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

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

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inverse oflaplace s/(s^2+25)

inverse laplace

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

cos (5 t)

Solution steps

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

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

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

\frac{d}{d t} (\frac{1}{5} sin (5 t)) = cos (5 t)

Evaluate

\frac{1}{5} sin (5 t)

at point

0 = 0

= cos (5 t)

\int_{0}^{\frac{π}{3}} sin^{4} (3 x) d x

ln (\frac{x + 3}{x - 1})

121 y^{^{''}} - 16 y = 0

\frac{\partial}{\partial x} ((x + y)^{2})

n = 1 \sum \infty \frac{1}{n ^{2} + 3 n + 2}

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