What is inverse oflaplace (2e^{-s})/((s+1/5)) ?

The answer to inverse oflaplace (2e^{-s})/((s+1/5)) is \H(t-1)*2e^{-(t-1)/5}

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inverse oflaplace (2e^{-s})/((s+1/5))

inverse laplace

\frac{2 e ^{- s}}{( s + \frac{1}{5} )}

H (t - 1) \cdot 2 e^{- \frac{t - 1}{5}}

Solution steps

L^{- 1} {\frac{2 e ^{- s}}{( s + \frac{1}{5} )}}

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{2}{s + \frac{1}{5}}} (t - 1)

L^{- 1} {\frac{2}{s + \frac{1}{5}}} : 2 e^{- \frac{t}{5}}

= H (t - 1) \cdot 2 e^{- \frac{t - 1}{5}}

y^{^{''}} - 5 y^{^{'}} - 14 y = 0

\frac{\partial}{\partial x} (6)

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

\frac{d}{d x} (9 x - 2)

\int 2 e^{x} 5 + e^{x} d x

What is inverse oflaplace (2e^{-s})/((s+1/5)) ?
The answer to inverse oflaplace (2e^{-s})/((s+1/5)) is \H(t-1)*2e^{-(t-1)/5}