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

The answer to inverse oflaplace 1/((s^2+4)^2) is 1/16 (sin(2t)-2tcos(2t))

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

inverse laplace

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

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

Solution steps

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

= L^{- 1} {\frac{1}{16} \cdot \frac{2 \cdot 2 ^{3}}{( s ^{2} + 2 ^{2} ) ^{2}}}

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)}

= \frac{1}{16} L^{- 1} {\frac{2 \cdot 2 ^{3}}{( s ^{2} + 2 ^{2} ) ^{2}}}

Use Inverse Laplace Transform table:

L^{- 1} {\frac{2 a ^{3}}{( s ^{2} + a ^{2} ) ^{2}}} = sin (a t) - a t cos (a t)

L^{- 1} {\frac{2 \cdot 2 ^{3}}{( s ^{2} + 2 ^{2} ) ^{2}}} = sin (2 t) - 2 t cos (2 t)

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

\frac{d}{d t} (5 t^{- 3})

\int_{- 1}^{1} e^{y} - y^{2} + 2 d y

\frac{\partial}{\partial y} (tan (3 x - 2 y + 1))

\frac{s + 7}{s ^{2} + 8 s + 25}

g (x) = (3 - 4 x)^{3}

What is inverse oflaplace 1/((s^2+4)^2) ?
The answer to inverse oflaplace 1/((s^2+4)^2) is 1/16 (sin(2t)-2tcos(2t))