What is laplacetransform 1(t-4) ?

The answer to laplacetransform 1(t-4) is 1/(s^2)-4/s

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laplacetransform 1(t-4)

laplace transform

1 (t - 4)

\frac{1}{s ^{2}} - \frac{4}{s}

Solution steps

L {1 \cdot (t - 4)}

Expand

1 (t - 4) : t - 4

= L {t - 4}

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 {t} - L {4}

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

L {4} : \frac{4}{s}

= \frac{1}{s ^{2}} - \frac{4}{s}

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

x \to 1 lim (2 x^{2} + x - 1)

\int 3 e^{x} + 7 sec^{2} (x) d x

\frac{d}{d x} (\frac{e ^{x}}{e ^{x} + e ^{y}})

What is laplacetransform 1(t-4) ?
The answer to laplacetransform 1(t-4) is 1/(s^2)-4/s