What is inverse oflaplace (50)/(100s+0.01s^2) ?

The answer to inverse oflaplace (50)/(100s+0.01s^2) is 1/2 \H(t)-0.5e^{-10000t}

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inverse oflaplace (50)/(100s+0.01s^2)

inverse laplace

\frac{50}{100 s + 0.01 s ^{2}}

\frac{1}{2} H (t) - 0.5 e^{- 10000 t}

Solution steps

L^{- 1} {\frac{50}{100 s + 0.01 s ^{2}}}

Take the partial fraction of

\frac{50}{100 s + 0.01 s ^{2}} : \frac{1}{2 s} - \frac{0.005}{0.01 s + 100}

= L^{- 1} {\frac{1}{2 s} - \frac{0.005}{0.01 s + 100}}

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{1}{2 s}} - L^{- 1} {\frac{0.005}{0.01 s + 100}}

L^{- 1} {\frac{1}{2 s}} : \frac{1}{2} H (t)

L^{- 1} {\frac{0.005}{0.01 s + 100}} : 0.5 e^{- 10000 t}

= \frac{1}{2} H (t) - 0.5 e^{- 10000 t}

\int \frac{4}{16 - x ^{2}} d x

t^{2} y^{^{''}} + 19 t y^{^{'}} + 82 y = 0

(9, - 7), (4, - 8)

\int_{0}^{9} \frac{1}{81 - x ^{2}} d x

\frac{\partial}{\partial x} (6 x y - 5 x^{2} y)

What is inverse oflaplace (50)/(100s+0.01s^2) ?
The answer to inverse oflaplace (50)/(100s+0.01s^2) is 1/2 \H(t)-0.5e^{-10000t}