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

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

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

inverse laplace

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

H (t) - e^{t} + e^{t} t

Solution steps

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

Take the partial fraction of

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

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

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

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

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

= H (t) - e^{t} + e^{t} t

\frac{d}{d x} (\frac{q ^{2} + 45 q + 875}{3}) = 0

f (x) = 8 x - 3 x^{2}, (- 3, - 51)

e^{3}

f (x) = - 3 x

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

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