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

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

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

inverse laplace

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

δ (t) - sin (t)

Solution steps

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

Take the partial fraction of

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

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

L^{- 1} {1} : δ (t)

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

= δ (t) - sin (t)

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

\int_{\frac{1}{2}}^{3} 16 x ln (2 x) d x

\frac{\partial}{\partial x} (ln (\frac{r}{ro}))

(x^{2} + y) \frac{d y}{d x} - y - 3 x^{2} = 2 y

\int x + 13 d x

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