What is inverse oflaplace (-0.1s)/(s^2-9.8) ?

The answer to inverse oflaplace (-0.1s)/(s^2-9.8) is -0.05e^{-3.13049…t}-0.05e^{3.13049…t}

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inverse oflaplace (-0.1s)/(s^2-9.8)

inverse laplace

\frac{- 0.1 s}{s ^{2} - 9.8}

- 0.05 e^{- 3.13049 \dots t} - 0.05 e^{3.13049 \dots t}

Solution steps

L^{- 1} {\frac{- 0.1 s}{s ^{2} - 9.8}}

Take the partial fraction of

\frac{- 0.1 s}{s ^{2} - 9.8} : - \frac{0.05}{s + 3.13049 \dots} - \frac{0.05}{s - 3.13049 \dots}

= L^{- 1} {- \frac{0.05}{s + 3.13049 \dots} - \frac{0.05}{s - 3.13049 \dots}}

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= - L^{- 1} {\frac{0.05}{s + 3.13049 \dots}} - L^{- 1} {\frac{0.05}{s - 3.13049 \dots}}

L^{- 1} {\frac{0.05}{s + 3.13049 \dots}} : 0.05 e^{- 3.13049 \dots t}

L^{- 1} {\frac{0.05}{s - 3.13049 \dots}} : 0.05 e^{3.13049 \dots t}

= - 0.05 e^{- 3.13049 \dots t} - 0.05 e^{3.13049 \dots t}

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

\int_{0}^{- 0.5} 150 x d x

(x^{2} + 5)^{2} (x - 5)

\int 6 x^{2} + 16 x - 10 d x

\frac{\partial}{\partial y} (sin (π (5 x - 2 y)))

What is inverse oflaplace (-0.1s)/(s^2-9.8) ?
The answer to inverse oflaplace (-0.1s)/(s^2-9.8) is -0.05e^{-3.13049…t}-0.05e^{3.13049…t}