What is inverse oflaplace e^{-s}((s^2+2))/(s^3) ?

The answer to inverse oflaplace e^{-s}((s^2+2))/(s^3) is \H(t-1)(1+(t-1)^2)

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inverse oflaplace e^{-s}((s^2+2))/(s^3)

inverse laplace

e^{- s} \frac{( s ^{2} + 2 )}{s ^{3}}

H (t - 1) (1 + (t - 1)^{2})

Solution steps

L^{- 1} {e^{- s} \frac{( s ^{2} + 2 )}{s ^{3}}}

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 - 1) L^{- 1} {\frac{s ^{2} + 2}{s ^{3}}} (t - 1)

L^{- 1} {\frac{s ^{2} + 2}{s ^{3}}} : H (t) + t^{2}

= H (t - 1) (H (t - 1) + (t - 1)^{2})

H (t - 1) = {01 t < 1 t \geq 1

= {0 1 + (t - 1)^{2} t < 1 t \geq 1

= H (t - 1) (1 + (t - 1)^{2})

\int \frac{1}{y + y ^{3}} d y

\frac{\partial}{\partial x} (1 (x - y) e^{1 y + 3 x^{2}})

f (x) = \frac{64}{( 1 + 4 x ) ^{3}}

y^{^{'}} = cos^{2} (x) cos^{2} (2 y)

\int \frac{x}{x ^{4} + 49} d x

What is inverse oflaplace e^{-s}((s^2+2))/(s^3) ?
The answer to inverse oflaplace e^{-s}((s^2+2))/(s^3) is \H(t-1)(1+(t-1)^2)