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

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

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

inverse laplace

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

e^{t} + e^{t} t

Solution steps

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

Take the partial fraction of

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

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

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

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

= e^{t} + e^{t} t

\frac{\partial}{\partial x} (e^{e^{x}})

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\int 8 x - 9 d x

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\int_{1}^{4} \frac{1}{w ^{2}} d w

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