What is inverse oflaplace (4se^{-3s})/((s^2+4)) ?

The answer to inverse oflaplace (4se^{-3s})/((s^2+4)) is \H(t-3)*4cos(2(t-3))

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inverse oflaplace (4se^{-3s})/((s^2+4))

inverse laplace

\frac{4 s e ^{- 3 s}}{( s ^{2} + 4 )}

H (t - 3) \cdot 4 cos (2 (t - 3))

Solution steps

L^{- 1} {\frac{4 e ^{- 3 s} s}{( s ^{2} + 4 )}}

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {e^{- a s} F (s)} = H (t - a) f (t - a)

Where

H (t)

is Heaviside step function

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

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

= H (t - 3) \cdot 4 cos (2 (t - 3))

\frac{d y}{d x}, y = 2 x - 5

n = 1 \sum \infty arctan (4 n)

\frac{2 s ^{2} + 1}{s + 3}

\int_{- 3}^{3} ∣ 1 - x ∣ d x

\frac{d}{d x} (sin (12 x) e^{7 x})

What is inverse oflaplace (4se^{-3s})/((s^2+4)) ?
The answer to inverse oflaplace (4se^{-3s})/((s^2+4)) is \H(t-3)*4cos(2(t-3))