What is inverse oflaplace e^{-2s} 1/(10s+2) ?

The answer to inverse oflaplace e^{-2s} 1/(10s+2) is \H(t-2) 1/10 e^{-(t-2)/5}

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inverse oflaplace e^{-2s} 1/(10s+2)

inverse laplace

e^{- 2 s} \frac{1}{10 s + 2}

H (t - 2) \frac{1}{10} e^{- \frac{t - 2}{5}}

Solution steps

L^{- 1} {e^{- 2 s} \frac{1}{10 s + 2}}

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}{10 s + 2}} (t - 2)

L^{- 1} {\frac{1}{10 s + 2}} : \frac{1}{10} e^{- \frac{t}{5}}

= H (t - 2) \frac{1}{10} e^{- \frac{t - 2}{5}}

\frac{d}{d x} (2 x^{\frac{3}{2}} + 7)

\int \frac{( ln ( x ) ) ^{11}}{x} d x

(arctan (x))^{^{'}}

\frac{d}{d x} (8 e^{x} + \frac{8}{3 x})

\int \frac{5}{3 x ^{2}} d x

What is inverse oflaplace e^{-2s} 1/(10s+2) ?
The answer to inverse oflaplace e^{-2s} 1/(10s+2) is \H(t-2) 1/10 e^{-(t-2)/5}