What is inverse oflaplace 2/(s+1)+(32s)/(s^2+9) ?

The answer to inverse oflaplace 2/(s+1)+(32s)/(s^2+9) is 2e^{-t}+32cos(3t)

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inverse oflaplace 2/(s+1)+(32s)/(s^2+9)

inverse laplace

\frac{2}{s + 1} + \frac{32 s}{s ^{2} + 9}

2 e^{- t} + 32 cos (3 t)

Solution steps

L^{- 1} {\frac{2}{s + 1} + \frac{32 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{32 s}{s ^{2} + 9}}

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

L^{- 1} {\frac{32 s}{s ^{2} + 9}} : 32 cos (3 t)

= 2 e^{- t} + 32 cos (3 t)

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

\int y^{\frac{1}{7}} d y

\frac{d}{d θ} (\frac{2}{sin ( θ )})

\frac{d y}{d x} = - \frac{y ^{2}}{1 + y ^{2}} e^{- x}, y (0) = 1

\int (x + 1) (x + 3)^{2} d x

What is inverse oflaplace 2/(s+1)+(32s)/(s^2+9) ?
The answer to inverse oflaplace 2/(s+1)+(32s)/(s^2+9) is 2e^{-t}+32cos(3t)