What is inverse oflaplace (e^{-0.1s})/(s+3) ?

The answer to inverse oflaplace (e^{-0.1s})/(s+3) is \H(t-0.1)e^{-3(t-0.1)}

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inverse oflaplace (e^{-0.1s})/(s+3)

inverse laplace

\frac{e ^{- 0.1 s}}{s + 3}

H (t - 0.1) e^{- 3 (t - 0.1)}

Solution steps

L^{- 1} {\frac{e ^{- 0.1 s}}{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 - 0.1) L^{- 1} {\frac{1}{s + 3}} (t - 0.1)

L^{- 1} {\frac{1}{s + 3}} : e^{- 3 t}

= H (t - 0.1) e^{- 3 (t - 0.1)}

\int \frac{sin ( x )}{1 + sin ( x )} d x

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

x \to \infty lim (\frac{2 x ^{2}}{x + 3})

\frac{\partial}{\partial x} (e^{a x} cos (b x + y))

\frac{d}{d x} (\frac{1 + x}{1 + y})

What is inverse oflaplace (e^{-0.1s})/(s+3) ?
The answer to inverse oflaplace (e^{-0.1s})/(s+3) is \H(t-0.1)e^{-3(t-0.1)}