What is inverse oflaplace 5/(s^2+25) ?

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

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inverse oflaplace 5/(s^2+25)

inverse laplace

\frac{5}{s ^{2} + 25}

sin (5 t)

Solution steps

L^{- 1} {\frac{5}{s ^{2} + 25}}

Use Inverse Laplace Transform table:

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

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

= sin (5 t)

\frac{\partial}{\partial x} (\frac{x - 7 y}{y ^{2} + 6 x ^{2}})

\int x^{2} + x d x

f (x) = 2 x^{2} + 9, at x = 1

\frac{d}{d y} (cos (y))

(- \frac{1}{x ^{2}})^{^{'}}

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