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

The answer to laplacetransform s^2+4s+13s^2+4s+13 is (28}{s^3}+8/(s^2)+\frac{13)/s

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laplacetransform s^2+4s+13s^2+4s+13

laplace transform

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

\frac{28}{s ^{3}} + \frac{8}{s ^{2}} + \frac{13}{s}

Solution steps

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

Refine

\frac{2}{s ^{3}} + 4 \frac{1}{s ^{2}} + 13 \frac{2}{s ^{3}} + 4 \frac{1}{s ^{2}} + \frac{13}{s} : \frac{28}{s ^{3}} + \frac{8}{s ^{2}} + \frac{13}{s}

= \frac{28}{s ^{3}} + \frac{8}{s ^{2}} + \frac{13}{s}

n = 0 \sum \infty (n + 1) x^{n}

\frac{d}{d x} (\frac{x ^{2} - x + 1}{x ^{2} + 1})

\frac{d y}{d x} + \frac{3}{3.5} y = 9.8

\frac{\partial}{\partial z} (z ln (x^{2} y cos (z)))

\frac{\partial}{\partial x} (sin (x) + x cos (x))

What is laplacetransform s^2+4s+13s^2+4s+13 ?
The answer to laplacetransform s^2+4s+13s^2+4s+13 is (28}{s^3}+8/(s^2)+\frac{13)/s