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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

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

\int \frac{8 x ^{2} - 4 x + 7}{4 x ^{3} + x ^{2} + 4 x + 1} d x

f (x) = - x^{2} - 4 x + 2, at x = - 1

\frac{d}{d x} (\frac{x ^{2} - 36}{x + 6})

\frac{d}{d x} (3^{x + 10})

f (x) = 2 sin (2 x)

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