What is inverse oflaplace-1.621s-0.462945095 ?

The answer to inverse oflaplace-1.621s-0.462945095 is -1.621\delta'(t)-0.462945095\delta(t)

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inverse oflaplace-1.621s-0.462945095

inverse laplace

- 1.621 s - 0.462945095

- 1.621 δ' (t) - 0.462945095 δ (t)

Solution steps

L^{- 1} {- 1.621 s - 0.462945095}

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)}

= - 1.621 L^{- 1} {s} - L^{- 1} {0.462945095}

L^{- 1} {s} : δ^{'} (t)

L^{- 1} {0.462945095} : 0.462945095 δ (t)

= - 1.621 δ^{^{'}} (t) - 0.462945095 δ (t)

\frac{\partial}{\partial φ} (ρ sin (φ) cos (θ))

y^{^{''}} - 4 y^{^{'}} + 5 y = 2020 e^{- t}

5 2 x^{3} + 8 x (2.2)

(- 7.1) (- 7.9)

(x - 2)^{5}

What is inverse oflaplace-1.621s-0.462945095 ?
The answer to inverse oflaplace-1.621s-0.462945095 is -1.621\delta'(t)-0.462945095\delta(t)