What is laplacetransform 3t^2+13t+13 ?

The answer to laplacetransform 3t^2+13t+13 is 6/(s^3)+(13}{s^2}+\frac{13)/s

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laplacetransform 3t^2+13t+13

laplace transform

3 t^{2} + 13 t + 13

\frac{6}{s ^{3}} + \frac{13}{s ^{2}} + \frac{13}{s}

Solution steps

L {3 t^{2} + 13 t + 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)}

= 3 L {t^{2}} + 13 L {t} + L {13}

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

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

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

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

Refine

3 \frac{2}{s ^{3}} + 13 \frac{1}{s ^{2}} + \frac{13}{s} : \frac{6}{s ^{3}} + \frac{13}{s ^{2}} + \frac{13}{s}

= \frac{6}{s ^{3}} + \frac{13}{s ^{2}} + \frac{13}{s}

e^{\frac{( x ^{5} )}{5}}

\frac{1}{( 1 - t ^{2} ) ^{\frac{1}{2}}}

f (x) = (\frac{x + 1}{x + 3})^{2}, at x = - 2

\int_{0}^{\infty} \frac{( y e ^{- \frac{y}{a}} )}{a} d y

y = x^{2} - 4 x, y = 2 x + 7

What is laplacetransform 3t^2+13t+13 ?
The answer to laplacetransform 3t^2+13t+13 is 6/(s^3)+(13}{s^2}+\frac{13)/s