1 ) ( ( s + 2 ) ^{2} - 1 )} : \frac{1}{2 ( s + 1 )} - \frac{2}{5 ( 2 s + 1 )} - \frac{3}{10 ( s + 3 )}

= L^{- 1} {\frac{1}{2 ( s + 1 )} - \frac{2}{5 ( 2 s + 1 )} - \frac{3}{10 ( 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}{2 ( s + 1 )}} - L^{- 1} {\frac{2}{5 ( 2 s + 1 )}} - L^{- 1} {\frac{3}{10 ( s + 3 )}}

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

L^{- 1} {\frac{2}{5 ( 2 s + 1 )}} : \frac{1}{5} e^{- \frac{t}{2}}

L^{- 1} {\frac{3}{10 ( s + 3 )}} : \frac{3}{10} e^{- 3 t}

= \frac{1}{2} e^{- t} - \frac{1}{5} e^{- \frac{t}{2}} - \frac{3}{10} e^{- 3 t}

f (x) = x^{3} - x, (- 1, 0)

\int \frac{1}{3 x + 8} d x

\int x^{2} + 16 d x

g (x) = (\frac{3 x ^{2} - 2}{2 x + 3})^{3}

5 x^{2} - 2 x

What is inverse oflaplace s/((2s+1)((s+2)^2-1)) ?
The answer to inverse oflaplace s/((2s+1)((s+2)^2-1)) is 1/2 e^{-t}-1/5 e^{-t/2}-3/10 e^{-3t}