What is laplacetransform-(e^{-3(t-1)})/3-1/3 ?

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

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laplacetransform-(e^{-3(t-1)})/3-1/3

laplace transform

- \frac{e ^{- 3 (t - 1)}}{3} - \frac{1}{3}

- \frac{1}{3 e ( s + 3 )} - \frac{1}{3 s}

Solution steps

L {- \frac{e ^{- 3 (t - 1)}}{3} - \frac{1}{3}}

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)}

= - \frac{1}{3} L {e^{- 3 t - 1}} - L {\frac{1}{3}}

L {e^{- 3 t - 1}} : \frac{1}{e ( s + 3 )}

L {\frac{1}{3}} : \frac{1}{3 s}

= - \frac{1}{3} \cdot \frac{1}{e ( s + 3 )} - \frac{1}{3 s}

Refine

- \frac{1}{3} \frac{1}{e ( s + 3 )} - \frac{1}{3 s} : - \frac{1}{3 e ( s + 3 )} - \frac{1}{3 s}

= - \frac{1}{3 e ( s + 3 )} - \frac{1}{3 s}

f (x) = \frac{6 x ^{3}}{4 - x}

x \to 0 lim (\frac{3 ^{x^{2}} - 1}{x})

\frac{1}{x + 6}

\int (- 2 x + 3) d x

3 x^{2} + 8

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