What is laplacetransform 2t^2-e^{-t} ?

The answer to laplacetransform 2t^2-e^{-t} is 4/(s^3)-1/(s+1)

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laplacetransform 2t^2-e^{-t}

laplace transform

2 t^{2} - e^{- t}

\frac{4}{s ^{3}} - \frac{1}{s + 1}

Solution steps

L {2 t^{2} - e^{- t}}

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)}

= 2 L {t^{2}} - L {e^{- t}}

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

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

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

Refine

2 \frac{2}{s ^{3}} - \frac{1}{s + 1} : \frac{4}{s ^{3}} - \frac{1}{s + 1}

= \frac{4}{s ^{3}} - \frac{1}{s + 1}

x \to \frac{π}{2} lim (cos^{2} (x))

f (x) = \frac{cos ( x ^{3} )}{e ^{x}}

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

\int e^{- x} cos (2 x) d x

x \to + 0 lim (x + ∣ x ∣)

What is laplacetransform 2t^2-e^{-t} ?
The answer to laplacetransform 2t^2-e^{-t} is 4/(s^3)-1/(s+1)