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

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

Popular >

inverse oflaplace 1/((z^2+1)^2)

inverse laplace

\frac{1}{( z ^{2} + 1 ) ^{2}}

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

Solution steps

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

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

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

\frac{\partial}{\partial x} (lo g_{10} (2 x))

e^{- \frac{1}{y ^{3}}}

\frac{d y}{d x} = \frac{2 y ^{4} + x ^{4}}{x y ^{3}}

\frac{d ^{3}}{d x ^{3}} (x^{5} e^{x})

\frac{d}{d x} (\frac{4 x ^{2}}{( 7 - 5 x ) ^{3}})

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