What is inverse oflaplace (0.1*m)/(s+0.1*m) ?

The answer to inverse oflaplace (0.1*m)/(s+0.1*m) is -0.01me^{-0.1mt}+0.1

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inverse oflaplace (0.1m)/(s+0.1m)

inverse laplace

\frac{0.1 \cdot m}{s + 0.1 \cdot m}

- 0.01 m e^{- 0.1 m t} + 0.1

Solution steps

L^{- 1} {\frac{0.1 m}{s + 0.1 m}}

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {s F (s)} = \frac{d}{d t} f (t) + f (0)

For

\frac{0.1 m}{s + 0.1 m} : F (s) = \frac{0.1}{s + 0.1 m}

= \frac{d}{d t} (L^{- 1} {\frac{0.1}{s + 0.1 m}}) + L^{- 1} {\frac{0.1}{s + 0.1 m}} (0)

L^{- 1} {\frac{0.1}{s + 0.1 m}} : 0.1 e^{- 0.1 m t}

\frac{d}{d t} (0.1 e^{- 0.1 m t}) = - 0.01 m e^{- 0.1 m t}

Evaluate

0.1 e^{- 0.1 m t}

at point

0 = 0.1

= - 0.01 m e^{- 0.1 m t} + 0.1

\frac{x}{x + 4}

\frac{1}{s + 1} - s - 1

y = x^{5}, (- 1, - 1)

\frac{d}{d x} (\frac{1}{2 x} + \frac{1}{4 x ^{\frac{3}{4}}})

\frac{\partial}{\partial y} (\frac{3 x - y}{5 x y + 9})

What is inverse oflaplace (0.1*m)/(s+0.1*m) ?
The answer to inverse oflaplace (0.1*m)/(s+0.1*m) is -0.01me^{-0.1mt}+0.1