What is inverse oflaplace-(e^{-s})/(((s^2+4)*s^2)) ?

The answer to inverse oflaplace-(e^{-s})/(((s^2+4)*s^2)) is -\H(t-1)(-1/8 sin(2(t-1))+(t-1)/4)

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inverse oflaplace-(e^{-s})/(((s^2+4)*s^2))

inverse laplace

- \frac{e ^{- s}}{( ( s ^{2} + 4 ) \cdot s ^{2} )}

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

Solution steps

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

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

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

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

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

\int 4 x y^{2} + 3 x^{2} d x

y = cos (a^{8} + x^{8})

\frac{d}{d x} (2 csc^{2} (x) cot (x))

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

\int \frac{1}{( x + 1 ) x ^{2} + 2 x - 15} d x

What is inverse oflaplace-(e^{-s})/(((s^2+4)*s^2)) ?
The answer to inverse oflaplace-(e^{-s})/(((s^2+4)*s^2)) is -\H(t-1)(-1/8 sin(2(t-1))+(t-1)/4)