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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

\frac{\partial}{\partial x} (sec^{2} (x))

\frac{\partial}{\partial x} (6 x^{2})

f (x) = 3 x \cdot cos (x), at x = π

\int 3 cos (n x) d x

\int \frac{x - 1}{2} d x

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