What is inverse oflaplace 6/s+8*2*1/(s^2+16) ?

The answer to inverse oflaplace 6/s+8*2*1/(s^2+16) is 6\H(t)+4sin(4t)

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inverse oflaplace 6/s+821/(s^2+16)

inverse laplace

\frac{6}{s} + 8 \cdot 2 \cdot \frac{1}{s ^{2} + 16}

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

Solution steps

L^{- 1} {\frac{6}{s} + 8 \cdot 2 \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}} + 8 \cdot 2 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) + 8 \cdot 2 \cdot \frac{1}{4} sin (4 t)

Refine

6 H (t) + 82 \frac{1}{4} sin (4 t) : 6 H (t) + 4 sin (4 t)

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

\int yz e^{x z} d x

f (x) = (3 x - 5) (x + 4), (2, 6)

\int_{3}^{\infty} \frac{x}{x ^{2} + 1} d x

x \to - 3 lim (\frac{5 x ^{2}}{x + 3})

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

What is inverse oflaplace 6/s+8*2*1/(s^2+16) ?
The answer to inverse oflaplace 6/s+8*2*1/(s^2+16) is 6\H(t)+4sin(4t)