What is laplacetransform 6(1-e^{-5t}) ?

The answer to laplacetransform 6(1-e^{-5t}) is 6/s-6/(s+5)

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laplacetransform 6(1-e^{-5t})

laplace transform

6 (1 - e^{- 5 t})

\frac{6}{s} - \frac{6}{s + 5}

Solution steps

L {6 (- e^{- 5 t} + 1)}

Expand

6 (1 - e^{- 5 t}) : 6 - 6 e^{- 5 t}

= L {6 - 6 e^{- 5 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)}

= L {6} - 6 L {e^{- 5 t}}

L {6} : \frac{6}{s}

L {e^{- 5 t}} : \frac{1}{s + 5}

= \frac{6}{s} - 6 \cdot \frac{1}{s + 5}

Refine

\frac{6}{s} - 6 \frac{1}{s + 5} : \frac{6}{s} - \frac{6}{s + 5}

= \frac{6}{s} - \frac{6}{s + 5}

\int c t an x d x

3^{x}

x \to \infty lim (5^{x} + 1)

x^{2}, 6 - x

\frac{1}{1 - x ^{3}}

What is laplacetransform 6(1-e^{-5t}) ?
The answer to laplacetransform 6(1-e^{-5t}) is 6/s-6/(s+5)