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

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

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inverse oflaplace (e^{-2s})/(s(s^2+1))

inverse laplace

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

H (t - 2) (1 - cos (t - 2))

Solution steps

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

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {e^{- a s} F (s)} = H (t - a) f (t - a)

Where

H (t)

is Heaviside step function

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

L^{- 1} {\frac{1}{s ( s ^{2} + 1 )}} : H (t) - cos (t)

= H (t - 2) (H (t - 2) - cos (t - 2))

H (t - 2) = {01 t < 2 t \geq 2

= {0 1 - cos (t - 2) t < 2 t \geq 2

= H (t - 2) (1 - cos (t - 2))

x^{2} y^{^{''}} + 0 y^{^{'}} - 2 y = x^{2}

y = \frac{x - 2}{x ^{2}}

f (x) = (x^{2} + 3)^{6}

x \to 3 lim (x^{2} - 1)

\int x (5 + 2 x)^{4} d x

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