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

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

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

inverse laplace

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

H (t - 2)

Solution steps

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

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

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

= H (t - 2)

\int \frac{1}{25 e ^{- 7 x} + e ^{7 x}} d x

\frac{\partial ^{2}}{\partial x \partial y} (ln (x^{2} + y^{3}))

y^{^{''}} + 36 y = 0, y (0) = 0, y^{^{'}} (0) = 1

2 - 3 x

\int 4 - y^{2} d y

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