What is laplacetransform-5(t-4) ?

The answer to laplacetransform-5(t-4) is -5/(s^2)+(20)/s

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laplacetransform-5(t-4)

laplace transform

- 5 (t - 4)

- \frac{5}{s ^{2}} + \frac{20}{s}

Solution steps

L {- 5 (t - 4)}

Expand

- 5 (t - 4) : - 5 t + 20

= L {- 5 t + 20}

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 {20}

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

L {20} : \frac{20}{s}

= - 5 \cdot \frac{1}{s ^{2}} + \frac{20}{s}

Refine

- 5 \frac{1}{s ^{2}} + \frac{20}{s} : - \frac{5}{s ^{2}} + \frac{20}{s}

= - \frac{5}{s ^{2}} + \frac{20}{s}

\frac{d}{d x} (4 x^{2 π})

3 x^{2} - 2 x + 5

\frac{d y}{d x} = 8 x y

x \to 0 lim (\frac{arcsin ( 3 x )}{arctan ( 5 x )})

\frac{d}{d x} (\frac{x ^{6}}{6} + \frac{6}{x ^{6}})

What is laplacetransform-5(t-4) ?
The answer to laplacetransform-5(t-4) is -5/(s^2)+(20)/s