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))

\int_{0}^{0.1} k x d x

\frac{\partial}{\partial x} (arctan (x y))

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

\frac{d}{d x} (cos (e^{x^{2} c o s (x)}))

n = 1 \sum \infty \frac{4 ^{n}}{5 ^{n}}

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))