What is inverse oflaplace 2/(s+1)+8a*s/(s^2+9) ?

The answer to inverse oflaplace 2/(s+1)+8a*s/(s^2+9) is 2e^{-t}+8acos(3t)

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inverse oflaplace 2/(s+1)+8a*s/(s^2+9)

inverse laplace

\frac{2}{s + 1} + 8 a \cdot \frac{s}{s ^{2} + 9}

2 e^{- t} + 8 a cos (3 t)

Solution steps

L^{- 1} {\frac{2}{s + 1} + 8 a \frac{s}{s ^{2} + 9}}

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{2}{s + 1}} + 8 a L^{- 1} {\frac{s}{s ^{2} + 9}}

L^{- 1} {\frac{2}{s + 1}} : 2 e^{- t}

L^{- 1} {\frac{s}{s ^{2} + 9}} : cos (3 t)

= 2 e^{- t} + 8 a cos (3 t)

\frac{d}{d x} ((1 - 2 x)^{5} (3 x - 2)^{6})

2 e^{2 t} + 8 t e^{2 t} + 4 t^{2} e^{2 t}

\frac{d}{d x} (\frac{ln ( x )}{2 x})

\frac{d}{d x} (\frac{6 - 7 x}{2 + x})

x \to 6 - lim (\frac{x ^{6}}{x - 6})

What is inverse oflaplace 2/(s+1)+8a*s/(s^2+9) ?
The answer to inverse oflaplace 2/(s+1)+8a*s/(s^2+9) is 2e^{-t}+8acos(3t)