What is laplacetransform 1+7t ?

The answer to laplacetransform 1+7t is 1/s+7/(s^2)

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laplacetransform 1+7t

laplace transform

1 + 7 t

\frac{1}{s} + \frac{7}{s ^{2}}

Solution steps

L {1 + 7 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} + 7 L {t}

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

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

= \frac{1}{s} + 7 \cdot \frac{1}{s ^{2}}

Refine

\frac{1}{s} + 7 \frac{1}{s ^{2}} : \frac{1}{s} + \frac{7}{s ^{2}}

= \frac{1}{s} + \frac{7}{s ^{2}}

\frac{\partial}{\partial x} (14.36 sin (23 x))

\int \frac{x ^{2}}{( 4 + x ^{3} ) ^{2}} d x

\int + x^{2} ln (x) d x

y (X) = \frac{5 X ^{3} + 2 X ^{2} - 1}{X ^{2}}

x \to \infty lim (\frac{x}{x})

What is laplacetransform 1+7t ?
The answer to laplacetransform 1+7t is 1/s+7/(s^2)