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

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

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

inverse laplace

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

t cos (t)

Solution steps

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

Take the partial fraction of

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

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

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

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

L^{- 1} {\frac{1}{s ^{2} + 1}} : sin (t)

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

= sin (t) - (sin (t) - t cos (t))

Simplify

sin (t) - (sin (t) - t cos (t)) : t cos (t)

= t cos (t)

\frac{d}{d x} (\frac{( a x + b y ( x ) )}{c x + d y})

\frac{d x}{d t} = 5 x (9 - x)

\frac{d}{d x} (a e^{x})

\int cos^{2} (\frac{1}{3}) x d x

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

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