What is inverse oflaplace 2+(4(s-4))/(s^2+16) ?

The answer to inverse oflaplace 2+(4(s-4))/(s^2+16) is 2\delta(t)+4(cos(4t)-sin(4t))

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inverse oflaplace 2+(4(s-4))/(s^2+16)

inverse laplace

2 + \frac{4 ( s - 4 )}{s ^{2} + 16}

2 δ (t) + 4 (cos (4 t) - sin (4 t))

Solution steps

L^{- 1} {2 + \frac{4 ( s - 4 )}{s ^{2} + 16}}

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= L^{- 1} {2} + 4 L^{- 1} {\frac{s - 4}{s ^{2} + 16}}

L^{- 1} {2} : 2 δ (t)

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

= 2 δ (t) + 4 (cos (4 t) - sin (4 t))

\frac{d}{d x} (- 3 sin (2 x))

g (t) = \frac{t - t}{t ^{\frac{1}{3}}}

\int \frac{e ^{x}}{e ^{x} + e ^{- x}} d x

n = 0 \sum \infty \frac{1}{2 ^{2^{n}}}

y = π^{3 x + 1}

What is inverse oflaplace 2+(4(s-4))/(s^2+16) ?
The answer to inverse oflaplace 2+(4(s-4))/(s^2+16) is 2\delta(t)+4(cos(4t)-sin(4t))