What is laplacetransform s^2+4s+13 ?

The answer to laplacetransform s^2+4s+13 is 2/(s^3)+4/(s^2)+(13)/s

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laplacetransform s^2+4s+13

laplace transform

s^{2} + 4 s + 13

\frac{2}{s ^{3}} + \frac{4}{s ^{2}} + \frac{13}{s}

Solution steps

L {s^{2} + 4 s + 13}

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 {s^{2}} + 4 L {s} + L {13}

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

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

L {13} : \frac{13}{s}

= \frac{2}{s ^{3}} + 4 \cdot \frac{1}{s ^{2}} + \frac{13}{s}

Refine

\frac{2}{s ^{3}} + 4 \frac{1}{s ^{2}} + \frac{13}{s} : \frac{2}{s ^{3}} + \frac{4}{s ^{2}} + \frac{13}{s}

= \frac{2}{s ^{3}} + \frac{4}{s ^{2}} + \frac{13}{s}

y^{^{'}} = 3 x + x y

\frac{d}{d x} ((4 x - 6)^{2})

9 x^{2}, \frac{1}{9} x

f (x) = tan (x + 4)

\frac{3 - x e ^{x}}{x + e ^{x}}

What is laplacetransform s^2+4s+13 ?
The answer to laplacetransform s^2+4s+13 is 2/(s^3)+4/(s^2)+(13)/s