What is laplacetransform 1+6t ?

The answer to laplacetransform 1+6t is 1/s+6/(s^2)

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laplacetransform 1+6t

laplace transform

1 + 6 t

\frac{1}{s} + \frac{6}{s ^{2}}

Solution steps

L {1 + 6 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} + 6 L {t}

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

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

= \frac{1}{s} + 6 \cdot \frac{1}{s ^{2}}

Refine

\frac{1}{s} + 6 \frac{1}{s ^{2}} : \frac{1}{s} + \frac{6}{s ^{2}}

= \frac{1}{s} + \frac{6}{s ^{2}}

\frac{d}{d x} (\frac{x ^{2} + 16}{x})

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

f (x) = \frac{40 x ^{5}}{e ^{x}}

\int \frac{e ^{x}}{5 + e ^{x}} d x

What is laplacetransform 1+6t ?
The answer to laplacetransform 1+6t is 1/s+6/(s^2)