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

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

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laplacetransform 1+e^{-t}

laplace transform

1 + e^{- t}

\frac{1}{s} + \frac{1}{s + 1}

Solution steps

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

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

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

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

\frac{4 x - 3}{x}

y^{^{''}} - \frac{3}{x} y^{^{'}} - \frac{4}{x ^{2}} y = 0

\frac{\partial}{\partial x} (17 x^{3} y^{2})

\frac{d}{d x} (e^{x} (5 x^{2} - 2 x + 3))

\frac{d y}{d x} = \frac{y sin ( x )}{1 - 2 y}

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