What is laplacetransform 5(t-1) ?

The answer to laplacetransform 5(t-1) is 5/(s^2)-5/s

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laplacetransform 5(t-1)

laplace transform

5 (t - 1)

\frac{5}{s ^{2}} - \frac{5}{s}

Solution steps

L {5 (t - 1)}

Expand

5 (t - 1) : 5 t - 5

= L {5 t - 5}

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)}

= 5 L {t} - L {5}

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

L {5} : \frac{5}{s}

= 5 \cdot \frac{1}{s ^{2}} - \frac{5}{s}

Refine

5 \frac{1}{s ^{2}} - \frac{5}{s} : \frac{5}{s ^{2}} - \frac{5}{s}

= \frac{5}{s ^{2}} - \frac{5}{s}

\frac{d}{d x} (\frac{A}{B + C e ^{x}})

\int e^{a x} d x

f (x) = \frac{1 - x ^{2}}{2 x + 1}

2 t x^{^{''}} - x^{^{'}} + \frac{1}{( x ^{^{'}} )} = 0

\int \frac{( x ^{3} )}{x ^{2} + 16} d x

What is laplacetransform 5(t-1) ?
The answer to laplacetransform 5(t-1) is 5/(s^2)-5/s