What is inverse oflaplace-(5s)/((s^2+1)^2) ?

The answer to inverse oflaplace-(5s)/((s^2+1)^2) is -(5tsin(t))/2

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inverse oflaplace-(5s)/((s^2+1)^2)

inverse laplace

- \frac{5 s}{( s ^{2} + 1 ) ^{2}}

- \frac{5 t sin ( t )}{2}

Solution steps

L^{- 1} {- \frac{5 s}{( s ^{2} + 1 ) ^{2}}}

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

= - L^{- 1} {\frac{5 s}{( s ^{2} + 1 ) ^{2}}}

L^{- 1} {\frac{5 s}{( s ^{2} + 1 ) ^{2}}} : \frac{5 t sin ( t )}{2}

= - \frac{5 t sin ( t )}{2}

\int (x + 1) (x + 3) d x

\frac{- 0.1 s}{s ^{2} - 9.8}

y = \frac{4}{3} (x + 1)^{\frac{3}{2}}

\int_{0}^{- 0.5} 150 x d x

(x^{2} + 5)^{2} (x - 5)

What is inverse oflaplace-(5s)/((s^2+1)^2) ?
The answer to inverse oflaplace-(5s)/((s^2+1)^2) is -(5tsin(t))/2