What is laplacetransform 1-u ?

The answer to laplacetransform 1-u is 1/s-1/(s^2)

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laplacetransform 1-u

laplace transform

1 - u

\frac{1}{s} - \frac{1}{s ^{2}}

Solution steps

L {1 - u}

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 {u}

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

L {u} : \frac{1}{s ^{2}}

= \frac{1}{s} - \frac{1}{s ^{2}}

y^{^{'}} = 4 - 0.75 y

n = 2 \sum \infty \frac{8}{n ^{2} - 1}

\frac{1}{x ^{2}} - x, x = 2, x = 3

\frac{\partial}{\partial u} (cos (u))

5 e^{5 x}

What is laplacetransform 1-u ?
The answer to laplacetransform 1-u is 1/s-1/(s^2)