What is laplacetransform t^2-e^{-9t}+5 ?

The answer to laplacetransform t^2-e^{-9t}+5 is 2/(s^3)-1/(s+9)+5/s

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laplacetransform t^2-e^{-9t}+5

laplace transform

t^{2} - e^{- 9 t} + 5

\frac{2}{s ^{3}} - \frac{1}{s + 9} + \frac{5}{s}

Solution steps

L {t^{2} - e^{- 9 t} + 5}

Use the linearity property of Laplace Transform:

For functions

f (t), g (t)

and constants

a, b : L {a \cdot f (t) + b \cdot g (t)} = a \cdot L {f (t)} + b \cdot L {g (t)}

= L {t^{2}} - L {e^{- 9 t}} + L {5}

L {t^{2}} : \frac{2}{s ^{3}}

L {e^{- 9 t}} : \frac{1}{s + 9}

L {5} : \frac{5}{s}

= \frac{2}{s ^{3}} - \frac{1}{s + 9} + \frac{5}{s}

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

\int sin (x) sec^{3} (x) d x

x y^{2} + 3 y^{2} - x^{2} \frac{d y}{d x} = 0

\int_{2}^{6} x d x

(\frac{x ^{2} + 3}{x ^{2} - 3})^{3}

What is laplacetransform t^2-e^{-9t}+5 ?
The answer to laplacetransform t^2-e^{-9t}+5 is 2/(s^3)-1/(s+9)+5/s