What is inverse oflaplace 3/(s(s-1)^3) ?

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

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inverse oflaplace 3/(s(s-1)^3)

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

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

f (x) = lo g_{10} (1 - x)

\int 30 x d x

\frac{d}{d x} (a - b (x - c)^{\frac{1}{3}})

\int (x + 1) e^{3 x} d x

\frac{d}{d x} (\frac{1}{2} tan^{2} (x))

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