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

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

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

inverse laplace

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

e^{2 t} + e^{2 t} \cdot 2 t + \frac{3 e ^{2 t} t ^{2}}{2}

Solution steps

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

Take the partial fraction of

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

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

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

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

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

= e^{2 t} + e^{2 t} \cdot 2 t + \frac{3 e ^{2 t} t ^{2}}{2}

\int \frac{5 x + 3}{x ^{2} - 9} d x

\int x^{3} sin (6 x) d x

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

\frac{d}{d x} (1 - \frac{1}{1 + x})

\frac{\partial}{\partial x} (\frac{4 x - 4 z}{3 y + 4 z})

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