What is inverse oflaplace {(2s-1)/(s^2(s+1)^3)} ?

The answer to inverse oflaplace {(2s-1)/(s^2(s+1)^3)} is 5\H(t)-t-5e^{-t}-e^{-t}*4t-(3e^{-t}t^2)/2

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inverse oflaplace {(2s-1)/(s^2(s+1)^3)}

inverse laplace

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

5 H (t) - t - 5 e^{- t} - e^{- t} \cdot 4 t - \frac{3 e ^{- t} t ^{2}}{2}

Solution steps

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

Take the partial fraction of

\frac{2 s - 1}{s ^{2} ( s + 1 ) ^{3}} : \frac{5}{s} - \frac{1}{s ^{2}} - \frac{5}{s + 1} - \frac{4}{( s + 1 ) ^{2}} - \frac{3}{( s + 1 ) ^{3}}

= L^{- 1} {\frac{5}{s} - \frac{1}{s ^{2}} - \frac{5}{s + 1} - \frac{4}{( s + 1 ) ^{2}} - \frac{3}{( 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{5}{s}} - L^{- 1} {\frac{1}{s ^{2}}} - L^{- 1} {\frac{5}{s + 1}} - L^{- 1} {\frac{4}{( s + 1 ) ^{2}}} - L^{- 1} {\frac{3}{( s + 1 ) ^{3}}}

L^{- 1} {\frac{5}{s}} : 5 H (t)

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

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

L^{- 1} {\frac{4}{( s + 1 ) ^{2}}} : e^{- t} \cdot 4 t

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

= 5 H (t) - t - 5 e^{- t} - e^{- t} \cdot 4 t - \frac{3 e ^{- t} t ^{2}}{2}

x \to 0 lim (\frac{x}{sin ( 4 ) ( x )})

\int e^{\frac{1}{2} t} d t

7 t^{7} + 9

x^{- m}

\int \frac{1}{6 x - x ^{2}} d x

What is inverse oflaplace {(2s-1)/(s^2(s+1)^3)} ?
The answer to inverse oflaplace {(2s-1)/(s^2(s+1)^3)} is 5\H(t)-t-5e^{-t}-e^{-t}*4t-(3e^{-t}t^2)/2