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

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

Popular >

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

inverse laplace

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

- sin (t) + t

Solution steps

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

Take the partial fraction of

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

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

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

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

= - sin (t) + t

x \to 0 + lim (arccot (x))

\int (2 x^{2} + 3 x - 3) (5 x + 3) d x

f (x) = \frac{x ^{4}}{8} - 8 x

x \to \infty lim (5 + \frac{1}{x})

\frac{d}{d x} ((x^{2} + 1)^{x})

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