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

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

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

inverse laplace

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

t - H (t) + e^{2 t}

Solution steps

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

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

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

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

= t - H (t) + e^{2 t}

\int \frac{x ^{2} + 2 x}{x ^{2} + 2 x + 1} d x

\frac{d}{d x} (x - \frac{3}{x})

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

y^{^{''}} - 9 y^{^{'}} + 18 y = 0

\frac{x}{y ^{13}} + B e^{y}

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