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

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

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

inverse laplace

\frac{1}{( 5 s ^{2} + s )}

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

\frac{d}{d x} (\frac{x ^{2} + 1}{x - 9})

\frac{\partial}{\partial x} (2 (x + y))

\int (1 + y) d y

\int \frac{1}{x ^{3}} e^{\frac{1}{x}} d x

\int v^{- 2} d v

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