What is inverse oflaplace (10e^{-s})/(5s+1) ?

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

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inverse oflaplace (10e^{-s})/(5s+1)

inverse laplace

\frac{10 e ^{- s}}{5 s + 1}

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

Solution steps

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

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{10}{5 s + 1}} (t - 1)

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

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

\frac{d}{d x} (a (x - sin (x)))

t an g e n t y = 4 x - x^{2} (1.3)

y^{^{'}} = x^{2} + x, y (1) = 3

\int sin^{3} (6 x) cos^{2} (6 x) d x

\int_{0}^{3} π (\frac{1}{x + 5})^{2} d x

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