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

f (x) = - 3 x

im pl i c i t \frac{d y}{d x}, 25 x = y^{4}

a re a 6 x^{2}, x^{2} + 9

t an g e n t f (x) = 3 x^{2} - 2 x

x \to a + lim (\frac{cos ( x ) ln ( x - a )}{ln ( e ^{x} - e ^{a} )})

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