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

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

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

inverse laplace

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

\frac{1}{8} (sin (2 t) - 2 t cos (2 t))

Solution steps

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

= L^{- 1} {\frac{1}{8} \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}{8} 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}{8} (sin (2 t) - 2 t cos (2 t))

p a r i t y x csc (x)

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

\int (e^{x} + 1) d x

\frac{d}{d x} (x^{4} + 12 x^{2} + 42)

\frac{d}{d x} (\frac{6 x}{2 + x ^{2}})

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