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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

L^{- 1} {\frac{2}{2 s + 1}} : e^{- \frac{t}{2}}

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

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

\int \frac{sin ( 1 + x )}{x} d x

\int_{x}^{5} sin (t^{3}) d t

\int tan^{4} (9 x) d x

f (x) = \frac{2}{x} - \frac{1}{x ^{2}}

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