What is laplacetransform 6e^{-3}-t^2+2t-8 ?

The answer to laplacetransform 6e^{-3}-t^2+2t-8 is 6/(e^3s)-2/(s^3)+2/(s^2)-8/s

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laplacetransform 6e^{-3}-t^2+2t-8

laplace transform

6 e^{- 3} - t^{2} + 2 t - 8

\frac{6}{e ^{3} s} - \frac{2}{s ^{3}} + \frac{2}{s ^{2}} - \frac{8}{s}

Solution steps

L {6 e^{- 3} - t^{2} + 2 t - 8}

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 {6 e^{- 3}} - L {t^{2}} + 2 L {t} - L {8}

L {6 e^{- 3}} : \frac{6}{e ^{3} s}

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

L {t} : \frac{1}{s ^{2}}

L {8} : \frac{8}{s}

= \frac{6}{e ^{3} s} - \frac{2}{s ^{3}} + 2 \cdot \frac{1}{s ^{2}} - \frac{8}{s}

Refine

\frac{6}{e ^{3} s} - \frac{2}{s ^{3}} + 2 \frac{1}{s ^{2}} - \frac{8}{s} : \frac{6}{e ^{3} s} - \frac{2}{s ^{3}} + \frac{2}{s ^{2}} - \frac{8}{s}

= \frac{6}{e ^{3} s} - \frac{2}{s ^{3}} + \frac{2}{s ^{2}} - \frac{8}{s}

x y^{^{'}} + 2 y = - 2 sin (x)

\int \frac{48}{( 6 x + 1 ) ^{9}} d x

\int 10 sin^{3} (x co) s^{7} x d x

\frac{4 x}{x}

\int \frac{x}{2 x - 6} d x

What is laplacetransform 6e^{-3}-t^2+2t-8 ?
The answer to laplacetransform 6e^{-3}-t^2+2t-8 is 6/(e^3s)-2/(s^3)+2/(s^2)-8/s