What is inverse oflaplace (s^2)/(s^4+2s^2+1) ?

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

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inverse oflaplace (s^2)/(s^4+2s^2+1)

inverse laplace

\frac{s ^{2}}{s ^{4} + 2 s ^{2} + 1}

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

Solution steps

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

Take the partial fraction of

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

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

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))

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

Simplify

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

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

2 - 7 x

\int \frac{1}{x ^{2}} - \frac{3}{x ^{4}} d x

2 t e^{3 t} + 4 t^{2} e^{- 5 t}

\frac{d}{d x} (\frac{1}{2 x - 3})

5 - 4 x

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