What is laplacetransform t^2-10t+31 ?

The answer to laplacetransform t^2-10t+31 is 2/(s^3)-(10}{s^2}+\frac{31)/s

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laplacetransform t^2-10t+31

laplace transform

t^{2} - 10 t + 31

\frac{2}{s ^{3}} - \frac{10}{s ^{2}} + \frac{31}{s}

Solution steps

L {t^{2} - 10 t + 31}

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 {t^{2}} - 10 L {t} + L {31}

L {t^{2}} : \frac{2}{s ^{3}}

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

L {31} : \frac{31}{s}

= \frac{2}{s ^{3}} - 10 \cdot \frac{1}{s ^{2}} + \frac{31}{s}

Refine

\frac{2}{s ^{3}} - 10 \frac{1}{s ^{2}} + \frac{31}{s} : \frac{2}{s ^{3}} - \frac{10}{s ^{2}} + \frac{31}{s}

= \frac{2}{s ^{3}} - \frac{10}{s ^{2}} + \frac{31}{s}

\int 1 + 4 a^{2} x^{2} d x

\int ln (x - 2) d x

y = \frac{1}{x ^{3}}, (1, 1)

f (x) = \frac{5 x}{x - 3}, (4, 20)

y = (x + 5)^{\frac{1}{x}}

What is laplacetransform t^2-10t+31 ?
The answer to laplacetransform t^2-10t+31 is 2/(s^3)-(10}{s^2}+\frac{31)/s