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}

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

n = 0 \sum \infty 1^{2 n}

\frac{\partial}{\partial z} (x cos (z))

\frac{d y}{d x} = y (\frac{1}{x} - y)

\frac{\partial}{\partial x} (3 y - 7)

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}