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

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

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

inverse laplace

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

e^{t} + e^{t} \cdot 4 t + e^{t} \cdot 2 t^{2}

Solution steps

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

Take the partial fraction of

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

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

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

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

L^{- 1} {\frac{4}{( s - 1 ) ^{3}}} : e^{t} \cdot 2 t^{2}

= e^{t} + e^{t} \cdot 4 t + e^{t} \cdot 2 t^{2}

\frac{\partial}{\partial x} (2 x^{2} y - 3)

\int x^{2} lo g_{10} (x) d x

f (x) = \frac{x ^{2} - 3 x + 2}{x ^{2} - 1}

y = 2.8414 x^{2} - 1.2187 x + 0.5399

\frac{d}{d x} ((x^{2} + x + 2)^{\frac{2}{3}})

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