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

The answer to inverse oflaplace s/((s^2+1)) is cos(t)

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

inverse laplace

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

cos (t)

Solution steps

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

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

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

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

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

Evaluate

sin (t)

at point

0 = 0

= cos (t)

y^{^{'}} = y cos (2 x), y (0) = 1.6

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

\int_{1}^{4} 5 - 2 t + 3 t^{2} d t

\frac{\partial}{\partial x} (2 x + 3 y + 4)

\int 2 x (x^{2} + 5)^{4} d x

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