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

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

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laplacetransform 1-e^{-5t}

laplace transform

1 - e^{- 5 t}

\frac{5}{s ( s + 5 )}

Solution steps

L {1 - 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 {1} - L {e^{- 5 t}}

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

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

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

Simplify

\frac{1}{s} - \frac{1}{s + 5} : \frac{5}{s ( s + 5 )}

= \frac{5}{s ( s + 5 )}

4 x^{2} e^{x}

\int \frac{1}{x + x x} d x

\int_{1}^{x} \frac{1}{t} d t

x^{6 x}

\frac{x ^{3}}{3} + x^{2} + 4 x - 10

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