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

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

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

inverse laplace

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

t - sin (t)

Solution steps

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

Take the partial fraction of

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

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

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}}} : t

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

= t - sin (t)

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

\frac{d}{d t} (t e^{4 t})

\int - 3 tan (3 y) d y

\frac{d}{d x} (\frac{x - y}{sin ( x ) - sin ( y )})

\frac{\partial}{\partial y} (x^{2} sin^{2} (yz))

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