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

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

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

laplace transform

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

\frac{1}{s} - \frac{1}{e ^{2} ( s + 1 )}

Solution steps

L {1 - 1 \cdot e^{(- t - 2)}}

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} - 1 \cdot e^{- 2} L {e^{- t}}

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

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

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

Refine

\frac{1}{s} - 1 e^{- 2} \frac{1}{s + 1} : \frac{1}{s} - \frac{1}{e ^{2} ( s + 1 )}

= \frac{1}{s} - \frac{1}{e ^{2} ( s + 1 )}

2 x - x^{2}, x^{2}

x = 2 - y^{2}, x = y

y^{^{''}} - 4 y = e^{2 x}

x \to 0 lim (sin (x^{x}))

\int \frac{1}{x ^{2} + 6 x + 58} d x

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