What is inverse oflaplace 1/(5s^2+0.2s) ?

The answer to inverse oflaplace 1/(5s^2+0.2s) is 5\H(t)-5e^{-0.04t}

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inverse oflaplace 1/(5s^2+0.2s)

inverse laplace

\frac{1}{5 s ^{2} + 0.2 s}

5 H (t) - 5 e^{- 0.04 t}

Solution steps

L^{- 1} {\frac{1}{5 s ^{2} + 0.2 s}}

Take the partial fraction of

\frac{1}{5 s ^{2} + 0.2 s} : \frac{5}{s} - \frac{25}{5 s + 0.2}

= L^{- 1} {\frac{5}{s} - \frac{25}{5 s + 0.2}}

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{5}{s}} - L^{- 1} {\frac{25}{5 s + 0.2}}

L^{- 1} {\frac{5}{s}} : 5 H (t)

L^{- 1} {\frac{25}{5 s + 0.2}} : 5 e^{- 0.04 t}

= 5 H (t) - 5 e^{- 0.04 t}

\frac{\partial}{\partial y} (e^{5 y} cos (- 5 x))

\int e^{x} sin (4 x) d x

\frac{d}{d x} (lo g_{3} (4 x + 1))

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

x \to \frac{π}{8} lim (tan (24 x))

What is inverse oflaplace 1/(5s^2+0.2s) ?
The answer to inverse oflaplace 1/(5s^2+0.2s) is 5\H(t)-5e^{-0.04t}