What is inverse oflaplace 8/((s+1)^2(s+2)) ?

The answer to inverse oflaplace 8/((s+1)^2(s+2)) is -8e^{-t}+e^{-t}*8t+8e^{-2t}

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inverse oflaplace 8/((s+1)^2(s+2))

inverse laplace

\frac{8}{( s + 1 ) ^{2} ( s + 2 )}

- 8 e^{- t} + e^{- t} \cdot 8 t + 8 e^{- 2 t}

Solution steps

L^{- 1} {\frac{8}{( s + 1 ) ^{2} ( s + 2 )}}

Take the partial fraction of

\frac{8}{( s + 1 ) ^{2} ( s + 2 )} : - \frac{8}{s + 1} + \frac{8}{( s + 1 ) ^{2}} + \frac{8}{s + 2}

= L^{- 1} {- \frac{8}{s + 1} + \frac{8}{( s + 1 ) ^{2}} + \frac{8}{s + 2}}

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{8}{s + 1}} + L^{- 1} {\frac{8}{( s + 1 ) ^{2}}} + L^{- 1} {\frac{8}{s + 2}}

L^{- 1} {\frac{8}{s + 1}} : 8 e^{- t}

L^{- 1} {\frac{8}{( s + 1 ) ^{2}}} : e^{- t} \cdot 8 t

L^{- 1} {\frac{8}{s + 2}} : 8 e^{- 2 t}

= - 8 e^{- t} + e^{- t} \cdot 8 t + 8 e^{- 2 t}

\int \frac{1}{( x - 4 ) ( x - 1 )} d x

\int_{2}^{3} \frac{7}{3 - x} d x

\frac{d}{d x} (\frac{( x ^{2} - 3 x - 4 )}{x})

x \to 0 lim (e^{a x})

\frac{t}{2 t - 9}

What is inverse oflaplace 8/((s+1)^2(s+2)) ?
The answer to inverse oflaplace 8/((s+1)^2(s+2)) is -8e^{-t}+e^{-t}*8t+8e^{-2t}