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

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

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

inverse laplace

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

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

Solution steps

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

\frac{d}{d x} (a (e^{b x} + b e^{b x} x))

\frac{\partial}{\partial x} (x (4 - x^{2}))

y^{^{''}} - 4 y^{^{'}} + 4 y = x^{2} e^{x}

y^{^{'}} - \frac{y}{x} = x cos (x)

\frac{d}{d x} (\frac{1}{4} x^{4} - 2 x^{2})

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