What is laplacetransform y^2+4y+20 ?

The answer to laplacetransform y^2+4y+20 is 2/(s^3)+4/(s^2)+(20)/s

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laplacetransform y^2+4y+20

laplace transform

y^{2} + 4 y + 20

\frac{2}{s ^{3}} + \frac{4}{s ^{2}} + \frac{20}{s}

Solution steps

L {y^{2} + 4 y + 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)}

= L {y^{2}} + 4 L {y} + L {20}

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

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

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

= \frac{2}{s ^{3}} + 4 \cdot \frac{1}{s ^{2}} + \frac{20}{s}

Refine

\frac{2}{s ^{3}} + 4 \frac{1}{s ^{2}} + \frac{20}{s} : \frac{2}{s ^{3}} + \frac{4}{s ^{2}} + \frac{20}{s}

= \frac{2}{s ^{3}} + \frac{4}{s ^{2}} + \frac{20}{s}

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What is laplacetransform y^2+4y+20 ?
The answer to laplacetransform y^2+4y+20 is 2/(s^3)+4/(s^2)+(20)/s