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

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

Solution steps

L^{- 1} {\frac{6}{s} + 16 (\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)

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

\frac{d}{d x} (cos (x) + 2 sin (x))

\int \frac{- 2 x}{x ^{2} - 1} d x

\frac{\partial}{\partial y} (8 y - (x + y)^{3})

2 sin^{4} (x)

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)