What is inverse oflaplace (2s)/((s^2+25)) ?

The answer to inverse oflaplace (2s)/((s^2+25)) is 2cos(5t)

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inverse oflaplace (2s)/((s^2+25))

inverse laplace

\frac{2 s}{( s ^{2} + 25 )}

2 cos (5 t)

Solution steps

L^{- 1} {\frac{2 s}{( s ^{2} + 25 )}}

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{2 s}{s ^{2} + 25} : F (s) = \frac{2}{s ^{2} + 25}

= \frac{d}{d t} (L^{- 1} {\frac{2}{s ^{2} + 25}}) + L^{- 1} {\frac{2}{s ^{2} + 25}} (0)

L^{- 1} {\frac{2}{s ^{2} + 25}} : \frac{2}{5} sin (5 t)

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

Evaluate

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

at point

0 = 0

= 2 cos (5 t)

0.1 e^{0.5061 (t)}

\int \frac{( ln ^{2} ( x ) )}{x} d x

\frac{d}{d x} (e^{- 2 x} cos (3 x) + sin (3 x))

\int x e^{2 x} sin (3 x) d x

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

What is inverse oflaplace (2s)/((s^2+25)) ?
The answer to inverse oflaplace (2s)/((s^2+25)) is 2cos(5t)