What is laplacetransform 2(1-e^{-5t}) ?

The answer to laplacetransform 2(1-e^{-5t}) is 2/s-2/(s+5)

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laplacetransform 2(1-e^{-5t})

laplace transform

2 (1 - e^{- 5 t})

\frac{2}{s} - \frac{2}{s + 5}

Solution steps

L {2 (- e^{- 5 t} + 1)}

Expand

2 (1 - e^{- 5 t}) : 2 - 2 e^{- 5 t}

= L {2 - 2 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 {2} - 2 L {e^{- 5 t}}

L {2} : \frac{2}{s}

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

= \frac{2}{s} - 2 \cdot \frac{1}{s + 5}

Refine

\frac{2}{s} - 2 \frac{1}{s + 5} : \frac{2}{s} - \frac{2}{s + 5}

= \frac{2}{s} - \frac{2}{s + 5}

x \to 2 lim (2 x^{2} - 8)

\frac{( x ^{2} - 4 ) d y}{d x} + 4 y = (x + 2)^{2}

- x^{2} + 5, x^{2} - 10 x + 17, 2, 3

f (x) = \frac{1}{( x ^{6} - \frac{7}{x} ) ^{2}}

- \frac{4}{( 4 x - 17 ) ^{\frac{3}{2}}}

What is laplacetransform 2(1-e^{-5t}) ?
The answer to laplacetransform 2(1-e^{-5t}) is 2/s-2/(s+5)