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

The answer to inverse oflaplace (s^2)/((s^2+1/4)) is \delta(t)-1/2 sin(t/2)

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

inverse laplace

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

δ (t) - \frac{1}{2} sin (\frac{t}{2})

Solution steps

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

Take the partial fraction of

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

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

L^{- 1} {1} : δ (t)

L^{- 1} {\frac{1}{4 s ^{2} + 1}} : \frac{1}{2} sin (\frac{t}{2})

= δ (t) - \frac{1}{2} sin (\frac{t}{2})

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

sin (3 x)

y = e^{x} + 1

\frac{d}{d x} (- 2 sin (2 x) + 2 cos (2 x))

\frac{x}{x - 8}

What is inverse oflaplace (s^2)/((s^2+1/4)) ?
The answer to inverse oflaplace (s^2)/((s^2+1/4)) is \delta(t)-1/2 sin(t/2)