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

The answer to inverse oflaplace 1/(s^2(s^2+1)^2) is (tcos(t)+2t-3sin(t))/2

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

inverse laplace

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

\frac{t cos ( t ) + 2 t - 3 sin ( t )}{2}

Solution steps

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

Take the partial fraction of

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

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

L^{- 1} {\frac{1}{s ^{2}}} : t

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

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

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

Simplify

t - sin (t) - \frac{1}{2} (sin (t) - t cos (t)) : \frac{t cos ( t ) + 2 t - 3 sin ( t )}{2}

= \frac{t cos ( t ) + 2 t - 3 sin ( t )}{2}

y = e^{x}, y = e^{2 x}, x = ln (2), x = ln (3)

x \to 0 + lim ((1 + 4 x)^{c o t (x)})

x^{2} + 1

y^{^{'}} = 2 y + 39

\frac{d}{d x} (\frac{x ^{\frac{3}{5}} x ^{3} + 2 x - 6}{( 4 x - 3 ) ^{8}})

What is inverse oflaplace 1/(s^2(s^2+1)^2) ?
The answer to inverse oflaplace 1/(s^2(s^2+1)^2) is (tcos(t)+2t-3sin(t))/2