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

The answer to inverse oflaplace 1/((s^2+4)^2*s) is -1/16 cos(2t)-(tsin(2t))/(16)+1/16 \H(t)

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

inverse laplace

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

- \frac{1}{16} cos (2 t) - \frac{t sin ( 2 t )}{16} + \frac{1}{16} H (t)

Solution steps

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

Take the partial fraction of

\frac{1}{( s ^{2} + 4 ) ^{2} s} : - \frac{s}{16 ( s ^{2} + 4 )} - \frac{s}{4 ( s ^{2} + 4 ) ^{2}} + \frac{1}{16 s}

= L^{- 1} {- \frac{s}{16 ( s ^{2} + 4 )} - \frac{s}{4 ( s ^{2} + 4 ) ^{2}} + \frac{1}{16 s}}

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{s}{16 ( s ^{2} + 4 )}} - L^{- 1} {\frac{s}{4 ( s ^{2} + 4 ) ^{2}}} + L^{- 1} {\frac{1}{16 s}}

L^{- 1} {\frac{s}{16 ( s ^{2} + 4 )}} : \frac{1}{16} cos (2 t)

L^{- 1} {\frac{s}{4 ( s ^{2} + 4 ) ^{2}}} : \frac{t sin ( 2 t )}{16}

L^{- 1} {\frac{1}{16 s}} : \frac{1}{16} H (t)

= - \frac{1}{16} cos (2 t) - \frac{t sin ( 2 t )}{16} + \frac{1}{16} H (t)

f (x) = x^{3} ln (2 x), at x = 5

x \to 1 lim (1 - (x - 1)^{2})

\frac{\partial}{\partial y} (e^{3 x} tan (y))

\int e^{x} cos (2 x) d x

\frac{d y}{d x} = k y^{2} ln (x), y (1) = - 5

What is inverse oflaplace 1/((s^2+4)^2*s) ?
The answer to inverse oflaplace 1/((s^2+4)^2*s) is -1/16 cos(2t)-(tsin(2t))/(16)+1/16 \H(t)