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

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

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

inverse laplace

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

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

Solution steps

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

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

L^{- 1} {\frac{8 s}{s ^{2} + 9}} : 8 cos (3 t)

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

\int_{1}^{\infty} \frac{1}{x ^{\frac{13}{12}}} d x

x \to 0 lim (\frac{e}{x})

\int \frac{1}{( x + y )} d x

\int (x^{3} - 9 x) d x

\int 4 cos^{3} (x) d x

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