What is inverse oflaplace (3s)/(s^2+4) ?

The answer to inverse oflaplace (3s)/(s^2+4) is 3cos(2t)

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inverse oflaplace (3s)/(s^2+4)

inverse laplace

\frac{3 s}{s ^{2} + 4}

3 cos (2 t)

Solution steps

L^{- 1} {\frac{3 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{3 s}{s ^{2} + 4} : F (s) = \frac{3}{s ^{2} + 4}

= \frac{d}{d t} (L^{- 1} {\frac{3}{s ^{2} + 4}}) + L^{- 1} {\frac{3}{s ^{2} + 4}} (0)

L^{- 1} {\frac{3}{s ^{2} + 4}} : \frac{3}{2} sin (2 t)

\frac{d}{d t} (\frac{3}{2} sin (2 t)) = 3 cos (2 t)

Evaluate

\frac{3}{2} sin (2 t)

at point

0 = 0

= 3 cos (2 t)

x + 6 y = 8, x + 8 = y^{2}

\frac{d}{d x} (- \frac{y}{x + 2 y})

\int \frac{2 x}{x ^{2} - 1} d x

\int \frac{x ^{2}}{6 - x ^{3}} d x

\frac{d}{d x} (x^{4} - 9 x^{2} + 16)

What is inverse oflaplace (3s)/(s^2+4) ?
The answer to inverse oflaplace (3s)/(s^2+4) is 3cos(2t)