What is laplacetransform-4(t-1) ?

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

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laplacetransform-4(t-1)

laplace transform

- 4 (t - 1)

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

Solution steps

L {- 4 (t - 1)}

Expand

- 4 (t - 1) : - 4 t + 4

= L {- 4 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)}

= - 4 L {t} + L {4}

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

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

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

Refine

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

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

\int_{0}^{2} e^{5 x} d x

x \to 0 lim (\frac{x + 1}{x})

\int \frac{10 x ^{2} - 9 x + 4}{x ^{3} + 4 x} d x

\frac{d}{d x} (\frac{7 x ^{2}}{( x ^{2} + x ) ^{3}})

n = 0 \sum \infty \frac{x ^{n}}{3 + 2 x}

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