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

The answer to inverse oflaplace 1/(s^2(s^2+4)) is t/4-1/8 sin(2t)

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

inverse laplace

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

\frac{t}{4} - \frac{1}{8} sin (2 t)

Solution steps

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

Take the partial fraction of

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

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

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

L^{- 1} {\frac{1}{4 s ^{2}}} : \frac{t}{4}

L^{- 1} {\frac{1}{4 ( s ^{2} + 4 )}} : \frac{1}{8} sin (2 t)

= \frac{t}{4} - \frac{1}{8} sin (2 t)

y = 3 x - 6, (2, 4)

\frac{d}{d x} (\frac{2 x ^{3} - 5 x ^{2} + 4 x}{( x - 1 ) ^{2}})

\int (- 2) x d x

\int_{0}^{1} x^{2} 6 x + 8 d x

What is inverse oflaplace 1/(s^2(s^2+4)) ?
The answer to inverse oflaplace 1/(s^2(s^2+4)) is t/4-1/8 sin(2t)