What is inverse oflaplace 1/(s^2+s/4) ?

The answer to inverse oflaplace 1/(s^2+s/4) is 4\H(t)-4e^{-t/4}

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inverse oflaplace 1/(s^2+s/4)

inverse laplace

\frac{1}{s ^{2} + \frac{s}{4}}

4 H (t) - 4 e^{- \frac{t}{4}}

Solution steps

L^{- 1} {\frac{1}{s ^{2} + \frac{s}{4}}}

Take the partial fraction of

\frac{1}{s ^{2} + \frac{s}{4}} : \frac{4}{s} - \frac{16}{4 s + 1}

= L^{- 1} {\frac{4}{s} - \frac{16}{4 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{4}{s}} - L^{- 1} {\frac{16}{4 s + 1}}

L^{- 1} {\frac{4}{s}} : 4 H (t)

L^{- 1} {\frac{16}{4 s + 1}} : 4 e^{- \frac{t}{4}}

= 4 H (t) - 4 e^{- \frac{t}{4}}

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

\int 3 x^{2} + 6 d x

x \to 5 lim (ln (x - 5))

\frac{d}{d x} (e + L (x))

What is inverse oflaplace 1/(s^2+s/4) ?
The answer to inverse oflaplace 1/(s^2+s/4) is 4\H(t)-4e^{-t/4}