What is inverse oflaplace e^{-3s}*5/(4s^2+16) ?

The answer to inverse oflaplace e^{-3s}*5/(4s^2+16) is \H(t-3) 5/8 sin(2(t-3))

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inverse oflaplace e^{-3s}*5/(4s^2+16)

inverse laplace

e^{- 3 s} \cdot \frac{5}{4 s ^{2} + 16}

H (t - 3) \frac{5}{8} sin (2 (t - 3))

Solution steps

L^{- 1} {e^{- 3 s} \frac{5}{4 s ^{2} + 16}}

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{5}{4 s ^{2} + 16}} (t - 3)

L^{- 1} {\frac{5}{4 s ^{2} + 16}} : \frac{5}{8} sin (2 t)

= H (t - 3) \frac{5}{8} sin (2 (t - 3))

4 x - x^{2} (1.3)

\frac{d}{d x} (ln (e^{x} + e^{- x}))

x \to 4 + lim (\frac{2}{( x - 4 ) ^{4}})

16 y^{^{''''}} = 256 y

\int ln (1 + \frac{1}{x ^{2}}) d x

What is inverse oflaplace e^{-3s}*5/(4s^2+16) ?
The answer to inverse oflaplace e^{-3s}*5/(4s^2+16) is \H(t-3) 5/8 sin(2(t-3))