What is inverse oflaplace (5s)/(s^2+4) ?

The answer to inverse oflaplace (5s)/(s^2+4) is 5cos(2t)

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inverse oflaplace (5s)/(s^2+4)

inverse laplace

\frac{5 s}{s ^{2} + 4}

5 cos (2 t)

Solution steps

L^{- 1} {\frac{5 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{5 s}{s ^{2} + 4} : F (s) = \frac{5}{s ^{2} + 4}

= \frac{d}{d t} (L^{- 1} {\frac{5}{s ^{2} + 4}}) + L^{- 1} {\frac{5}{s ^{2} + 4}} (0)

L^{- 1} {\frac{5}{s ^{2} + 4}} : \frac{5}{2} sin (2 t)

\frac{d}{d t} (\frac{5}{2} sin (2 t)) = 5 cos (2 t)

Evaluate

\frac{5}{2} sin (2 t)

at point

0 = 0

= 5 cos (2 t)

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

\frac{d}{d x} (∣ x ∣^{2})

\int x e^{4 - x^{2}} d x

x \to \frac{π}{6} lim (\frac{sin ( x )}{x})

y^{^{''}} + 64 y = \frac{32}{3} cos (7 t)

What is inverse oflaplace (5s)/(s^2+4) ?
The answer to inverse oflaplace (5s)/(s^2+4) is 5cos(2t)