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

The answer to inverse oflaplace (s^2)/(s+1) is \delta'(t)-\delta(t)+e^{-t}

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

inverse laplace

\frac{s ^{2}}{s + 1}

δ' (t) - δ (t) + e^{- t}

Solution steps

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

Take the partial fraction of

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

= L^{- 1} {s - 1 + \frac{1}{s + 1}}

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

L^{- 1} {s} : δ^{'} (t)

L^{- 1} {1} : δ (t)

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

= δ^{^{'}} (t) - δ (t) + e^{- t}

x y^{^{'}} = y + 2 x^{2} sin (x), y (π) = 0

n = 1 \sum \infty \frac{e ^{n}}{n ^{2}}

\frac{\partial}{\partial x} (2 x^{3} + 3 y^{3} - 12 x y)

\int \frac{1}{( x ^{2} - 6 x ) ^{\frac{3}{2}}} d x

\frac{\partial}{\partial x} (x y E (x, y) (x, y)^{\frac{x}{y}})

What is inverse oflaplace (s^2)/(s+1) ?
The answer to inverse oflaplace (s^2)/(s+1) is \delta'(t)-\delta(t)+e^{-t}