What is laplacetransform 10(t-1) ?

The answer to laplacetransform 10(t-1) is (10}{s^2}-\frac{10)/s

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laplacetransform 10(t-1)

laplace transform

10 (t - 1)

\frac{10}{s ^{2}} - \frac{10}{s}

Solution steps

L {10 (t - 1)}

Expand

10 (t - 1) : 10 t - 10

= L {10 t - 10}

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)}

= 10 L {t} - L {10}

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

L {10} : \frac{10}{s}

= 10 \cdot \frac{1}{s ^{2}} - \frac{10}{s}

Refine

10 \frac{1}{s ^{2}} - \frac{10}{s} : \frac{10}{s ^{2}} - \frac{10}{s}

= \frac{10}{s ^{2}} - \frac{10}{s}

(2 x - 5 y + 2) d x + (1 - 6 y - 5 x) d y = 0

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

\frac{d}{d x} (- 16 x^{2} + 64 x)

\int \frac{x ^{- \frac{1}{3}}}{1 + x ^{\frac{1}{6}}} d x

\int \frac{3 x + 5}{x ^{2} + 2 x + 10} d x

What is laplacetransform 10(t-1) ?
The answer to laplacetransform 10(t-1) is (10}{s^2}-\frac{10)/s