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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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