What is inverse oflaplace (2s^2+5s+5)/((s+1)^3) ?

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

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inverse oflaplace (2s^2+5s+5)/((s+1)^3)

inverse laplace

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

2 e^{- t} + e^{- t} t + e^{- t} t^{2}

Solution steps

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

Take the partial fraction of

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

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

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

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

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

= 2 e^{- t} + e^{- t} t + e^{- t} t^{2}

(\frac{5}{x}), at x = 4

\int 3 e^{x} + 5 sin (x) d x

y = (x^{3} - 25 x)^{8}, (5, 0)

x \to 4 lim (x^{3} + 4 x)

\frac{d y}{d x}, x = cos (2 y)

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