What is inverse oflaplace (4s)/(s^2+4) ?

The answer to inverse oflaplace (4s)/(s^2+4) is 4cos(2t)

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inverse oflaplace (4s)/(s^2+4)

inverse laplace

\frac{4 s}{s ^{2} + 4}

4 cos (2 t)

Solution steps

L^{- 1} {\frac{4 s}{s ^{2} + 4}}

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{4 s}{s ^{2} + 4} : F (s) = \frac{4}{s ^{2} + 4}

= \frac{d}{d t} (L^{- 1} {\frac{4}{s ^{2} + 4}}) + L^{- 1} {\frac{4}{s ^{2} + 4}} (0)

L^{- 1} {\frac{4}{s ^{2} + 4}} : 2 sin (2 t)

\frac{d}{d t} (2 sin (2 t)) = 4 cos (2 t)

Evaluate

2 sin (2 t)

at point

0 = 0

= 4 cos (2 t)

y^{^{''}} - 3 y^{^{'}} - 2 y = 0

x^{5} - x^{3}

y^{^{'}} = \frac{4 x}{y}

x \to 1 lim ((x) (x - 1)^{- 1})

\int_{0}^{1} x^{3} (1 - x)^{2} d x

What is inverse oflaplace (4s)/(s^2+4) ?
The answer to inverse oflaplace (4s)/(s^2+4) is 4cos(2t)