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

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

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inverse oflaplace 6/s+8a*s/(s^2+16)

inverse laplace

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

6 H (t) + 8 a cos (4 t)

Solution steps

L^{- 1} {\frac{6}{s} + 8 a \frac{s}{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 a L^{- 1} {\frac{s}{s ^{2} + 16}}

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

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

= 6 H (t) + 8 a cos (4 t)

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

h

\frac{d}{d x} (\frac{1 - sin ( x )}{1 + sin ( x )})

\int e^{e x} d x

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