What is inverse oflaplace 6/s+16*1/(s^2+16) ?

The answer to inverse oflaplace 6/s+16*1/(s^2+16) is 6\H(t)+4sin(4t)

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inverse oflaplace 6/s+16*1/(s^2+16)

inverse laplace

\frac{6}{s} + 16 \cdot \frac{1}{s ^{2} + 16}

6 H (t) + 4 sin (4 t)

Solution steps

L^{- 1} {\frac{6}{s} + 16 \cdot \frac{1}{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} {\frac{6}{s}} + 16 L^{- 1} {\frac{1}{s ^{2} + 16}}

L^{- 1} {\frac{6}{s}} : 6 H (t)

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

= 6 H (t) + 16 \cdot \frac{1}{4} sin (4 t)

Refine

6 H (t) + 16 \frac{1}{4} sin (4 t) : 6 H (t) + 4 sin (4 t)

= 6 H (t) + 4 sin (4 t)

f (x) = \frac{2}{4 - x}, at x = - 1

(sin^{2} (x))^{^{'}}

\frac{d}{d x} (4 x^{\frac{1}{3}})

5 x - x^{2}, 0

h \to 0 lim (h)

What is inverse oflaplace 6/s+16*1/(s^2+16) ?
The answer to inverse oflaplace 6/s+16*1/(s^2+16) is 6\H(t)+4sin(4t)