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

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

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

inverse laplace

\frac{e ^{- s}}{s ^{2} + s}

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

Solution steps

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

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

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

= H (t - 1) (H (t - 1) - e^{- (t - 1)})

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

= {0 1 - e^{- t + 1} t < 1 t \geq 1

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

(sin (x^{3}))^{^{'}}

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

f (x) = x^{3} - 7 x + 6, at x = - 1

\int \frac{1}{( x ^{3} + x )} d x

\int \frac{x - 4}{( x + 2 ) ^{2} + 16} d x

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