What is laplacetransform-2t+8 ?

The answer to laplacetransform-2t+8 is -2/(s^2)+8/s

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laplacetransform-2t+8

laplace transform

- 2 t + 8

- \frac{2}{s ^{2}} + \frac{8}{s}

Solution steps

L {- 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)}

= - 2 L {t} + L {8}

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

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

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

Refine

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

= - \frac{2}{s ^{2}} + \frac{8}{s}

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

\frac{d}{d x} (- \frac{2 x}{( 9 + x ^{2} ) ^{2}})

\frac{d x}{d t} = 12 - \frac{x}{200 + 3 t}

f (x) = x^{3} - 3 x^{2} - 9 x + 27, at x = 4

\frac{\partial}{\partial x} (\frac{3 x + y ^{2}}{x y})

What is laplacetransform-2t+8 ?
The answer to laplacetransform-2t+8 is -2/(s^2)+8/s