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}}}

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {s F (s)} = \frac{d}{d t} f (t) + f (0)

For

\frac{5 s}{( s ^{2} + 1 ) ^{2}} : F (s) = \frac{5}{( s ^{2} + 1 ) ^{2}}

= \frac{d}{d t} (L^{- 1} {\frac{5}{( s ^{2} + 1 ) ^{2}}}) + L^{- 1} {\frac{5}{( s ^{2} + 1 ) ^{2}}} (0)

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

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

Evaluate

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

at point

0 = 0

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

5 x^{2} - sin (π x)

y^{^{'}} + 4 y = e^{- 4 t}, y (0) = 2

y = \frac{3 + csc ( x )}{7 - csc ( x )}

\frac{2 x}{( x ^{2} - 5 ) ^{3}}, at x = 0

\int \frac{1}{x ^{3} x ^{2} - 81} d x

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