What is inverse oflaplace 8/((z^2+4)^2) ?

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

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inverse oflaplace 8/((z^2+4)^2)

inverse laplace

\frac{8}{( z ^{2} + 4 ) ^{2}}

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

Solution steps

L^{- 1} {\frac{8}{( z ^{2} + 4 ) ^{2}}}

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

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

\int (x^{2} + 8 x) cos (x) d x

\frac{d}{d x} (x^{4})

\frac{d}{d x} ((- x^{2} + x) (- 2 x^{2} + 4 x - 1))

cos (x^{2}) (2 x)

n = 0 \sum \infty \frac{2 n}{e ^{n}}

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